FINANCE 601, Fall 2019
Mid-Term Examinations, Fall 2017 and Fall 2018
(Appendix A.2.2-A.2.5 of the Main Courseware, pp. 469-490)
Answers and Solutions
Instructor: C.C.Y. Kwan
Remarks:
The answers provided below are usually in much greater detail than what are re-
quired for the purpose of grading. However, for questions where clear explanations are required
as indicated, listing of some equations or analytical expressions without explanations is deemed
inadequate. So is the inclusion of materials that are not even asked. The grader’s task is to assess
objectively how well you understand the concepts involved, by reading your written answers.
A.2.2 First Mid-Term Examination, 2017
1. Question: (21 marks) You have recently used the online mortgage calculators from TD
Canada Trust and from HSBC Canada to help you make mortgage loan decisions. For a
$400; 000 mortgage loan amortized over 25 years at a stated annual interest rate of 6%; you
prefer equal weekly repayments among the several repayment options o¤ered, with the …rst
repayment date being exactly one week after obtaining the loan. You have noticed that the
required weekly repayments as indicated by the two banks are di¤erent.
You are able to replicate exactly the dollar amounts of TD Canada Trust’s weekly repayments,
provided that the stated annual interest rate is compounded every 26 weeks in a 365-day year
and that there are 1; 304 equal weekly repayments in total.
However, to replicate HSBC
Canada’s weekly repayments requires that each year be treated as having exactly 52 weeks.
Thus, the stated 6% interest rate here is a stated 52-week interest rate that is compounded
every 26 weeks, and there are only 1; 300 (= 25
52) equally weekly repayments in total.
(a) (6 marks) For a 365-day year, determine the e¤ective weekly interest rate for borrowing
from each of the above two banks.
Determine also the corresponding continuously
compounded annual interest rate, if the lender is TD Canada Trust. For the intended
precision, each answer here ought to be of the form x.xxxxxx%.
1
(b) (10 marks) Determine the weekly repayments as required by each of the above two
banks, as indicated by their online mortgage calculators.
For the intended precision,
each answer here ought to be rounded to the nearest hundredth of a dollar.
(c) (5 marks) Suppose that the lender is TD Canada Trust. What will be the principal
outstanding immediately after you have made 156 weekly repayments? How much will
be the interest component in the 157-th weekly repayment? For the intended precision,
each answer here ought to be rounded to the nearest hundredth of a dollar.
{Hint: It is known that, for any positive real number r and any positive integer n;
1
1
1
+
+
+
2
1 + r (1 + r)
(1 + r)3
+
1
1
1
=
n
(1 + r)
r
(1 + r) n :
where r is any positive real number and n is any positive integer.}
Answer:
(a) In the case of TD Canada Trust, the annual frequency of compounding is
365
= 2:00549451:
26 7
As the e¤ective 26-week interest rate is
6%
0:06
=
= 0:02991781;
2:00549451
2:00549451
the e¤ective weekly interest rate is
(1 + 0:02991781)1=26
1 = 0:00113445 = 0:113445%:
Then, as the e¤ective annual interest rate is
(1 + 0:00113445)365=7
1 = 0:06090251;
the continuously compounded annual interest rate is
ln(1 + 0:06090251) = 0:05911997 = 5:911997%:
In the case of HSBC Canada, a year is treated as having exactly 52 weeks, and thus the
e¤ective 26-week interest rate is
6%
= 3%:
2
2
The e¤ective weekly interest rate is
(1 + 0:03)1=26
1 = 0:00113752 = 0:113752%:
(b) For a $400; 000 mortgage loan, the weekly repayment W can be computed from
$400; 000 = W
1
(1 + r) n
;
r
where r is the e¤ective weekly interest rate and n is the number of weekly repayments.
In the case of TD Canada Trust, as r = 0:113445%; n = 1; 304; and
1
(1 + r) n
1
=
r
(1:00113445) 1;304
= 680:52073;
0:00113445
the repayment each week is
W =
$400; 000
= $587:79:
680:52073
In the case of HSBC Canada, as r = 0:113752%; n = 1; 300; and
1
(1 + r) n
1
=
r
(1:00113752) 1;300
= 678:57316;
0:00113752
the repayment each week is
W =
$400; 000
= $589:47:
678:57316
(c) In the case of TD Canada Trust, immediately after the 156-th weekly repayments, there
are still
1; 304
156 = 1; 148
weekly repayments to be made. Thus, the principal outstanding at the time is
W
1
(1:00113445) 1;148
= $587:79
0:00113445
641:63824 = $377; 145:45:
The interest component in the 157-th weekly repayment is
0:113445% = $427:85:
$377; 145:45
3
2. Question: (21 marks) According to recent bond quotation data from Financial Post, a bond
issued by Loblaws Companies Ltd., which pays semi-annual coupons at an annual coupon rate
of 6:05%; will mature in June 9, 2034. The Canadian market convention for bond quotations
is followed here. It is implicit that the bond has a $100 face value. For the settlement date
of September 29, 2017, the quoted annual yield to maturity is 4:47%:
There are 183 days in the semi-annual period (from June 9, 2017 to December 9, 2017)
containing the settlement date, and there are 71 days between the settlement date and the
next coupon date (from September 29, 2017 to December 9, 2017). As of the settlement date,
there will still be 34 semi-annual coupons to be paid before the bond matures.
(a) (15 marks) Find the accrued interest and the quoted price of the bond for the settlement
date, given that the full price of the bond on the settlement date is
(1 + r)
C
(1 + r) n
F
+
r
(1 + r)n
1
:
Here, on the settlement date, r is the e¤ective bond yield each coupon period, C is
the per-period coupon, F is the face value, n is the number of coupon payments before
the bond matures, and
is the proportion of a semi-annual period based on which the
accrued interest is computed. Indicate whether the quoted price is a full price or a clean
price according to the market convention. The accuracy of each numerical result that
you provide here is required to be to the nearest 1=1; 000 of a dollar.
(b) (6 marks) Macaulay’s duration, when stated in the number of coupon periods, is the
proportionality constant D in
dP
=
P+ C
D
dr
;
1+r
where P is the clean price of the bond and all other symbols are as de…ned in Part (a).
Modi…ed duration, as measured in the number of coupon periods, is D=(1 + r): For
the settlement date of September 29, 2017, Macaulay’s duration of the above corporate
bond is 11:088 years. This duration measure is known to have been contaminated by
a saw-toothed time pattern. Using the results from Part (a), correct the corresponding
modi…ed duration for the same date, also as measured in years, in order to eliminate the
saw-toothed time pattern.
4
Answer:
(a) The full price of the bond is P + C; which can be computed directly from
P + C = (1 + r)
C
(1 + r) n
F
+
r
(1 + r)n
1
:
Here, we have
F = $100;
n = 34 semi-annual periods,
4:47%
= 0:02235;
2
6:05%($100)
C=
= $3:025;
2
r=
and
=
183 71
112
=
= 0:61202 proportion of a semi-annual period.
183
183
The accrued interest is
C = 0:61202($3:025) = $1:851 (rounded to the nearest 1=1; 000 of a dollar).
The full price is
$100
(1 + 0:02235) 34
+
0:02235
(1 + 0:02235)34
23:64012636 + $100 0:471643176)
1
P + C = (1 + 0:02235)0:61202 $3:025
= 1:013619988
($3:025
= 1:013619988
$118:6756998
= $120:292 (rounded to the nearest 1=1; 000 of a dollar).
The quoted price is a clean price,
P = $120:292
$1:851 = $118:441 (rounded to the nearest 1=1; 000 of a dollar).
(b) Macaulay’s duration D and the modi…ed duration D(M ) are de…ned in
dP
=
P+ C
D
5
dr
=
1+r
D(M ) dr:
Given that Macaulay’s duration D is 11:088 years, which is the same as
2(11:088) = 22:176 semi-annual periods,
the modi…ed duration is
D(M ) =
D
22:176
=
= 21:6912 semi-annual periods.
1+r
1 + 0:02235
To correct the modi…ed duration to eliminate the saw-toothed time pattern, we multiply
it with the factor
P+ C
$120:292
=
= 1:01563:
P
$118:441
The modi…ed duration after the correction is
P+ C
P
D(M ) = 1:01563
21:6912 semi-annual periods
= 22:030 semi-annual periods = 11:015 years.
Notice that the correction to D(M ) can also be made, by applying directly the multiplicative factor
P+ C
P
1
= 1:01563
1+r
1
= 0:99343
1:02235
to Macaulay’s duration
D = 11:088 years.
That is,
0:99343
11:088 years = 11:015 years.
3. Question: (21 marks) For a basic model of bond valuation, let c be a bond’s semi-annual
coupon as a proportion of the bond’s face value and r be investors’ required return each
semi-annual period for holding the bond. It is known that the price of the bond with n full
semi-annual periods before maturity, as a proportion of the bond’s face value, is
pn = c
Xn
1
1
1
+
=c
n
k=1 (1 + r)k
(1 + r)
(1 + r) n
1
+
:
r
(1 + r)n
It is also known that Macaulay’s duration of the bond, as measured in the number of semiannual periods, is
P
P
c nk=1 1=(1 + r)k q + [1=(1 + r)n ] n
c nk=1 k=(1 + r)k + n=(1 + r)n
P
Dn = Pn
=
;
c k=1 1=(1 + r)k + 1=(1 + r)n
[c nk=1 1=(1 + r)k ] + [1=(1 + r)n ]
6
where
Pn
k=(1 + r)k
q = Pk=1
:
n
k
k=1 1=(1 + r)
(a) (13 marks) Consider the case where n > 1: With n and c held constant, establish the
sign of the …rst derivative of q with respect to r (i.e., dq=dr): Then, verify whether Dn
increases or decreases with increasing r: Justify your answer analytically.
[Hint: It is known that, according to the Cauchy-Schwarz inequality,
Xn
k=1
a2k
Xn
k=1
b2k
Xn
k=1
2
ak b k
;
for any real numbers a1 ; a2 ; : : : ; an and b1 ; b2 ; : : : ; bn : Strict inequality holds if a1 ; a2 ; : : : ; an
cannot be duplicated exactly by hb1 ; hb2 ; : : : ; hbn ; where h is a constant.]
(b) (8 marks) For the special case where c = r; if r is held constant, does Dn increases or
decreases with increasing n; or neither? Justify your answer analytically, by simplifying
Dn+1
Dn to an expression that con…rms its sign.
Answer:
(a) To establish the sign of the …rst derivative of q with respect to r; for
Pn
k=(1 + r)k
;
q = Pk=1
n
k
k=1 1=(1 + r)
where r > 0 is a given real number and n > 1 is a given integer, we …rst let
x=
1
1+r
for notational convenience. Notice that 0 < x < 1: We can write
Pn
kxk
:
q = Pk=1
n
k
k=1 x
As
d k
kx = k 2 xk 1
dx
and
d k
x = kxk 1 ;
dx
7
we can also write
Pn
Pn
k 1
k
2 k 1
kx
k
x
k=1 kx
k=1
k=1
Pn
2
( k=1 xk )
Pn
P
P
Pn
n
n
k
2 k
k
k
x
k
x
kx
k=1
k=1
k=1
k=1 kx
=
Pn
2
x ( k=1 xk )
Pn
Pn
Pn
k 2
2 k
k
k=1 kx
k=1 k x
k=1 x
=
P
2
x ( nk=1 xk )
i
i hP
hP
Pn
n
n
k=2
k=2 2
k=2 2
kxk=2
kx
x
k=1 x
k=1
k=1
=
P
2
x ( nk=1 xk )
Now, let
Pn
Pn
k
k=1 x
dq
=
dx
2
:
ak = xk=2 and bk = kxk=2 ; for k = 1; 2; : : : ; n:
By using the Cauchy-Schwarz inequality, we have
Xn
k=1
Xn
a2k
k=1
Xn
b2k
k=1
2
ak b k
:
Strict inequality holds if a1 ; a2 ; : : : ; an cannot be duplicated exactly by hb1 ; hb2 ; : : : ; hbn ;
where h is a constant. As a1 = b1 ; a2 = b2 =2; : : : ; an = bn =n; such a constant h cannot
be found. Thus, strict inequality holds here.
As
Xn
x
k=1
2
xk
in the denominator of the expression of dq=dx is positive, it follows from
hX n
k=1
xk=2
2
in the numerator there that
i hX n
k=1
kxk=2
2
i
>
hX n
k=1
xk=2
kxk=2
i2
dq
> 0:
dx
Given how x is de…ned, we have
dx
d
=
dr
dr
1
1+r
=
1
<0
(1 + r)2
and then
dq
dq dx
=
< 0;
dr
dx dr
thus con…rming that the sign of the …rst derivative of q with respect to r is negative.
8
We now verify whether Dn increases or decreases with increasing r; when n and c are
held constant. As
Dn =
c
Pn
k
q + [1=(1 + r)n ] n
k=1 k=(1 + r)
P
;
[c nk=1 1=(1 + r)k ] + [1=(1 + r)n ]
Dn is a weighted average of q and n: Notice that q < n: The weight that n receives is
c
1=(1 + r)n
1
:
= Pn
k
n
c k=1 (1 + r)n k + 1
k=1 1=(1 + r) + 1=(1 + r)
Pn
The higher the bond yield r; the lower is the weight that n receives and the higher is
the weight that q receives. Further, we have already established that a higher r makes
q lower. Thus, a higher r will result in a lower Dn :
(b) for the special case where c = r; the denominator of the expression
P
c nk=1 k=(1 + r)k + n=(1 + r)n
Dn = Pn
c k=1 1=(1 + r)k + 1=(1 + r)n
is 1: This is because
Xn
1
r
1=(1 + r)k + 1=(1 + r)n = r
k=1
(1 + r) n
+ (1 + r) n = 1:
r
The bond sells at par.
We can now work with
Dn = r
and
Dn+1 = r
As
Dn+1
=
=
=
k=1
Xn+1
k=1
k=(1 + r)k + n=(1 + r)n
k=(1 + r)k + (n + 1)=(1 + r)n+1 :
Xn+1
Xn
k
k
n+1
r
+
k
k
k=1 (1 + r)
k=1 (1 + r)
(1 + r)n+1
r(n + 1)
n+1
n
+
n+1
n+1
(1 + r)
(1 + r)
(1 + r)n
(1 + r)(n + 1)
n
n+1
(1 + r)
(1 + r)n
n+1
n
n
(1 + r)
(1 + r)n
1
;
(1 + r)n
Dn = r
=
Xn
n
(1 + r)n
which is always positive, we con…rm that Dn increases with increasing n if c = r:
9
4. Question: (21 marks) McMaster Corporation always pays its quarterly dividends on January
15, April 15, July 15, and October 15 each year. The four quarterly dividends remain constant
for any given year. The …rm has a tradition of maintaining a 2% annual growth in dividends.
Suppose that today is January 15, 2018. As a shareholder, you have received a $100 quarterly
dividend today. The dividend is deposited to a savings account at your bank, which pays a
1:6% e¤ective quarterly interest rate.
The next three dividends for 2018 will also be $100 each quarter. For 2019, the four dividends
will be $102 (= $100
(= $102
1:02) each quarter.
For 2020, the four dividends will be $104:04
1:02) each quarter. The same growth pattern will continue in all subsequent years.
All future dividends will also be deposited at the same savings account.
For computational convenience, let us label today as t = 0 and all subsequent dividend
payment dates as t = 1; 2; 3; 4; 5; : : : The dividend increases occur at t = 4; 8; 12; 16; 20; : : :
(a) (11 marks) Determine the account balance as of January 15, 2028, which is exactly
10 years from today (that is, t = 40); immediately after the corresponding quarterly
dividend has been deposited at the savings account.
(b) (10 marks) How long does it take for the account balance to exceed $6; 500? The
answer must include the year, the month, and the day. Besides the speci…c year, the
date must be among January 15, April 15, July 15, and October 15, immediately after
the corresponding quarterly dividend has been deposited at the savings account.
{Hint: It is known that, for a growing annuity with 0
1
1+g
(1 + g)2
+
+
+
1 + r (1 + r)2 (1 + r)3
+
g < r;
(1 + g)n 1
1
=
1
(1 + r)n
r g
and
1+g
1+r
n
n
(1 + g)n 1
1+r
1 + g (1 + g)2
1+g
+
+
=
1+
+
1
;
2
n
1
1 + r (1 + r)
(1 + r)
r g
1+r
where n is a positive integer. For g = 0; the above expressions can be simpli…ed as
1
1
1
+
+
+
2
1 + r (1 + r)
(1 + r)3
1
1
=
1
n
(1 + r)
r
(1 + r) n
1
1+r
=
1
n
1
(1 + r)
r
(1 + r) n ;
+
and
1+
1
1
+
+
1 + r (1 + r)2
+
10
respectively.}
Answer:
(a) Method 1:
The e¤ective quarterly interest rate rq is 1:6%: The value of the four
quarterly dividends received at t = 0; 1; 2; and 3; as measured at t = 0; is
$100
1 + rq
1
rq
(1 + rq ) 4 =
$100(1:016)
1
0:016
1
= $390:65:
(1:016)4
Given a g = 2% growth rate, the value of the next four dividends received at t = 4; 5; 6;
and 7; as measured at t = 4; is
$390:65
1:02:
The value of the subsequent four dividends received at t = 8; 9; 10; and 11; as measured
at t = 8; is
$390:65
(1:02)2 :
As the same growth pattern continues, the four dividends received at t = 36; 37; 38; and
39; as measured at t = 36; is
$390:65
(1:02)9 :
The e¤ective annual interest rate is
ra = (1 + rq )4
1 = (1:016)4
1 = 0:06555245:
The value of all dividends received at t = 0; 1; 2; 3; : : : ; 39; as measured at t = 0; is
1+g
(1 + g)2
(1 + g)9
+
+
+
1 + ra (1 + ra )2
(1 + ra )9
"
#
10
1 + ra
1+g
1
ra g
1 + ra
"
#
10
1:06555245
1:02
1
0:06555245 0:02
1:06555245
$390:65 1 +
= $390:65
= $390:65
= $3; 234:5664:
The corresponding future value, as measured at t = 40; is
$3; 234:5664
(1 + ra )10 = $3; 234:5664
11
(1:06555245)10 = $6; 103:30:
The account balance at t = 40 (which is January 15, 2028), immediately after that date’s
deposit, is
$6; 103:30 + $100
(1 + g)10 = $6; 103:30 + $100
(1:02)10
= $6; 103:30 + $121:90 = $6; 225:20:
Method 2:
We still use rq = 0:016 and ra = 0:06555245: The value of dividend
payments at t = 0; 4; 8; 12; : : : ; 36; as measured at t = 0; is
(1 + g)2
1+g
(1 + g)9
+
$100 1 +
+
+
1 + ra (1 + ra )2
(1 + ra )9
"
#
10
1 + ra
1+g
= $100
1
ra g
1 + ra
"
#
10
1:06555245
1:02
= $100
1
0:06555245 0:02
1:06555245
= $828:00:
The value of dividend payments at t = 1; 5; 9; 13; : : : ; 37; as measured at t = 1; is also
$828:00: So are the corresponding values of dividend payments at t = 2; 6; 10; 14; : : : ; 38;
as measured at t = 2; and at t = 3; 7; 11; 15; : : : ; 39; as measured at t = 3:
The value of these four dollar amounts, as measured at t = 0; is
1
1
1
+
+
2
1 + rq (1 + rq )
(1 + rq )3
(1 + rq )[1 (1 + rq ) 4 ]
rq
(1:016)[1 (1:016) 4 ]
0:016
3:90650
$828:00 1 +
= $828:00
= $828:00
= $828:00
= $3; 234:5664:
The corresponding future value, as measured at t = 40; is
$3; 234:5664
(1 + ra )10 = $3; 234:5664
(1:06555245)10 = $6; 103:30:
The account balance at t = 40 (which is January 15, 2028), immediately after that date’s
12
deposit, is
$6; 103:30 + $100
(1 + g)10 = $6; 103:30 + $100
(1:02)10
= $6; 103:30 + $121:90 = $6; 225:20:
Method 3: We still use rq = 0:016 and ra = 0:06555245: The di¤erence between this
method and Method 2 is that we …rst compute instead the value of dividend payments
at t = 0; 4; 8; 12; : : : ; 36; 40; as measured at t = 0: As the dividend payment at t = 40 is
included, we have instead
1+g
(1 + g)2
(1 + g)10
+
+
+
1 + ra (1 + ra )2
(1 + ra )10
"
#
11
1+g
1 + ra
1
ra g
1 + ra
"
#
11
1:02
1:06555245
1
0:06555245 0:02
1:06555245
$100 1 +
= $100
= $100
= $100
8:925991 = $892:60:
We can still use the same $828:00 (from Method 2) for each remaining set of dividend
payments, as measured at t = 1; 2; and 3: The total value of the four sets of dividend
payments, as measured at t = 0; is
1
1
1
+
+
2
1 + rq (1 + rq )
(1 + rq )3
1 (1 + rq ) 3
= $892:60 + $828:00
rq
1 (1:016) 3
= $892:60 + $828:00
0:016
= $892:60 + $828:00 2:9065 = $3; 299:17:
$892:60 + $828:00
The corresponding future value, as measured at t = 40; is
$3; 299:17
(1 + ra )10 = $3; 299:17
(1:06555245)10 = $6; 225:20:
This is the account balance as of January 15, 2028.
13
(b) Let k be the required number of quarters, subsequent to January 15, 2028, for the account
balance to exceed $6; 500: It can be solved from
$6; 225:20
(1 + rq )k + $121:90
1
(1 + rq ) k
rq
(1 + rq )k > $6; 500:
The inequality can be simpli…ed as
$6; 225:20
(1 + rq )k +
$121:90
rq
[(1 + rq )k
1] > $6; 500:
Letting
x = (1 + rq )k ;
and noting that
rq = 0:016
we have
6; 225:20x +
121:90
(x
0:016
1) > 6; 500;
which leads to
13; 843:95x > 14; 118:75
and then
x > 1:01985:
From
(1 + rq )k = (1:016)k > 1:01985;
we have
k ln(1:016) > ln(1:01985)
and then
k>
ln(1:01985)
= 1:238:
ln(1:016)
As k is an integer, the smallest k is 2: Thus, as soon as the July 15, 2028 dividend is
deposited, the account balance will exceed $6; 500:
For the purpose of grading, it is acceptable to attempt di¤erent integers for k in
$6; 225:20
= $6; 225:20
$121:90
rq
$121:90
(1:016)k +
0:016
(1 + rq )k +
14
[(1 + rq )k
1]
[(1:016)k
1]:
For k = 1; the result is $6; 446:70; for k = 2; the result is $6; 671:74: Thus, as $6; 671:74 >
$6; 500; July 15, 2028 is the answer.
A.2.3 First Mid-Term Examination, 2018
1. Question: (21 marks) You have recently used the online mortgage calculators from TD
Canada Trust and from HSBC Canada to help you make mortgage loan decisions. For a
$360; 000 mortgage loan amortized over 25 years at a stated annual interest rate of 6%; you
prefer equal repayments every two weeks among the several repayment options o¤ered, with
the …rst repayment date being exactly two weeks after obtaining the loan. You have noticed
that the required bi-weekly repayments as indicated by the two banks are di¤erent.
You are able to replicate exactly the dollar amounts of TD Canada Trust’s bi-weekly repayments, provided that the stated annual interest rate of 6% for the loan is compounded every
26 weeks in a 365-day year and that there are 652 equal bi-weekly repayments in total. However, to replicate HSBC Canada’s bi-weekly repayments requires that each year be treated as
having exactly 52 weeks. Thus, the stated 6% interest rate here is a stated 52-week interest rate that is compounded every 26 weeks, and there are only 650 (= 25
26) bi-weekly
repayments in total.
(a) (6 marks) For a 365-day year, determine the e¤ective bi-weekly interest rate for borrowing from each of the above two banks. Determine also the corresponding continuously
compounded quarterly interest rate, if the lender is TD Canada Trust. For the intended
precision, each answer here ought to be of the form x.xxxxxx%.
(b) (10 marks) Determine the bi-weekly repayments as required by each of the above two
banks, as indicated by their online mortgage calculators.
For the intended precision,
each answer here ought to be rounded to the nearest hundredth of a dollar.
(c) (5 marks) Suppose that the lender is TD Canada Trust. What will be the principal
outstanding immediately after you have made 78 bi-weekly repayments?
will be the interest component in the 79-th bi-weekly repayment?
How much
For the intended
precision, each answer here ought to be rounded to the nearest hundredth of a dollar.
15
Answer:
(a) In the case of TD Canada Trust, the annual frequency of compounding is
365
= 2:00549451:
26 7
As the e¤ective 26-week interest rate is
6%
0:06
=
= 0:02991781;
2:00549451
2:00549451
the e¤ective bi-weekly interest rate is
(1 + 0:02991781)1=13
1 = 0:00227019 = 0:227019%:
Then, as the e¤ective annual interest rate is
(1 + 0:00227019)365=14
1 = 0:06090251 = 6:090251%;
the continuously compounded annual interest rate is
ln(1 + 0:06090251) = 0:05911997 = 5:911997%;
and continuously compounded quarterly interest rate is
1
1
ln(1 + 0:06090251) = (0:05911997) = 0:01477999 = 1:477999%:
4
4
In the case of HSBC Canada, a year is treated as having exactly 52 weeks, and thus the
e¤ective 26-week interest rate is
6%
= 3%:
2
The e¤ective bi-weekly interest rate is
(1 + 0:03)1=13
1 = 0:00227634 = 0:227634%:
(b) For a $360; 000 mortgage loan, the bi-weekly repayment W can be computed from
$360; 000 = W
16
1
(1 + r) n
;
r
where r is the e¤ective bi-weekly interest rate and n is the number of bi-weekly repayments. In the case of TD Canada Trust, as r = 0:227019%; n = 652; and
1
(1 + r) n
1
=
r
(1:00227019) 652
= 340:067469;
0:00227019
the repayment every two weeks is
W =
$360; 000
= $1; 058:61:
340:067469
In the case of HSBC Canada, as r = 0:227634%; n = 650; and
1
1
(1 + r) n
=
r
(1:00227634) 650
= 339:093719;
0:00227634
the repayment every two weeks is
W =
$360; 000
= $1; 061:65:
339:093719
(c) In the case of TD Canada Trust, immediately after the 78-th weekly repayments, there
are still
652
78 = 574
bi-weekly repayments to be made. Thus, the principal outstanding at the time is
W
1
(1:00227019) 574
= $1; 058:61
0:00227019
320:637247 = $339; 430:91:
The interest component in the 79-th bi-weekly repayment is
$339; 430:91
0:227019% = $770:57:
2. Question: (21 marks) In a bond market, the bid price is what a buyer is willing to pay
for a bond, while the ask price is what a seller is willing to accept for the same bond. The
corresponding annual yields to maturity are the bid yield and the ask yield.
According to
some recent bond quotes, which follow the Canadian bond market convention, a bond issued
by Loblaws Companies Ltd. pays semi-annual coupons at an annual coupon rate of 5:900%:
17
The bond will mature in January 18, 2036. It is implicit that the bond has a $100 face value.
For the settlement date of September 26, 2018, the quoted bid yield is 4:441%:
There are 184 days in the semi-annual period (from July 18, 2018 to January 18, 2019)
containing the settlement date, and there are 114 days between the settlement date and the
next coupon date (from September 26, 2018 to January 18, 2019). As of the settlement date,
there will still be 35 semi-annual coupons to be paid before the bond matures.
(a) (15 marks) Find the quoted bid price of the bond, which is a clean price, and the accrued
interest as of the settlement date, given that the full price of the bond can be expressed
as
(1 + r)
C
(1 + r) n
F
+
r
(1 + r)n
1
:
Here, on the settlement date, r is the e¤ective bond yield each coupon period, C is
the per-period coupon, F is the face value, n is the number of coupon payments before
the bond matures, and
is the proportion of a semi-annual period based on which the
accrued interest is computed. The accuracy of each numerical result here is required to
be to the nearest 1=1; 000 of a dollar.
(b) (6 marks) For the same bond and for the same settlement date of September 26, 2018, it
is known that the quoted ask price and ask yield are $118:308 and 4:379%; respectively.
As of the settlement date, Macaulay’s duration based on the quoted ask price is 11:511
years. What is the corrected value of the corresponding modi…ed duration, also measured
in years, once its saw-toothed time pattern has been eliminated. The accuracy of the
numerical result here is required to be to the nearest 1=1; 000 of a year.
Answer:
(a) The full price of the bond is P + C; which can be computed directly from
P + C = (1 + r)
C
1
(1 + r) n
F
+
r
(1 + r)n
Here, we have
F = $100;
n = 35 semi-annual periods,
18
:
4:441%
= 0:022205;
2
5:900%($100)
C=
= $2:950;
2
r=
and
=
184 114
70
=
= 0:3804 proportion of a semi-annual period.
184
184
The accrued interest is
C = 0:3804($2:950) = $1:122 (rounded to the nearest 1=1; 000 of a dollar).
The full price (where the quoted price is a bid price) is
(1 + 0:022205) 35
$100
+
0:022205
(1 + 0:022205)35
24:1554455 + $46:3628332)
P + C = (1 + 0:022205)0:3804 $2:950
= 1:00838936
($2:950
1
= $118:608 (rounded to the nearest 1=1; 000 of a dollar).
The quoted bid price is,
P = $118:608
$1:122 = $117:486 (rounded to the nearest 1=1; 000 of a dollar).
(b) The quoted ask price is
P = $118:308;
and the e¤ective semi-annual yield (which is an ask yield) is
r=
4:379%
= 2:1895% = 0:021895:
2
Macaulay’s duration D and the modi…ed duration D(M ) are de…ned in
dP
=
P+ C
D
dr
=
1+r
D(M ) dr:
Given that
D = 11:511 years,
which is the same as
D=2
11:511 semi-annual periods = 23:022 semi-annual periods,
19
the modi…ed duration is
D(M ) =
D
23:022
=
= 22:5287 semi-annual periods.
1+r
1 + 0:021895
To correct the modi…ed duration to eliminate the saw-toothed time pattern, we multiply
it with the factor
P+ C
$118:308 + $1:122
=
= 1:00948:
P
$118:308
The modi…ed duration after the correction is
P+ C
P
D(M ) = 1:00948
22:5287 semi-annual periods
= 22:742 semi-annual periods = 11:371 years.
Notice that the correction to D(M ) can also be made, by applying directly the multiplicative factor
P+ C
P
1
= 1:00948
1+r
1
= 0:98786
1:021895
to Macaulay’s duration
D = 11:511 years.
That is,
11:511 years = 11:371 years.
0:98786
3. Question: (21 marks) For a basic model of bond valuation, let c be a bond’s semi-annual
coupon as a proportion of the bond’s face value and r be investors’required return each semiannual period for holding the bond. The price of the bond with n full semi-annual periods
before maturity, as a proportion of the bond’s face value, is
pn = c
Xn
1
1
+
:
k
k=1 (1 + r)
(1 + r)n
Here, n is a positive integer. Macaulay’s duration of the bond, as measured in the number
of semi-annual periods, is
c
wn
Dn =
=
pn
where
Pn
k
k=1 1=(1 + r)
q + [1=(1 + r)n ] n
pn
Pn
k=(1 + r)k
q = Pk=1
:
n
k
k=1 1=(1 + r)
20
;
It is known that, for n > 1;
Dn+1
pn (c + pn ) cwn
;
pn (c + pn )
Dn =
pn 1 (c + pn 1 )
an 1 =
and
pn (c + pn )
cwn 1 =
Xn 1 c2 (n
cwn =
k=1
k) + 2c
(1 + r)k
(n
an 1
;
(1 + r)n 1
2)c +
1
;
(1 + r)n 1
an
1
an 1 + c(pn
=
n
(1 + r)
(1 + r)n
1) +
c r
:
(1 + r)n
(a) (11 marks) With both n and c held constant for the case where n > 1; determine the
sign of the …rst derivative of q with respect to r: Then, verify whether Dn increases or
decreases with increasing r; or neither. Justify your answer analytically.
(b) (10 marks) With both c and r held constant, consider the special case where c
r: Does
Dn increase or decrease with increasing n; or neither? Justify your answer analytically,
by means of an induction proof. In the induction proof, indicate clearly the base case
and the induction hypothesis, before going through the corresponding induction step.
Answer:
(a) To establish the sign of the …rst derivative of q with respect to r; for
Pn
k=(1 + r)k
q = Pk=1
;
n
k
k=1 1=(1 + r)
where r > 0 is a given real number and n > 1 is a given integer, we …rst let
x=
1
1+r
for notational convenience. Notice that 0 < x < 1: We can write
Pn
kxk
q = Pk=1
:
n
k
k=1 x
As
d k
kx = k 2 xk 1
dx
21
and
d k
x = kxk 1 ;
dx
we can also write
Now, let
Pn
k=1 x
k
Pn
Pn
2 k 1
k=1 k x
k=1 kx
2
k
k
Pn
k 1
k=1 kx
Pn
( k=1 x )
Pn
Pn
Pn
Pn
k
k
2 k
k
k=1 kx
k=1 kx
k=1 k x
k=1 x
=
P
2
x ( nk=1 xk )
Pn
Pn
Pn
k
2 k
k 2
x
k
x
k=1
k=1
k=1 kx
=
P
2
x ( nk=1 xk )
hP
i hP
i
Pn
n
n
k=2 2
k=2 2
k=2
x
kx
kxk=2
k=1
k=1
k=1 x
=
P
2
x ( nk=1 xk )
dq
=
dx
2
:
ak = xk=2 and bk = kxk=2 ; for k = 1; 2; : : : ; n:
By using the Cauchy-Schwarz inequality, we have
Xn
k=1
Xn
a2k
k=1
Xn
b2k
k=1
2
ak b k
:
Strict inequality holds if a1 ; a2 ; : : : ; an cannot be duplicated exactly by hb1 ; hb2 ; : : : ; hbn ;
where h is a constant. As a1 = b1 ; a2 = b2 =2; : : : ; an = bn =n; such a constant h cannot
be found. Thus, strict inequality holds here.
As
x
Xn
k=1
xk
2
in the denominator of the expression of dq=dx is positive, it follows from
hX n
i hX n
i hX n
i2
k=2 2
k=2 2
k=2
k=2
x
kx
>
x
kx
k=1
k=1
k=1
in the numerator there that
dq
> 0:
dx
Given how x is de…ned, we have
dx
d
=
dr
dr
1
1+r
=
1
<0
(1 + r)2
and then
dq
dq dx
=
< 0;
dr
dx dr
22
thus con…rming that the sign of the …rst derivative of q with respect to r is negative.
We now verify whether Dn increases or decreases with increasing r; when n and c are
held constant. As
Dn =
c
Pn
k
q + [1=(1 + r)n ] n
k=1 k=(1 + r)
P
;
[c nk=1 1=(1 + r)k ] + [1=(1 + r)n ]
Dn is a weighted average of q and n: Notice that q < n: The weight that n receives is
c
1=(1 + r)n
1
:
= Pn
k
n
c k=1 (1 + r)n k + 1
k=1 1=(1 + r) + 1=(1 + r)
Pn
The higher the bond yield r; the lower is the weight that n receives and the higher is
the weight that q receives. Further, we have already established that a higher r makes
q lower. Thus, a higher r will result in a lower Dn :
(b) For the special case where c
r; the following is an induction proof to con…rm that Dn
increases with increasing n: What is to be proven is
Dn+1
for n
Dn > 0;
1: For the induction proof, the base case is n = 1: As D1 = 1 and 1 < D2 < 2;
we have
D2
D1 > 0
Dn
Dn 1 > 0:
and thus the base case is established.
By the induction hypothesis, we have
We already know that
Dn
Dn 1 =
pn 1 (c + pn 1 ) cwn 1
:
pn 1 (c + pn 1 )
As
pn 1 (c + pn 1 ) > 0;
we must have
pn 1 (c + pn 1 )
23
cwn 1 > 0:
Further, we know from
an 1
= pn 1 (c + pn 1 )
(1 + r)n 1
cwn 1
that
an 1 > 0:
As
Dn+1
Dn =
pn (c + pn ) cwn
pn (c + pn )
and
pn (c + pn ) > 0;
the sign of Dn+1
Dn is the same as the sign of pn (c + pn )
cwn :
Given that
pn (c + pn )
we need c
cwn =
r to ensure that pn
1
an 1 + c(pn
(1 + r)n
1: That is, if c
1) +
c r
;
(1 + r)n
r; the bond will not be selling at
a discount. If so, given that we already have an 1 > 0; the sum of the three additive
terms inside the square brackets on the right hand side of the above must be positive.
With
pn (c + pn )
cwn > 0;
we have
Dn+1
Dn > 0;
thus completing the induction proof.
4. Question: (21 marks) McMaster Corporation always pays its quarterly dividends on March
15, June 15, September 15, and December 15 each year. The four quarterly dividends remain
constant for any given year. The company has a tradition of maintaining a 2% annual growth
in dividends. Suppose that today is March 15, 2019. As a shareholder, you have received a
$100 quarterly dividend today.
The next three dividends for 2019 will also be $100 each quarter. For 2020, the four dividends
will be $102 (= $100
1:02) each quarter.
24
For 2021, the four dividends will be $104:04
(= $102
1:02) each quarter. The same 2% growth pattern will continue in all subsequent
years.
You have just opened two savings accounts at a bank, with each one paying a 1:5% e¤ective
quarterly interest rate. Let us label them as account A and account B. You will deposit the
dividends received on March 15, June 15, and September 15, including today’s dividend, to
account A. You will also deposit half the dividends received on December 15 to account A
and the remaining half to account B.
For computational convenience, let us label today as t = 0 and all subsequent dividend
payment dates as t = 1; 2; 3; 4; 5; : : : The dividend increases occur at t = 4; 8; 12; 16; 20; : : :
(a) As of March 15, 2029, which is exactly 10 years from today (that is, t = 40); immediately
after the corresponding quarterly dividend has been deposited, determine the following:
(a1) (9 marks) The total balance of the two accounts combined.
(a2) (5 marks) The balance of account B.
(b) (7 marks) How long does it take for the total balance of the two accounts combined to
exceed $6; 400? The answer must include the year, the month, and the day. Besides the
speci…c year, the date must be among March 15, June 15, September 15, and December
15, immediately after the corresponding quarterly dividend has been deposited.
{Hint: It is known that, for a growing annuity with 0
1
1+g
(1 + g)2
+
+
+
1 + r (1 + r)2 (1 + r)3
and
1 + g (1 + g)2
1+
+
+
1 + r (1 + r)2
g < r;
(1 + g)n 1
1
+
=
1
n
(1 + r)
r g
(1 + g)n 1
1+r
+
=
1
n
1
(1 + r)
r g
n
1+g
1+r
1+g
1+r
;
n
;
where n is a positive integer.}
Answer:
(a1) For this part of the answer, the two accounts are combined for computational purposes.
25
Method 1:
The e¤ective quarterly interest rate rq is 1:5%: The value of the four
quarterly dividends received at t = 0; 1; 2; and 3; as measured at t = 0; is
$100
1 + rq
1
rq
(1 + rq ) 4 =
$100(1:015)
1
0:015
1
= $391:22:
(1:015)4
Given a g = 2% growth rate, the value of the next four dividends received at t = 4; 5; 6;
and 7; as measured at t = 4; is
$391:22
1:02:
The value of the subsequent four dividends received at t = 8; 9; 10; and 11; as measured
at t = 8; is
$391:22
(1:02)2 :
As the same growth pattern continues, the four dividends received at t = 36; 37; 38; and
39; as measured at t = 36; is
$391:22
(1:02)9 :
The e¤ective annual interest rate is
ra = (1 + rq )4
1 = (1:015)4
1 = 0:06136355:
The value of all dividends received at t = 0; 1; 2; 3; : : : ; 39; as measured at t = 0; is
1+g
(1 + g)9
(1 + g)2
+
+
+
1 + ra (1 + ra )2
(1 + ra )9
"
#
10
1 + ra
1+g
1
ra g
1 + ra
"
#
10
1:06136355
1:02
1
0:06136355 0:02
1:06136355
$391:22 1 +
= $391:22
= $391:22
= $3; 292:7612:
The corresponding future value, as measured at t = 40; is
$3; 292:7612
(1 + ra )10 = $3; 292:7612
(1:06136355)10 = $5; 973:13:
The total balance of the two accounts at t = 40 (which is March 15, 2029), immediately
after that date’s deposit, is
$5; 973:13 + $100
(1 + g)10 = $5; 973:13 + $100
(1:02)10
= $5; 973:13 + $121:90 = $6; 095:03:
26
Method 2:
We still use rq = 0:015 and ra = 0:06136355: The value of dividend
payments at t = 0; 4; 8; 12; : : : ; 36; as measured at t = 0; is
1+g
(1 + g)2
(1 + g)9
+
+
+
1 + ra (1 + ra )2
(1 + ra )9
"
#
10
1 + ra
1+g
1
ra g
1 + ra
"
#
10
1:06136355
1:02
1
0:06136355 0:02
1:06136355
$100 1 +
= $100
= $100
= $841:66:
The value of dividend payments at t = 1; 5; 9; 13; : : : ; 37; as measured at t = 1; is also
$841:66: So are the corresponding values of dividend payments at t = 2; 6; 10; 14; : : : ; 38;
as measured at t = 2; and at t = 3; 7; 11; 15; : : : ; 39; as measured at t = 3:
The value of these four dollar amounts, as measured at t = 0; is
1
1
1
+
+
2
1 + rq (1 + rq )
(1 + rq )3
(1 + rq )[1 (1 + rq ) 4 ]
rq
(1:015)[1 (1:015) 4 ]
0:015
3:91220
$841:66 1 +
= $841:66
= $841:66
= $841:66
= $3; 292:7612:
The corresponding future value, as measured at t = 40; is
$3; 292:7612
(1 + ra )10 = $3; 292:7612
(1:06136355)10 = $5; 973:13:
The total balance of the two accounts combined at t = 40 (which is March 15, 2029),
immediately after that date’s deposit, is
$5; 973:13 + $100
(1 + g)10 = $5; 973:13 + $100
(1:02)10
= $5; 973:13 + $121:90 = $6; 095:03:
27
Method 3: We still use rq = 0:015 and ra = 0:06136355: The di¤erence between this
method and Method 2 is that we …rst compute instead the value of dividend payments
at t = 0; 4; 8; 12; : : : ; 36; 40; as measured at t = 0: As the dividend payment at t = 40 is
included, we have instead
1+g
(1 + g)2
(1 + g)10
$100 1 +
+
+
+
1 + ra (1 + ra )2
(1 + ra )10
"
#
11
1 + ra
1+g
= $100
1
ra g
1 + ra
"
#
11
1:06136355
1:02
= $100
1
0:06136355 0:02
1:06136355
= $100
9:08863327 = $908:86:
We can still use the same $841:66 (from Method 2) for each remaining set of dividend
payments, as measured at t = 1; 2; and 3: The total value of the four sets of dividend
payments, as measured at t = 0; is
1
1
1
+
+
1 + rq (1 + rq )2 (1 + rq )3
1 (1 + rq ) 3
= $908:86 + $841:66
rq
1 (1:015) 3
= $908:86 + $841:66
0:015
= $908:86 + $841:66 2:91220 = $3; 359:96:
$908:86 + $841:66
The corresponding future value, as measured at t = 40; is
$3; 359:96
(1 + ra )10 = $3; 299:17
(1:06136355)10 = $6; 095:03:
This is the total balance of the two accounts combined as of March 15, 2029.
(a2) Let ra be the e¤ective annual interest rate. We have
ra = (1 + rq )4
1 = (1:015)4
1 = 0:06136355:
The 10 deposits to account B are at t = 3; 7; 11; : : : ; 39: The deposits can be characterized
as a growing annuity. Their total value as measured at t = 1 is
(
"
#)
10
1
1+g
1
= $50 24:17587 0:328014
$50
ra g
1 + ra
= $396:502:
28
As measured at t = 0; it is
$396:5016
(1 + rq ) = $396:5016
(1:015) = $402:449:
As measured at t = 40; it is
$402:449
(1 + rq )40 = $402:449
(1 + ra )10 = $402:449
1:814 = $730:05:
This is the total balance of account B as of March 15, 2029.
(b) For this part of the answer, the two accounts are combined for computational purposes.
Let k be the required number of quarters, subsequent to March 15, 2029, for the account
balance to exceed $6; 400: It can be solved from
$6; 095:03
(1 + rq )k + $121:90
1
(1 + rq ) k
rq
(1 + rq )k > $6; 400:
The inequality can be simpli…ed as
$6; 095:03
(1 + rq )k +
$121:90
rq
[(1 + rq )k
1] > $6; 400:
Letting
x = (1 + rq )k ;
and noting that
rq = 0:015
we have
6; 095:03x +
121:90
(x
0:015
1) > 6; 400;
which leads to
14; 221:70x > 14; 526:67
and then
x > 1:02144:
From
(1 + rq )k = (1:015)k > 1:02144;
we have
k ln(1:015) > ln(1:02144)
29
and then
k>
ln(1:02144)
= 1:425:
ln(1:015)
As k is an integer, the smallest k is 2: Thus, as soon as the September 15, 2029 dividend
is deposited, the total balance of the two accounts combined will exceed $6; 400:
Notice that, for the purpose of grading, it is acceptable to attempt di¤erent integers for
k in
$6; 095:03
= $6; 095:03
$121:90
rq
$121:90
(1:015)k +
0:015
(1 + rq )k +
[(1 + rq )k
1]
[(1:015)k
1]:
For k = 1; the result is $6; 308:35; for k = 2; the result is $6; 524:88: Thus, as $6; 524:88 >
$6; 400; September 15, 2029 is the answer.
A.2.4 Second Mid-Term Examination, 2017
1. Question: (17 marks in total) The two parts of this question are unrelated.
(a) (9 marks in total) McMaster Inc. has an $60 million, 30-year bond issue outstanding,
carrying an 8 percent e¤ective annual interest rate.
years ago.
This bond was issued exactly 5
The bond indenture has a call provision, allowing the company to retire
the entire bond issue at a 9 percent call premium, which is not tax-deductible.
The
company’s tax rate is 30 percent. Investment dealers have assured the company that it
can easily sell additional $60 million worth of 25-year bonds that carry a lower e¤ective
annual interest rate of 6 percent. Predictions are that interest rates are unlikely to fall
below 6 percent in the foreseeable future. Issuing costs of the new bonds amount to $3
million. Such costs are tax deductible, and the corresponding tax bene…ts are assumed
to be immediately available to the company.
(a1) Is it advisable for the company to refund the $60 million existing bonds, based on
the information provided?
Show all computations as required for a decision.
(6
marks)
(a2) Will the decision be any di¤erent, if McMaster is a tax-exempt institution instead?
Show also all computations as required for a decision. (3 marks)
30
[Hint: It is known that
1
1
1
+
+
+
2
1 + r (1 + r)
(1 + r)3
+
1
1
=
n
(1 + r)
(1 + r) n
;
r
for any positive real number r and any positive integer n:]
(b) (8 marks in total) Hamilton Corporation has the following capital structure, based on
market values, that it wishes to maintain in the future:
debt, 30%; preferred stock,
20%; common stock, 50%: New debt …nancing is available in the form of perpetual 6
percent annual coupon bonds that can be sold at face value, and issuing and underwriting
expenses can be ignored. New preferred shares with annual dividends of $1 per share
will net the company $14 per share. New common shares can be sold to the public at
$16 per share, with 4 percent after-tax issuing costs. Equity investors expect to earn a
9:6% annual return for investing in the company’s common stock. The …rm’s tax rate
is 30 percent. Compute the following:
(b1) The …rm’s cost of debt. (2 marks)
(b2) The …rm’s cost of preferred equity. (2 marks)
(b3) The …rm’s cost of common equity. (2 marks)
(b4) The …rm’s weighted average cost of capital. (2 marks)
Answer:
(a1) Let M = million. The call premium is
0:09($60M ) = $5:4M:
The after-tax issuing cost is
$3M (1
0:30) = $2:1M:
The total of the above is
$5:4M + $2:1M = $7:5M:
Currently, bondholders’required return is
6% = 0:06:
31
The …rm’s cost of debt is
0:06(1
0:30) = 0:042:
Interest savings are
$60M (0:08
0:06)(1
0:30) = $0:84M
each year for 25 years. The present value of interest savings is
"
25 #
1
1
1+0:042
$0:84M
= $0:84M (15:29701) = $12:8495M;
0:042
which is greater than the cost of $7:5M: Conclusion: it is advisable to refund the $60
million existing bonds.
(a2) Let M = million. The call premium is
0:09($60M ) = $5:4M:
As the tax rate is zero, the issuing cost is
$3M:
The total of the above is
$5:4M + $3M = $8:4M:
Currently, bondholders’required return is
6% = 0:06:
The …rm’s cost of debt is
0:06:
Interest savings are
$60M (0:08
0:06) = $1:2M
each year for 25 years. The present value of interest savings is
"
25 #
1
1
1+0:06
$1:2M
= $1:2M (12:78336) = $15:34003M;
0:06
which is greater than the cost of $8:4M: Conclusion: it is advisable to refund the $60
million existing bonds.
32
(b1) The …rm’s cost of debt is
kb = (1
0:3)0:06 = 0:042 = 4:2%:
(b2) The …rm’s cost of preferred equity is
kp =
$1
= 0:07143 = 7:1429%:
$14
(b3) The …rm’s cost of common equity is
ke =
$16
$16(1 0:04)
9:6% = 10%:
(b4) The …rm’s weighted average cost of capital is
W ACC = 0:3kb + 0:20kp + 0:5ke
= 0:3(0:042) + 0:20(0:071429) + 0:5(0:10)
= 0:076886 = 7:6886%:
2. Question: (21 marks in total) MFIN Inc., whose …scal year has just ended, is considering
the replacement of an equipment belonging to an asset class under a declining-balance capital
cost allowance (CCA) rate of 30%: The existing equipment, which will have no resale value in
10 years, can be sold for $15; 000 today. The new equipment, which currently costs $60; 000;
has a 10-year life.
The resale value 10 years from now will be $5; 000: To facilitate the
operations of the new equipment, some additional equipments belonging to an asset class of a
20% CCA rate are also required, for a total purchase price of $10; 000 today. Such additional
equipments, which will have no resale value in 10 years, are intended to be kept forever by
the company. The new equipment will enable MFIN to reduce its before-tax operating costs
by $13; 000 each year (at year end) for the …rst 5 years and $8; 000 each year (at year end)
for the next 5 years.
It is unclear whether MFIN will have net dispositions or net acquisitions for the same
asset class, when the new equipment is disposed 10 years from now. Thus, both scenarios
must be considered. MFIN’s weighted average cost of capital is 12%; and its tax rate is 25%:
Perform an incremental analysis, with keeping the existing equipment as the reference, to
decide whether to replace the existing equipment now.
In order to reach a decision, …rst
consider the six speci…c present-value components (as incremental values) in parts (a)-(f)
33
below. In each case, indicate whether it is a positive or negative component, for the purpose
of computing the net present values in part (g) that follows.
(a) The net purchase price of the new equipment and the additional equipments. (2 mark)
(b) The present value of CCA tax shields, if the new equipment and the additional equipments are to be kept forever. (4 marks)
(c) The present value of salvage. (3 marks)
(d) The present value of the loss of CCA tax shields due to salvage, under each of the two
scenarios; i.e., net dispositions and net acquisitions.
Be speci…c as to which present
value is for which scenario. (4 marks)
(e) The present value of the reduction of after-tax operating costs for the …rst 5 years. (2
marks)
(f) The present value of the reduction of after-tax operating costs for the second 5 years. (3
marks)
Then, answer the following question:
(g) What is MFIN’s decision based on the net present value of replacing the existing equipment, relative to keeping it, under the scenario of net dispositions of the new equipment
10 years from now? (3 marks)
{Hint: The present value of tax shields from declining-balance CCA is
CdT
2(d + k)
2+k
1+k
;
and the present value factor of an annuity for n years is [1
(1 + k) n ]=k: It is your respon-
sibility to determine which of the following ought to be used for computing the present value
of the loss of CCA tax shields due to salvage:
1
Sn dT
n
(1 + k) 2(d + k)
2+k
1+k
or
1
(1 + k)n
Sn dT
d+k
:
Here, C and Sn are the dollar values of the assets involved, d is the CCA rate, T is the
corporate tax rate, and k is the company’s weighted average cost of capital.}
Answer:
34
(a) This is a negative component,
$60; 000
$10; 000 + $15; 000 =
$55; 000:
(b) This is a positive component. For the new equipment itself, it is
($60; 000
$15; 000)(0:3)(0:25)(2 + 0:12)
= $7; 605:23:
2(0:3 + 0:12)(1 + 0:12)
For the additional equipments, it is
($10; 000)(0:2)(0:25)(2 + 0:12)
= $1; 478:79:
2(0:2 + 0:12)(1 + 0:12)
The total is
$7; 605:23 + $1; 478:79 = $9; 084:02:
(c) This is a positive component,
$5; 000
10
1
1 + 0:12
= $5; 000
0:3219732 = $1; 609:87:
(d) In either case, this is a negative component. If net disposition (i.e., dispositions exceeding acquisitions), it is
1
$5; 000(0:3)(0:25)
=
10
(1 + 0:12)
0:3 + 0:12
$0:3219732
892:8571 =
$287:48:
If net acquisition (i.e., acquisitions exceeding dispositions) instead, it is
1
$5; 000(0:3)(0:25)(2 + 0:12)
=
10
(1 + 0:12)
2(0:3 + 0:12)(1 + 0:12)
(e) This is a positive component,
"
1
$13; 000(1 0:25)
5#
1
1+0:12
0:12
(f) This is a positive component,
"
1
1
[$8;
000(1
0:25)]
(1 + 0:12)5
= $9; 750
5#
1
1+0:12
0:12
$0:3219732
845:0255 =
$272:08:
3:604776 = $35; 146:57:
= $0:567427 6; 000 3:604776 = $12; 272:68:
(g) Given the information in the question, all present-value components are accounted for;
a decision can be made by using the results in (a)-(f).
For net dispositions, the net
present value of replacing the existing equipment is
$55; 000 + $9; 084:02 + $1; 609:87
$287:48 + $35; 146:57 + $12; 272:68 = $2; 825:66 > 0:
Thus, the decision is to replace the existing equipment.
35
3. Question: (22 marks in total) Under the assumption of frictionless short sales, consider a
portfolio selection problem based on three securities, for which
3-element column vector of expected returns and V is a 3
Let M be a 3 2 matrix, where the …rst column is
0:07 0:10 0:12
=
1
M ) 1M 0V
1
is a
3 covariance matrix of returns.
and each element of the second column is
1
1: The 3-element column vector of e¢ cient portfolio weights is x = V
for which x0 = r 0 (M 0 V
0
M (M 0 V
1
M ) 1 r;
and x0 = 1; where is a 3-element column vector of
ones. Here, the prime stands for matrix transposition, and r is a 2-element column vector,
where the …rst element is the expected return of the e¢ cient portfolio involved and the second
element is 1: The covariance of returns of any two portfolios with portfolio weight (column)
vectors x and y is x0 V y or, equivalently, y 0 V x; where x0 = y 0 = 1: The portfolio weight
1
( 0V
=
?
?
(column) vector of the global minimum variance portfolio o is xo = V
1
) 1 ; where
x0o = 1: It is known that
0
(M V
1
1
M)
V
1
=
3:45620
0:28191
0:28191
0:02490
=
356:1533 119:0316 50:1009
; V
(a) Find the two missing elements in the 3
1
1
0
M (M V
0
; and 0 V
2 matrix V
1
M)
1
1
9:0708 14:5575
0:5133
1:0920
= 525:2858:
M (M 0 V
1
M ) 1 above, as
indicated by the two question marks there. (2 marks)
(b) Given that the determinant of V is 1:4870 10 7 ; that the correlation of returns between
securities 2 and 3 is 0:15; and that the lower triangle of V is
2
3
0:0025
4 0:0005 0:0064
5;
0:0010
?
0:0100
can a risk-free portfolio be constructed by using the three securities? Explain clearly
with computational support. Then, …nd the missing element in the lower triangle of V ;
as indicated by the question mark there. (5 marks)
(c) An e¢ cient portfolio h with expected return of
3:45620
0:28191
0:28191
0:02490
0:10
1
h = 0:10 is constructed.
=
0:063708
0:003293
Given that
;
determine the correlation of returns between portfolio h and the global minimum variance portfolio. (5 marks)
(d) For portfolio construction based on the three securities, by maximizing the slope of the
tangent line on the ( ; )-plane, determine the range of possible values of the -intercept
36
0
;
of the tangent line that ensures the success of the intended task of slope maximization.
(5 marks)
(e) Suppose that, for a tangency portfolio p based on the three securities, the corresponding
tangent line has an intercept of r = 0:05 on the axis of expected return. Let
p and
p
be the expected return and the standard deviation of returns of portfolio p; respectively.
It is known that
z=
r
p
xp = V
2
p
p = 0:099016; and z
0
1
(
4:406187 6:381977 5:793544
r )=
= 16:581708; for which x0p = 1: Determine
p:
0
;
(5 marks)
Answer:
(a) From
V
1
1
0
M (M V
M)
1
=
a
b
9:0708 14:5575
0:5133
1:0920
0
;
we have
a + 9:0708 + 14:5575 = 0
and
b
0:5133
1:0920 = 1:
a=
23:6283
It follows that
and
b = 2:6053:
(b) Each leading principal minor is a determinant.
The …rst leading principal minor is
0:0025: The second leading principal minor is
0:0025(0:0064)
0:0005(0:0005) = 0:00001575:
The third leading principal minor is given to be 1:4870
principal minors are positive, the 3
10 7 : As all three leading
3 covariance matrix of returns is positive de…nite
according to Sylvestor’s criterion. Thus, regardless of how investment funds are allocate
among the three securities, the corresponding variance of returns of the portfolio is always
positive.
37
Given that the correlation of returns between securities 2 and 3 is 0:15; the missing
element in the lower triangle of V ; which is the covariance of returns between the two
securities, is
p
p
0:15 0:0064 0:0100 = 0:15
0:08
0:10 = 0:0012:
(c) The variance of returns of the global minimum variance portfolio is
2
o
= x0o V xo = x0o V V 1 ( 0 V
1
=
= 0:00190373:
525:2858
1
) 1 = x0o ( 0 V
1
1
) 1 = ( 0V
) 1
The corresponding standard deviation of returns is
o = 0:043632:
The variance of returns of portfolio k is
2
k
= x0k V xk = r 0k (M 0 V
= r 0k (M 0 V
=
0:10 1
=
0:10 1
1
1
1
M ) 1M 0V
M ) 1M 0V
3:45620
0:28191
1
M (M 0 V
0:28191
0:02490
0:063708
0:003293
V V
1
1
M (M 0 V
1
M ) 1 r k = r 0k (M 0 V
0:10
1
= 0:0063708
M ) 1 rk
1
M ) 1 rk
0:003293 = 0:003078:
The corresponding standard deviation of returns is
k = 0:05548:
The covariance of returns between portfolio k and the global minimum variance portfolio
is
ok
=
0
0
ko = xk V xo = xk V V
=
1
= 0:00190373:
525:2858
1
( 0V
1
1
) 1 = x0k ( 0 V
) 1 = ( 0V
Thus, their correlation of returns is
ko
o k
=
0:00190373
= 0:78644:
0:043632 0:05548
(d) Given
V
1
=
356:1533 119:0316 50:1009
38
0
; and 0 V
1
= 525:2858;
1
) 1
the global minimum variance portfolio weight (column) vector is
1
356:1533 119:0316 50:1009
525:2858
0
0:67802 0:22660 0:09538 :
=
0
xo =
The corresponding expected return is
x0o
2
3
0:07
0:67802 0:22660 0:09538 4 0:10 5
=
0:12
= 0:04746 + 0:02266 + 0:01145
= 0:08157:
Thus, the intercept of the tangent line must be less than 0:08157; that is, the permissible
range is from minus in…nity to any positive value that is less than 0:08157:
(e) Given
r
p
z=
2
p
xp ;
we have
r
p
z1 + z2 + z3 =
2
p
(x1 + x2 + x3 ) =
r
p
2
p
;
because the three portfolio weights sum to one. Given also
z1 + z2 + z3 = z 0 = 16:581708
and
z1 + z2 + z3 =
r
p
2
p
=
0:099016
0:05
2
p
=
0:049016
2
p
;
we have
2
p =
0:049016
= 0:002956
16:581708
and then
p = 0:05437:
4. Question: (24 marks in total) The two parts of this question are unrelated.
(a) (12 marks in total) Consider a frictionless capital market with a …xed supply of n risky
securities, where all investors are assumed to have homogeneous expectations and to be
able to lend or borrow at the same risk-free interest rate Rf : Each investor constructs
39
a portfolio p by maximizing ( p
Rf )= p ; where
p and
p are the portfolio’s expected
return and standard deviation of returns, respectively. The only constraint is that the n
portfolio weights, x1 ; x2 ; : : : ; xn ; must sum to 1: It is known that the optimization result
can be deduced from
i
where
Rf =
Xn
Rf
p
2
p
ij xj ;
j=1
i is the expected return of security i; and
for i = 1; 2; : : : ; n;
ij is the covariance of returns between
securities i and j:
(a1) Before market equilibrium is established, what does
clearly with analytical justi…cation. (3 marks)
Pn
j=1
ij xj represent?
Explain
(a2) Under this optimization setting for individual investors, show analytically and explain clearly how the security market line of the capital asset pricing model can be
reached. In particular, explain clearly how homogeneous expectations by investors
are used for facilitating the attainment of market equilibrium in such an analytical
setting. (3 marks).
ei and R
em be the random returns
(a3) For a capital market with n risky securities, let R
of security i and the market portfolio m; respectively.
Let
2
i
2
m
and
be the
corresponding variance of returns. Suppose that there is a linear relationship
ei =
R
i+
e
ei ;
i Rm + e
for i = 1; 2; : : : ; n;
where i and i are parameters and eei is random noise with a zero expected value
em and eei are uncorrelated. Suppose also that eei and eej are uncorrelated,
and R
for i 6= j: Such a linear relationship also holds for any portfolio p based on the n
securities; that is,
ep =
R
where
ep =
R
Xn
i=1
ei ;
xi R
p =
Xn
i=1
p+
xi i ;
e
ep ;
p Rm + e
p =
Xn
i=1
xi i ; and eep =
Xn
i=1
xi eei :
Here, x1 ; x2 ; : : : ; xn is any set of portfolio weights satisfying the condition of
Xn
i=1
40
xi = 1:
(a3i) Decompose the variance of returns of security i into systematic and unsystematic
components, in terms of
2
ei :
m ; and the variance of e
i;
analytically. (3 marks)
Justify your answer
(a3ii) Suppose that an equally-weighted portfolio based on k securities is formed, for
k = 2; 3; 4; : : : ; n: For each k; the corresponding portfolio weights are
x1 = x2 =
= xk =
1
:
k
As k increases, does the unsystematic component of the variance of portfolio returns tend to increase, tend to decrease, or tend to remain unchanged? Explain
clearly with analytical justi…cation. (3 marks)
(b) (12 marks in total) Consider a capital market with frictionless short sales, in which great
many risky securities are available for investing. Assume that all investors in the market
have homogeneous expectations. The derivation of the arbitrage pricing model starts
with a security return generating equation.
In the special case where there are only
four mean-removed economic factors, labeled as Fe1 ; Fe2 ; Fe3 ; and Fe4 ; the security return
generating equation for each security i of the n securities considered is
ei = ai + bi1 Fe1 + bi2 Fe2 + bi3 Fe3 + bi4 Fe4 + eei :
R
ei is the random return of security i; eei is random noise with a zero mean, and ai ;
Here, R
bi1 ; bi2 ; bi3 ; and bi4 are parameters.
(b1) Explain clearly what each ai represents and whether its value known in advance.
(2 marks)
(b2) Suppose that an investor allocates dollar amounts w1 ; w2 ; : : : ; wn to securities 1; 2; : : : ; n;
respectively, under the conditions that the investment requires no cash outlays and
does not respond to any of Fe1 ; Fe2 ; Fe3 ; and Fe4 : Establish a set of analytical conditions for a self-…nanced risk-free portfolio. Explain clearly whether the set of
w1 ; w2 ; : : : ; wn is unique or there are in…nitely many ways to assign w1 ; w2 ; : : : ; wn
to the n securities, for the purpose of forming a self-…nanced risk-free portfolio.
Explain clearly whether the sum
Xn
i=1
wi eei
is exactly zero or is only approximately zero in this analytical setting. (6 marks)
41
(b3) Then, explain clearly why, in the absence of arbitrage pro…ts, a linear relationship
of the form
i =
0+
can be established.
3 ; and
Here,
2 bi2 +
3 bi3 +
4 bi4 ;
for i = 1; 2; : : : ; n;
i is the expected return of security i; and
4 are parameters, which are common for all securities.
clearly how
and
1 bi1 +
0 and
2 can be determined.
0;
1;
2;
Finally, explain
However, the determination of
1;
3;
4 ; for the purpose of completing the model derivation, is not required here.
(4
marks)
Answer: Part (a) of the question has two related components. Part (b) of the question also
has two related components. The answers provided below, especially in the concepts involved,
are in much greater detail than what are required for the purpose of grading.
However, to
earn full component marks, your answers must include clear and valid explanations. Listing
of equations or analytical expressions without explanations is deemed inadequate. Neither is
the inclusion of materials that are not even asked. The grader has the task of assessing how
well you understand the concepts involved, by reading your written answers.
(a1) We …rst write
Xn
j=1
ij xj =
Xn
j=1
ei ; R
ej )xj ;
Cov(R
ei and R
ej are random returns of securities i and j; respectively. By de…nition,
where R
h
ei ; R
ej ) = E (R
ei
Cov(R
e
i )(Rj
i
)
j
is the covariance of returns between securities i and j: Here, i and j are the expected
h
i
e
e
returns of the two securities, and E (Ri
i )(Rj
j ) represents the expected value
ei
of (R
e
j ); which is a random variable.
i )(Rj
Taking the expected value of a random
variable is a linear operation. Thus, we can write
Xn
j=1
ij xj =
Xn
j=1
ei ;
e i ; xj R
ej ) = Cov R
Cov(R
The …rst step above is by recognizing that
h
e i ; xj R
ej ) = E (R
ei
Cov(R
42
e
i )(xj Rj
Xn
j=1
i
xj j ) :
ej :
xj R
The second step above is by interchanging the order of two linear operations, which
involve a summation and an expected value.
Next, let us construct a portfolio p by maximizing ( p
ep =
R
Xn
j=1
Rf )= p : Recognizing that
ej
xj R
is the random return of portfolio p; we can also write
Xn
j=1
e e
ij xj = Cov(Ri ; Rp );
ei and R
ep :
which is the covariance between R
ei ; R
ep ); the covariance between R
ei and R
ep ; simply as
(a2) Labeling Cov(R
ip =
The set of equations in
i
Rf =
j=1
Xn
Rf
p
Xn
2
p
j=1
ip ; we can write
ij xj :
ij xj ;
for i = 1; 2; : : : ; n;
becomes
i
Rf =
Rf
p
2
p
ip ;
for i = 1; 2; : : : ; n;
ip
;
2
p
for i = 1; 2; : : : ; n:
or, equivalently,
i = Rf +
p
Rf
Now, consider a frictionless capital market where all investors have homogeneous expectations. In this frictionless capital market, short sellers are not required to provide
margin deposits for any shorted securities. They also have immediate access to the short
sale proceeds. The condition of homogeneous expectations is the simplifying assumption
that all investors agree in their assessments of the available information about the n risky
securities.
Further, all investors can lend or borrow risk-free at the same interest rate Rf : Being able
to lend or borrow at the same interest rate allows investors to allocate their investment
funds between portfolio p and the risk-free security in di¤erent ways, depending on the
investors’ risk-return preferences.
If an investor chooses to borrow risk-free, both the
expected return and the corresponding risk will be higher than those of portfolio p:
43
In such a market, all investors consider portfolio p as the best portfolio based on the
n risky securities. They all want to hold such a portfolio. As a result, the aggregate
demand on each security will a¤ect its price and thus the corresponding expected return.
Security prices will adjust, until the supply of each security equals its aggregate demand.
When market equilibrium is reached, each investor’s portfolio p becomes the market
portfolio m; because all investors will allocate their investment funds in the same way.
Then, the set of equations
i = Rf +
p
Rf
ip
;
2
p
for i = 1; 2; : : : ; n:
i = Rf +
p
Rf
im
;
2
m
for i = 1; 2; : : : ; n:
becomes
Here,
im
2
m
is the beta coe¢ cient of security i; which is
i:
The equation
i = Rf +
p
Rf
i
is the security market line of the capital asset pricing model.
The assumption of homogeneous expectations is highly important for establishing an
equilibrium pricing relationship here. Without such an assumption, a security that is
considered desirable for portfolio investments by some investors may not also be considered desirable by some other investors.
As a result, with the portfolio choices by
individual investors being di¤erent, it is unclear what e¤ects their actions in aggregate
will have on security price movements in the market.
(a3i) To answer this part of the question, we use a basic property of the variance of the sum of
random variables. To illustrate, if x
e and ye are two random variables, then we can write
Here, we start with
V ar(e
x + ye) = V ar(e
x) + V ar(e
y ) + 2Cov(e
x; ye):
ei =
R
i+
e
ei ;
i Rm + e
for i = 1; 2; : : : ; n;
where i and i are parameters and eei is random noise with a zero expected value and
em and eei are uncorrelated. Notice that, for each security i;
R
V ar(ai ) = 0;
44
because ai is a parameter, not a random variable. Notice also that
em ; eei ) = Cov(e
em ) = 0;
Cov(R
ei ; R
em and eei are uncorrelated. For each security i; as
because R
can be reduced to
ei ) = V ar( i +
V ar(R
e
ei )
i Rm + e
ei ) = V ar( i R
em + eei );
V ar(R
em + eei ) in terms of its individual components will have
and as the expression of V ar( i R
zero covariance terms, we can write directly
em ) + V ar(e
ei ) = V ar( i R
ei ) =
V ar(R
2
e
ei ):
i V ar(Rm ) + V ar(e
Notice that the variance of “a constant times a random variable” is the square of the
em ); which is also labeled
constant times the variance of the random variable. Here, V ar(R
as
2
ei ); which is known as
m ; is the variance of returns of the market portfolio, and V ar(e
the residual variance of security i and often labeled as
2
ei ; captures the part of the risk
of security i that has nothing to do with risk of the market portfolio. Thus,
2
em )
V ar(R
is the systematic component of the variance of returns of security i; and V ar(e
ei ) is the
unsystematic component of the variance of returns of security i:
(a3ii) As
ei =
R
i+
e
ei ;
i Rm + e
for i = 1; 2; : : : ; n;
The random return of an equally-weighted portfolio based on k securities is
where
and
ek =
R
k +
e
ek ;
k Rm + e
Xk
ek = 1
ei ;
R
R
i=1
k
1 Xk
k =
i;
i=1
k
1 Xk
=
i;
k
i=1
k
eek =
45
1 Xk
eei :
i=1
k
Given that
em ; eei ) = 0
Cov(R
and
Cov(e
ei ; eej ) = 0; for i 6= j;
ek can be decomposed as
the variance of R
ek ) = V ar( k R
em + eek ) =
V ar(R
As
eek =
we have
2
e
ek ):
k V ar(Rm ) + V ar(e
1 Xk
eei ;
i=1
k
1 Xk
1 1 Xk
V
ar(e
e
)
=
V ar(e
ei ) ;
i
i=1
i=1
k2
k k
which is 1=k of the average of the residual variances of the k securities.
V ar(e
ek ) =
Thus, as k
increases, V ar(e
ek ) tends to decrease.
(b1) Given
ei = ai + bi1 Fe1 + bi2 Fe2 + bi3 Fe3 + bi4 Fe4 + eei ;
R
ei can be expressed as
the expected value of R
ei ) = ai + bi1 E(Fe1 ) + bi2 E(Fe2 ) + bi3 E(Fe3 ) + bi4 E(Fe4 ) + E(e
E(R
ei ):
As Fe1 ; Fe2 ; Fe3 ; and Fe4 have been mean-removed and as eei is random noise with a zero
mean, all expected values on the right hand side of this equation are zeros; i.e.,
E(Fe1 ) = 0;
E(Fe2 ) = 0;
E(Fe3 ) = 0;
E(Fe4 ) = 0;
and E(e
ei ) = 0:
Thus, we have
e i ) = ai :
E(R
ei or, equivalently, the expected return of security i;
That is, ai is the expected value of R
commonly labeled as
i:
46
In this analytical setting, we do NOT know the value of each
i in advance.
Knowing
its value in advance will defeat the whole purpose of …nding a pricing relationship, given
the available information that is provided by the security return generating equations.
(b2) As the return generating equation for each security i is linear, we expect
to bi1 ; bi2 ; bi3 ; and bi4 linearly.
i to be related
However, the exact linear relationship, which allows
us to determine the value of each
i ; for i = 1; 2; : : : ; n; will have to be established by
using an arbitrage reasoning. A derivation of the arbitrage pricing model is all about
establishing the value of each
i ; given the available information that is provided by the
security return generating equations.
To achieve a self-…nanced investment, we assign values to w1 ; w2 ; : : : ; wn ; satisfying the
condition of
Xn
i=1
wi = w1 + w2 +
+ wn = 0:
In a frictionless capital market, the proceeds from short selling securities are immediately
available. Thus, we can always have positive and negative values among w1 ; w2 ; : : : ; wn
to satisfy the above condition.
This self-…nance investment is intended to be insensitive to the randomness in each of
Fe1 ; Fe2 ; Fe3 ; and Fe4 : Such a feature can be achieved, as long as we set w1 ; w2 ; : : : ; wn in
such a way that the conditions
Xn
i=1
Xn
+ wn bn1 = 0;
wi bi2 = w1 b12 + w2 b22 +
+ wn bn2 = 0;
Xn
wi bi3 = w1 b13 + w2 b23 +
+ wn bn3 = 0;
Xn
wi bi4 = w1 b14 + w2 b24 +
+ wn bn4 = 0
i=1
i=1
and
wi bi1 = w1 b11 + w2 b21 +
i=1
are also satis…ed. As this capital market has great many securities, there are in…nitely
many ways to assign w1 ; w2 ; : : : ; wn to satisfy the above four conditions.
From an
algebraic perspective, as there are 5 equations with n variables, which are w1 ; w2 ; : : : ; wn ;
there will be in…nitely many solutions, as long as n > 5:
Given the security return generating equations
ei =
R
e
e
e
e
ei ;
i + bi1 F1 + bi2 F2 + bi3 F3 + bi4 F4 + e
47
for i = 1; 2; : : : ; n;
the random return of this self-…nanced investment is
Xn
Xn
Xn
Xn
ei =
wi R
wi i +
wi bi1 Fe1 +
wi bi2 Fe2
i=1
i=1
i=1
i=1
Xn
Xn
Xn
+
wi bi3 Fe3 +
wi bi4 Fe4 +
wi eei :
i=1
It reduces to
Xn
i=1
i=1
ei =
wi R
Xn
i=1
wi i +
i=1
Xn
i=1
wi eei ;
because we have assigned w1 ; w2 ; : : : ; wn to satisfy the above …ve conditions.
As there are great many securities in the market, the law of large numbers in statistics
gives us
Xn
i=1
A direct implication is that
Xn
i=1
wi eei
0:
Xn
ei
wi R
wi i :
i=1
That is, the return of this self-…nanced portfolio is nearly risk-free, because there is no
randomness on the right hand side.
In the derivation of the arbitrage pricing model,
it is treated as risk-free. A crucial point here is that there are in…nitely many ways to
establish a self-…nanced risk-free investment.
In the absence of arbitrage pro…ts, we must have
Xn
wi i = 0:
i=1
(b3) Given the return generating equation, which is linear, we expect
i to be related to bi1 ;
bi2 ; bi3 ; and bi4 linearly. The linear relationship can be written as
where the coe¢ cients
0;
i =
0+
1;
2;
1 bi1 +
3 and
2 bi2 +
3 bi3 +
4 bi4 ;
4 have yet to be determined.
Given that each
of the sums
Xn
Xn
Xn
Xn
Xn
Xn
wi i ;
wi ;
wi bi1 ;
wi bi2 ;
wi bi3 ; and
wi bi4
i=1
i=1
i=1
i=1
i=1
is zero, a linear combination of them is also zero. Thus, we can write
Xn
Xn
Xn
Xn
wi i
wi
wi bi1
wi bi2
0
1
2
i=1
i=1
i=1
i=1
Xn
Xn
wi bi3
wi bi4
3
4
i=1
Xn i=1
=
wi ( i
0
1 bi1
2 bi2
3 bi3
4 bi4 ) = 0:
i=1
48
i=1
There are in…nitely many way for us to assign w1 ; w2 ; : : : ; wn ; for the purpose of achieving
a self-…nanced risk-free investment. Regardless of how we assign w1 ; w2 ; : : : ; wn ; we must
have the same pricing relationship when relating each
i to the available information that
the n security generating equations provide. Thus, we must have
1 bi1
0
i
2 bi2
3 bi3
4 bi4 = 0;
for i = 1; 2; : : : ; n;
or, equivalently,
i =
0+
Here, the coe¢ cients
To determine
for i = 1; 2; : : : ; n:
1 bi1 +
2 bi2 +
3 bi3 +
1;
3 ; and
4 have yet to be determined.
0;
2;
4 bi4 ;
0 ; we consider a risk-free security i:
we must have bi1 = bi2 = bi3 = bi4 = 0 and
As this speci…c security i has no risk,
i = Rf ; which is the risk-free interest rate.
Thus, we can write
Rf =
which leads to
To determine
i =
0+
0+
1
2
0+
3
0+
4
0;
0 = Rf :
2 ; we consider a portfolio 2 that responds to factor 2 only.
Let x1 ; x2 ; : : : ; xn
be a set of portfolio weights satisfying the conditions of
Xn
xi bi2 = 1
Xn
xi bi3 =
i=1
and
Xn
i=1
xi bi1 =
i=1
Xn
i=1
xi bi4 = 0;
Notice that there are in…nitely many ways to construct such a portfolio. If the expected
return of portfolio 2; which is
2
Xn
= Rf +
1
= Rf +
2:
i=1
2 ; is known, we can determine
xi bi1 +
2
Xn
i=1
xi bi2 +
The result is
2 =
A.2.5 Second Mid-Term Examination, 2018
49
2
Rf :
3
Xn
i=1
2 from
xi bi3 +
4
Xn
i=1
xi bi4
1. (16 marks in total) The two parts of this question are unrelated.
(a) McMaster Corporation, whose …scal year has just ended, is considering whether to re…nance $30 million of its existing preferred stock that carries a 11% dividend rate. The
stock can be called at a 6% premium of its par value. This premium is not tax-deductible
for the …rm. Given the current market condition, there should not be any di¢ culty selling a $30 million worth of new preferred issue with a 10% dividend rate. The new issue,
however, will require $1:2 million in ‡otation costs, which include underwriting fees, legal
fees, and registration fees. Such costs are to be amortized over 5 years for tax purposes.
McMaster’s tax rate is 30%: Should the …rm re…nance the preferred issue? (7 marks)
(b) (9 marks in total) MFIN Corporation has the following capital structure, based on
market values, that it wishes to maintain in the future:
debt, 30%; preferred stock,
20%; common stock, 50%: New debt …nancing is available in the form of perpetual
6% annual coupon bonds that can be sold at face value, and issuing and underwriting
expenses can be ignored. New preferred shares with annual dividends of $1:10 per share
will net the company $16 per share. Common shares can be sold to the public at $16
per share, with 4% after-tax issuing costs.
The company’s common stock has a beta
coe¢ cient of 0:8; the current e¤ective annual yield on long-term government bonds is
4%; and the return on the market portfolio is expected to be 12%: The …rm’s tax rate
is 30%: Compute the following:
(b1) The company’s cost of debt. (2 marks)
(b2) The company’s cost of preferred equity. (2 marks)
(b3) The company’s cost of common equity. (3 marks)
(b4) The company’s weighted average cost of capital. (2 marks)
Answer:
(a) Let M = 1; 000; 000; which is one million. The call premium is
$30M
6% = $1:8M:
Flotation costs are
$1:2M;
50
as given. The total cost for re…nancing the preferred stock is
$1:8M + $1:2M = $3M;
which is a present value. Perpetual annual savings in dividend payments are
$30M
(11%
10%) = $0:3M:
The …rm’s tax rate is 30%: Annual tax savings due to ‡otation costs for 5 years are
1
5
$1:2M
(30%) = $0:072M:
The present value of the total savings above is
$0:3M
+ $0:072M
0:10
(1:10) 5
= $3M + $0:2729M = $3:2729M:
0:10
1
As the present value of total savings ($3:2729M ) is greater than the total cost for re…nancing the preferred stock ($3M ); the decision is to re…nance the preferred stock.
(b) The …rm’s cost of debt is
kb = (1
0:3)0:06 = 0:042 = 4:2%:
The …rm’s cost of preferred equity is
kp =
1:1
= 0:06875 = 6:875%:
16
To …nd the …rm’s cost of common equity ke ; we …rst …nd investors’ expected return,
labeled as re ; via the CAPM; that is,
re = rf + (rm
rf )
= 0:04 + (0:12
0:04)(0:8) = 0:104 = 10:4%:
Here, rm is the expected return of the market portfolio, rf is the risk-free interest rate,
and
is the beta coe¢ cient of the …rm’s common stock.
The …rm’s cost of common
equity ke is re multiplied by the factor of “share price to net proceeds.” That is,
ke =
16
1
re =
16(1 0:04)
0:96
0:104 = 0:10833 = 10:833%:
Thus, the …rm’s weighted average cost of capital is
W ACC = 0:3kb + 0:2kp + 0:5ke
= 0:3(0:042) + 0:2(0:06875) + 0:5(0:10833)
= 0:08052 = 8:052%:
51
2. (22 marks in total) The …scal year of McMaster Corporation of Canada has just ended. In
this question, CCA stands for declining-balance capital cost allowances.
(a) The Canadian corporate tax rate is T; and McMaster’s weighted average cost of capital
is k: McMaster is considering the purchase of a new machine now for $C: The machine
belongs to an asset class under a CCA rate d: For the scenario that the machine is to be
kept forever by McMaster, derive an expression of the present value of CCA tax shields
in terms of C; d; k; and T: (6 marks)
(b) (16 marks in total) McMaster’s tax rate is 25%; and its weighted average cost of capital
is 10%: The company is considering the purchase of a new machine, known as machine
A; to replace an existing machine, for improving its operational e¢ ciency. Both machine
A and the existing machine are in an asset class with a 30% CCA rate. The existing
machine, which will have no resale value in 10 years, can be sold for $16; 000 today.
Machine A; which currently costs $50; 000; has a 10-year life. To facilitate its operations,
an auxiliary machine, known as machine B; will also have to be purchased 4 years from
now for $14; 000 when it becomes available. Machine B; which belongs to an asset class
with a 20% CCA rate and will have no resale value 10 years from now, is intended to
be kept forever by the company. The replacement of the existing machine will enable
McMaster to reduce its before-tax operating costs by $4; 000 each year (at year end) for
the …rst 4 years, and by $12; 000 each year (at year end) for the next 6 years. The resale
value of machine A; 10 years from now, will be $5; 000: It is expected that, when the
machine A is disposed 10 years from now, McMaster will have net acquisitions for the
same asset class.
Perform an incremental analysis, with keeping the existing machine as the reference,
to decide whether to replace the existing machine now. In order to reach a decision, …rst
consider the six speci…c present value (PV) components, as incremental values, in parts
(b1)-(b6) below. In each case, indicate whether it is a positive or negative component,
for the purpose of computing the net present values in part (b7). Each numerical answer
ought to be rounded to the nearest hundredth of a dollar.
(b1) The PV of the incremental purchase prices of the two new machines. (2 marks)
(b2) The PV of the incremental CCA tax shields, if both new machines are to be kept
forever. (4 marks)
52
(b3) The PV of salvage of machine A: (2 marks)
(b4) The PV of the loss of CCA tax shields due to salvage. (2 marks)
(b5) The PV of the after-tax operating cost reductions for the …rst 4 years. (2 marks)
(b6) The PV of the after-tax operating cost reductions for the next 6 years. (2 marks)
(b7) What is McMaster decision based on the net present value of replacing the existing
machine, relative to keeping it? (2 marks)
Answer:
(a) For the task, we start with the following:
j
t= 0
Year
t
1
2
3
4
..
.
j
1
j
2
j
3
j
n
j
j
n+1 n+2
Undepreciated Capital Cost, UCC ($); Maximum CCA
at t 1
Rate for Year t
C
d=2
C(1 d=2)
d
C(1 d=2)(1 d)
d
2
C(1 d=2)(1 d)
d
..
..
.
.
years
Maximum CCA ($)
for Year t
Cd=2
C(1 d=2)d
C(1 d=2)(1 d)d
C(1 d=2)(1 d)2 d
..
.
As
P V of CCA
Cd=2 Cd(1 d=2) Cd(1 d=2)(1 d) Cd(1 d=2)(1
+
+
+
1+k
(1 + k)2
(1 + k)3
(1 + k)4
#)
(
"
2
Cd
1
1 d=2
1 d
1 d
=
+
+
1+
+
1+k 2
1+k
1+k
1+k
#
"
Cd 1
1 d=2
1
=
+
1 d
1+k 2
1+k
1
1+k
=
Cd
1 1 d=2
+
1+k 2
d+k
Cd
2+k
=
1 + k 2(d + k)
Cd
2+k
=
;
2(d + k) 1 + k
=
we have
P V of CCA tax shields =
(b) Here are the details:
53
CdT
2(d + k)
2+k
1+k
:
d)2
+
(b1) The PV of the incremental purchase prices of the two new machines is
$50; 000
$16; 000 +
$14; 000
= $43; 562:19:
(1 + 0:1)4
This is a negative component.
(b2) If both new machines are to be kept forever, the PV of the incremental CCA tax
shields is
$14; 000(0:2)(0:25)(2 + 0:1)
$16; 000)(0:3)(0:25)(2 + 0:1)
+
2(0:3 + 0:1)(1 + 0:1)
2(0:2 + 0:1)(1 + 0:1)(1 + 0:1)4
= $6; 085:227 + $1; 521:257 = $7; 606:48:
($50; 000
This is a positive component.
(b3) The PV of salvage of machine A is
$5; 000
= $1; 927:72:
(1 + 0:1)10
This is a positive component.
(b4) The PV of the loss of CCA tax shields due to salvage is
$5; 000(0:3)(0:25)(2 + 0:1)
= $345:02:
2(0:3 + 0:1)(1 + 0:1)(1 + 0:1)10
This is a negative component.
(b5) The PV of the after-tax operating cost reductions for the …rst 4 years is
$4; 000(1
1
0:25)
(1 + 0:1) 4
= $9; 509:60:
0:1
This is a positive component.
(b6) The PV of the after-tax operating cost reductions for the next 6 years is
$12; 000(1
0:25)
1
(1 + 0:1) 6
1
= $26; 772:31:
0:1
(1 + 0:1)4
This is a positive component.
(b7) The net present value of replacing the existing machine is
$43; 562:19 + $7; 606:48 + $1; 927:72
$345:02 + $9; 509:60 + $26; 772:31
= 1; 908:91 > 0:
Thus, the decision is to replace the existing machine.
54
3. (20 marks in total) Under the assumption of frictionless short sales, consider a portfolio
selection problem based on three securities, for which
column vector of expected returns and V is a 3
be a 3
2 matrix, where the …rst column is
0:07 0:10 0:12
=
is a 3-element
3 covariance matrix of returns. Let M
and each element of the second column is
1: The 3-element column vector of e¢ cient portfolio weights is x = V
for which x0 = 1; where
0
is a 3-element column vector of ones.
1
M (M 0 V
1
M ) 1 r;
Here, the prime stands
for matrix transposition, and r is a 2-element column vector, where the …rst element is the
expected return of the e¢ cient portfolio involved and the second element is 1: The covariance
of returns of any two portfolios with portfolio weight (column) vectors x and y is x0 V y or,
equivalently, y 0 V x; where x0 = y 0 = 1: The portfolio weight (column) vector of the global
1
( 0V
1
0:28540
0:02516
;
minimum variance portfolio o is xo = V
) 1 ; for which x0o = 1: Here is some
partial information:
(M 0 V
1
M) 1 =
V
1
=
3:49825
0:28540
360:4713 124:5313 49:0091
0
; and 0 V
1
= 534:0118:
(a) Given that the determinant of V is 1:4936 10 7 ; that the correlation of returns between
securities 2 and 3 is 0:15; and that the upper triangle of V is
2
3
0:0025 0:0004 0:0010
4
5;
0:0064
?
0:0100
can a risk-free portfolio be constructed by using the three securities? Explain clearly
with computational support.
Then, …nd the missing element in the upper triangle of
V ; as indicated by the question mark there. (5 marks)
(b) An e¢ cient portfolio k with expected return of
k = 0:10 is constructed.
It is known
that
(M 0 V
1
M) 1
0:10
1
=
0:064421
0:003383
:
Determine the correlation of returns between portfolio k and the global minimum variance
portfolio. (5 marks)
(c) Suppose that, for a tangency portfolio p based on the three securities, the corresponding
tangent line has an intercept of r = 0:05 on the axis of expected return. Let
p and
p
be the expected return and the standard deviation of returns of portfolio p; respectively.
55
It is known that
z=
r
p
xp = V
2
p
p = 0:098533; and z
0
1
(
4:665238 6:440814 5:760578
r )=
= 16:866631; for which x0p = 1: Determine
p:
0
;
(5 marks)
(d) For portfolio construction based on the three securities, by maximizing the slope of the
tangent line on the ( ; )-plane, determine the range of possible values of the -intercept
of the tangent line that ensures the success of the intended task of slope maximization.
(5 marks)
Answer:
(a) Each leading principal minor is a determinant.
The …rst leading principal minor is
0:0025: The second leading principal minor is
0:0025(0:0064)
0:0004(0:0004) = 0:00001584:
The third leading principal minor is given to be 1:4936
principal minors are positive, the 3
10 7 : As all three leading
3 covariance matrix of returns is positive de…nite
according to Sylvestor’s criterion. Thus, regardless of how investment funds are allocate
among the three securities, the corresponding variance of returns of the portfolio is always
positive.
Given that the correlation of returns between securities 2 and 3 is 0:15; the missing
element in the lower triangle of V ; which is the covariance of returns between the two
securities, is
p
p
0:15 0:0064 0:0100 = 0:15
0:08
0:10 = 0:0012:
(b) The variance of returns of the global minimum variance portfolio is
2
o
= x0o V xo = x0o V V 1 ( 0 V
1
=
= 0:00187262:
534:0118
1
) 1 = x0o ( 0 V
The corresponding standard deviation of returns is
o =
p
0:00187262 = 0:043274:
56
1
) 1 = ( 0V
1
) 1
The variance of returns of portfolio k is
2
k
1
= x0k V xk = r 0k (M 0 V
= r 0k (M 0 V
=
0:10 1
=
0:10 1
1
1
M ) 1M 0V
1
M ) 1M 0V
3:49825
0:28540
M (M 0 V
0:28540
0:02516
0:064421
0:003383
V V
1
1
1
M (M 0 V
M ) 1 r k = r 0k (M 0 V
0:10
1
= 0:0064421
M ) 1 rk
1
M ) 1 rk
0:003383 = 0:003059:
The corresponding standard deviation of returns is
p
k =
0:003059 = 0:05531:
The covariance of returns between portfolio k and the global minimum variance portfolio
is
ok
1
=
0
0
ko = xk V xo = xk V V
=
1
= 0:00187262:
534:0118
1
( 0V
1
) 1 = x0k ( 0 V
) 1 = ( 0V
Thus, their correlation of returns is
ko
=
o k
0:00187262
= 0:78242:
0:043274 0:05531
(c) Given
r
p
z=
2
p
xp ;
we have
r
p
z1 + z2 + z3 =
2
p
(x1 + x2 + x3 ) =
r
p
2
p
;
because the three portfolio weights sum to one. Given also
z1 + z2 + z3 = z 0 = 16:866631
and
z1 + z2 + z3 =
r
p
2
p
=
0:098533
0:05
2
p
we have
2
p =
and then
p =
0:048533
= 0:0028774
16:866631
p
0:0028774 = 0:05364:
57
=
0:048533
2
p
;
1
) 1
(d) Given
V
1
=
360:4713 124:5313 49:0091
0
; and 0 V
1
= 534:0118;
the global minimum variance portfolio weight (column) vector is
1
360:4713 124:5313 49:0091
534:0118
0
0:67503 0:23320 0:09178 :
=
xo =
The corresponding expected return is
x0o
0
2
3
0:07
0:67503 0:23320 0:09178 4 0:10 5
=
0:12
= 0:04725 + 0:02332 + 0:01101
= 0:08158:
Thus, the intercept of the tangent line must be less than 0:08158; that is, the permissible
range is from minus in…nity to any positive value that is less than 0:08158:
4. (26 marks in total) The four parts of this question are unrelated.
(a) The input parameters for implementing a portfolio selection model for n risky securities
include an n-element vector of expected returns and an n n covariance matrix of returns.
As the true values of these input parameters are unknown, they are often estimated by
using past return observations. For the estimation, let Rit be the return of security i in
month t; for i = 1; 2; : : : ; n and t = 1; 2; : : : ; T: Verify analytically and explain clearly
whether the sample covariance matrix of returns is always positive semi-de…nite, always
positive de…nite, or neither. (7 marks)
ei and R
em be the random
(b) For a frictionless capital market with n risky securities, let R
returns of security i and the market portfolio m; respectively.
Let
2
i and
2
m be the
corresponding variances of returns. Suppose that there is a linear relationship
where
i and
ei =
R
i+
e
ei ;
i Rm + e
for i = 1; 2; : : : ; n;
e
ei is random noise with a zero expected value and Rm
i are parameters and e
and eei are uncorrelated. This linear relationship leads to decomposition of the variance
of security returns into systematic and unsystematic components; that is,
2
i =
2 2
i m+
58
2
ei ;
for i = 1; 2; : : : ; n;
where
2
ei : Suppose also that eei and eej are uncorrelated for i 6= j and
ei is the variance of e
2
e1 =
2
e2 =
2
e3 =
=
2
en :
The random return of any portfolio p based on the n securities is
ep =
R
Xn
i=1
ei ;
xi R
where x1 ; x2 ; : : : ; xn are portfolio weights satisfying the condition of
Xn
i=1
xi = 1:
Now, an equally-weighted portfolio based on k securities is formed, where 1 < k < n:
For each k; the corresponding portfolio weights are
x1 = x2 =
= xk = 1=k and xk+1 = xk+2 =
= xn = 0:
Justify analytically and explain clearly, as k increases, whether the unsystematic component of the variance of portfolio returns increases, decreases, or remains constant. (5
marks)
(c) Use Sharpe’s approach to derive the security market line of the capital asset pricing
model. What assumptions are required in the derivation? Explain clearly.
(7 marks)
(d) (7 marks in total) Consider a frictionless capital market where there are great many
risky securities for investing. Assume that all investors in the market have homogeneous
expectations. The derivation of the arbitrage pricing model starts with a security return
generating equation.
In the special case where there are only three mean-removed
economic factors, labeled as Fe1 ; Fe2 ; and Fe3 ; the security return generating equation for
each security i of the n securities considered is
ei = ai + bi1 Fe1 + bi2 Fe2 + bi3 Fe3 + eei :
R
ei is the random return of security i; eei is random noise with a zero mean, and ai ;
Here, R
bi1 ; bi2 ; and bi3 are parameters.
(d1) Suppose that an investor allocates dollar amounts w1 ; w2 ; : : : ; wn to securities 1; 2; : : : ; n;
respectively, under the conditions that the investment requires no cash outlays and
does not respond to any of Fe1 ; Fe2 ; and Fe3 : Establish a set of analytical conditions for
59
a self-…nanced risk-free portfolio. Explain clearly whether the set of w1 ; w2 ; : : : ; wn is
unique or there are in…nitely many ways to assign w1 ; w2 ; : : : ; wn to the n securities,
for the purpose of forming a self-…nanced risk-free portfolio. Then, explain clearly
whether the sum
Xn
i=1
wi eei
is exactly zero or is only approximately zero in this analytical setting. (4 marks)
(d2) Explain clearly why, in the absence of arbitrage pro…ts, a linear relationship of the
form
i =
0+
can be established. Here,
1 bi1 +
2 bi2 +
3 bi3 ;
for i = 1; 2; : : : ; n;
i is the expected return of security i; and
3 are parameters, which are common for all securities.
0;
1;
2 ; and
(3 marks)
Answer:
(a) Let
1 XT
Rit
t=1
T
be the sample estimate of the expected return i ; for i = 1; 2; : : : ; n; and
1 XT
bij =
Rit Ri Rjt Rj
t=1
T 1
Ri =
be the sample covariance of returns between securities i and j; for i; j = 1; 2; : : : ; n;
including i = j: The sample covariance matrix Vb is an n n matrix with each (i; j)element being bij : Let also
Rit Ri
uit = p
T 1
T matrix U ; for i = 1; 2; : : : ; n and t = 1; 2; : : : ; T: With
be the (i; t)-element of an n
p
the term T 1 in the expression of uit above being common for all i and all t; we can
write
2
(R11 R1 )
6
1
6 (R21 R2 )
U =p
6
..
T 14
.
(Rn1 Rn )
2
(R11 R1 )
6
1
6 (R12 R1 )
U0= p
6
..
T 14
.
(R1T R1 )
60
3
R1 )
R2 ) 7
7
7;
..
..
..
5
.
.
.
(Rn2 Rn )
(RnT Rn )
3
(R21 R2 )
(Rn1 Rn )
(R22 R2 )
(Rn2 Rn ) 7
7
7;
..
..
...
5
.
.
(R2T R2 )
(RnT Rn )
(R12
(R22
R1 )
R2 )
(R1T
(R2T
and then
UU0 =
1
T
1
2 PT
6
4
t=1
PT
t=1
R1t
Rnt
R1
..
.
R1t
Rn
R1t
PT
R1
t=1
...
PT
R1
t=1
R1t
R1
..
.
Rnt
Rn
Rnt
Rn
Rnt
Rn
The matrix product x0 Vb x; for any n-element column vector x; is therefore
3
7
5 = Vb :
x0 U U 0 x = (U 0 x)0 (U 0 x):
Now, let
w = U 0x
be a T -element column vector and label its elements as w1 ; w2 ;
x0 Vb x = w0 w =
XT
t=1
; wT : It follows that
wt2 ;
which is never negative. With x and, consequently, w being arbitrary, the positive semide…niteness of the sample covariance matrix is con…rmed. For the sample covariance
matrix Vb to be positive de…nite, causes for Vb to have a zero determinant must be ruled
out.
Speci…cally, none of the n securities considered can be risk-free; otherwise, the
corresponding Vb will have a zero determinant. Further, no two securities considered
can be perfectly correlated in their returns, and none of the n securities considered can be
replicated exactly by a linear combination of any of the remaining securities. Without
ruling out such cases, we can only con…rm that the sample covariance matrix is always
positive semi-de…nite.
(b) As
ei =
R
i+
e
ei ;
i Rm + e
for i = 1; 2; : : : ; n;
the random return of an equally-weighted portfolio based on k of the n securities is
where
ek =
R
k +
e
ek ;
k Rm + e
Xk
ei ;
ek = 1
R
R
i=1
k
1 Xk
=
k
i;
i=1
k
1 Xk
=
k
i;
i=1
k
61
and
1 Xk
eei :
i=1
k
eek =
Given that
em ; eei ) = 0
Cov(R
and
Cov(e
ei ; eej ) = 0; for i 6= j;
ek can be decomposed as
the variance of R
ek ) = V ar( k R
em + eek ) =
V ar(R
2
e
ek ):
k V ar(Rm ) + V ar(e
This is because all covariance terms involved are zeros. Here, V ar(e
ek ) is the unsystematic
component of the variance of portfolio returns.
em ) as
To match the notation in the question, we can write V ar(R
2
ei ) as
m ; and V ar(e
2
ei :
Given that
2
e1 =
2
e2 =
2
e3 =
=
2
en ;
we can use a common positive constant A to represent each of these variance terms. It
follows from
eek =
that
V ar(e
ek ) =
1 Xk
eei
i=1
k
1 Xk
V ar(e
ei ):
i=1
k2
Again, this is because all covariance terms involved are zeros. We can then write
V ar(e
ek ) =
1 Xk
i=1
k2
2
ei =
1
A
(kA)
=
:
k2
k
Thus, as k increases, V ar(e
ek ) always decreases. This is an example that illustrates a
fundamental portfolio concept; speci…cally, the unsystematic component of the variance
of returns can be diversi…ed away in portfolio settings.
(c) Apart from the alternative assumptions of multivariate probability distribution of security returns or quadratic utility functions of investors to justify the mean-variance
approach, we need some more assumptions to derive the capital asset pricing model An
additional assumption as required for the derivation is a frictionless capital market, where
short sales of any risky securities are frictionless; that is, there are no transaction costs in
62
security trading, and the short-sale proceeds are immediately available to the short seller
who also provides no cash deposits when short selling a security. Another assumption
is that risk-free lending or borrowing is available at the same interest rate. Further, all
investors are assumed to have homogeneous expectations on the expected returns, the
variances of returns, and the covariances of returns for all securities in the market.
Sharpe’s derivation of the security market line (SML) is as follows: Consider a frictionless
capital market where risk-free lending and borrowing is available at the same interest
rate Rf : Under the assumption of homogeneous expectations, all investors will choose
the market portfolio m in market equilibrium; they will allocate their investment funds
between the risk-free security and the market portfolio.
Suppose that, as a minor
departure from market equilibrium, there is a very small excess demand for security
i: Given this excess demand, let us construct a portfolio p by allocating the available
investment funds between portfolio m and security i: The expected return of portfolio
p is
p = (1
") m + " i ;
where " is the proportion of investment funds for the investment in security i: The
expected returns of securities i and m are
i and
p ; respectively.
The standard deviation
of returns of portfolio p is
q
(1
p =
where
2
i and
")2 2m + "2 2i + 2(1
")" im ;
2
m are the variances of returns of securities i and m; respectively, and
is their covariance of returns. Using di¤erential calculus tools, we can write
@ p
@
=
[(1
@"
@"
") m + " i ] =
i
m
and
q
@ p
@
=
(1 ")2 2m + "2 2i + 2(1 ")" im
@"
@"
1
=
2(1 ") 2m + 2" 2i + 2(1 2") im
2 p
1
=
(1 ") 2m + " 2i + (1 2") im :
p
63
im
At " = 0 (when there is no excess demand for security i); we have
p j"=0
=
m;
@ p
=
@" "=0
and
2
m
im
;
m
@ p =@" "=0
(
= i
@ p =@"j"=0
im
d p
=
d p "=0
m) m
:
2
m
Equating d p =d p "=0 and the slope of the capital market line, which is ( m
Rf )= m ;
leads to
( i
im
m) m
=
2
m
Rf
m
:
m
Then, with
( i
2
m) m = ( m
Rf )( im
i = Rf + ( m
Rf )
2
m );
it follows that
im
:
2
m
De…ning
i =
im
;
2
m
we have
i = Rf + ( m
Rf ) i ; for i = 1; 2; : : : ; n:
This is the security market line (SML) of the capital asset pricing model (CAPM).
An alternative derivation without using di¤erential calculus tools is as follows: Since
the excess demand for security i is very small, the additional funds to be allocated to
it, as captured by "; will also be very small.
That is, portfolios p and m di¤ers only
marginally in the invested amounts in security i: The di¤erence in the expected returns
of the two portfolios is
4 =
p
m = (1
") m + " i
4 = "( i
m ):
m:
That is,
The variance of returns of portfolio p is
2
p = (1
")2 2m + "2 2i + 2(1
64
")" im :
The di¤erence in the variances of returns of the two portfolios is
2
p
2
m =
Given that " is very small, as 2
2
p
"(2
") 2m + "2 2i + 2(1
")" im :
"
2; "2
0; and 1
1; we can write
m )( p +
m) =
2
m = ( p
"
2" 2m + 2" im :
Since the two portfolios are very close to each other, we have
p+
m
2 m:
Then, the di¤erence between the standard deviations of the two portfolios is
4 =
m =
p
2" 2m + 2" im
:
2 m
That is,
4 =
2
m)
"( im
:
m
Thus, we have
(
4
= i
4
im
m) m
;
2
m
which provides the risk-return trade-o¤ for having a portfolio that departs slightly from
the market portfolio m:
For this departure to be worthwhile, portfolio p has to be an e¢ cient portfolio.
All
e¢ cient portfolios, including portfolio m; are on the capital market line. Given that the
slope of the CML is ( m
Rf )= m ;we can write
4
(
= i
4
im
m) m
=
2
m
Rf
m
;
m
which leads to the SML.
(d) To answer the questions in (d1) and (d2) below, we …rst establish what each ai in the
security return generating equation represents. Given
ei = ai + bi1 Fe1 + bi2 Fe2 + bi3 Fe3 + eei ;
R
ei can be expressed as
the expected value of R
ei ) = ai + bi1 E(Fe1 ) + bi2 E(Fe2 ) + bi3 E(Fe3 ) + E(e
E(R
ei ):
65
As Fe1 ; Fe2 ; and Fe3 have been mean-removed and as eei is random noise with a zero mean,
all expected values on the right hand side of this equation are zeros; i.e.,
E(Fe1 ) = 0;
E(Fe2 ) = 0;
E(Fe3 ) = 0;
and E(e
ei ) = 0:
Thus, we have
e i ) = ai :
E(R
ei or, equivalently, the expected return of security i;
That is, ai is the expected value of R
commonly labeled as
expect
i :As the return generating equation for each security i is linear, we
i to be related to bi1 ; bi2 ; and bi3 linearly.
However, the exact linear relationship,
which allows us to determine the value of each
i ; for i = 1; 2; : : : ; n; will have to be
established by using an arbitrage reasoning. A derivation of the arbitrage pricing model
is all about establishing the value of each
i ; given the available information that is
provided by the security return generating equations.
(d1) To achieve a self-…nanced investment, we assign values to w1 ; w2 ; : : : ; wn ; satisfying
the condition of
Xn
i=1
wi = w1 + w2 +
+ wn = 0:
In a frictionless capital market, the proceeds from short selling securities are immediately available.
Thus, we can always have positive and negative values among
w1 ; w2 ; : : : ; wn to satisfy the above condition.
This self-…nance investment is intended to be insensitive to the randomness in each
of Fe1 ; Fe2 ; and Fe3 : Such a feature can be achieved, as long as we set w1 ; w2 ; : : : ; wn
in such a way that the conditions
Xn
wi bi1 = w1 b11 + w2 b21 +
+ wn bn1 = 0;
Xn
wi bi2 = w1 b12 + w2 b22 +
+ wn bn2 = 0;
Xn
wi bi3 = w1 b13 + w2 b23 +
+ wn bn3 = 0
i=1
i=1
and
i=1
66
are also satis…ed.
As this capital market has great many securities, there are in-
…nitely many ways to assign w1 ; w2 ; : : : ; wn to satisfy the above three conditions.
From an algebraic perspective, as there are 4 equations with n variables, which are
w1 ; w2 ; : : : ; wn ; there will be in…nitely many solutions, as long as n > 4:
Given the security return generating equations
ei =
R
e
e
e
ei ;
i + bi1 F1 + bi2 F2 + bi3 F3 + e
for i = 1; 2; : : : ; n;
the random return of this self-…nanced investment is
Xn
i=1
ei =
wi R
It reduces to
Xn
Xn
Xn
wi i +
wi bi1 Fe1 +
wi bi2 Fe2
i=1
i=1
i=1
Xn
Xn
+
wi bi3 Fe3 +
wi eei :
i=1
Xn
i=1
i=1
ei =
wi R
Xn
i=1
wi i +
Xn
i=1
wi eei ;
because we have assigned w1 ; w2 ; : : : ; wn to satisfy the above four conditions.
nAs there are great many securities in the market, the law of large numbers in
statistics gives us
Xn
i=1
A direct implication is that
Xn
i=1
ei
wi R
wi eei
0:
Xn
i=1
wi i :
That is, the return of this self-…nanced portfolio is nearly risk-free, because there is
no randomness on the right hand side.
model, it is treated as risk-free.
In the derivation of the arbitrage pricing
A crucial point here is that there are in…nitely
many ways to establish a self-…nanced risk-free investment.
In the absence of arbitrage pro…ts, we must have
Xn
i=1
wi i = 0:
(d2) Given the return generating equation, which is linear, we expect
bi1 ; bi2 ; and bi3 linearly. The linear relationship can be written as
i =
0+
67
1 bi1 +
2 bi2 +
3 bi3 ;
i to be related to
where
0;
2 ; and
1;
Xn
wi i ;
Xn
wi i
i=1
3 are parameters.
Xn
i=1
wi ;
Xn
i=1
Given that each of the sums
wi bi1 ;
Xn
i=1
wi bi2 ; and
Xn
i=1
wi bi3
is zero, a linear combination of them is also zero. Thus, we can write
=
Xni=1
i=1
0
wi ( i
Xn
1
1 bi1
2 bi2
i=1
0
Xn
wi
i=1
wi bi1
2
3 bi3 ) = 0:
Xn
i=1
wi bi2
3
Xn
i=1
wi bi3
There are in…nitely many way for us to assign w1 ; w2 ; : : : ; wn ; for the purpose of
achieving a self-…nanced risk-free investment. Regardless of how we assign w1 ; w2 ; : : : ; wn ;
we must have the same pricing relationship when relating each
i to the available
information that the n security generating equations provide. Thus, we must have
i
1 bi1
0
2 bi2
3 bi3 = 0;
for i = 1; 2; : : : ; n;
or, equivalently,
i =
Here, the parameters
0+
1 bi1 +
0;
1;
2 bi2 +
2 ; and
68
3 bi3 ;
for i = 1; 2; : : : ; n:
3 have yet to be determined.