Actuarial Society of India
Examinations
May 2006
ST6 – Finance and Investment B
Indicative Solution
ST6
0506
Sol 1a)
Role of Arbitrageurs:
1. Arbitrageurs perform the important role of keeping the cash market in a security
and the derivatives market in line
2. Arbitrageurs look for arbitrage opportunities i.e., price inconsistencies which
would give them an opportunity to make a risk-free profit
3. To do this, they set up equal and opposite positions in different markets
4. The activities of the arbitrageurs increase/reduce the demand for securities that
are under/overpriced relative to other securities, causing the market price to
increase/ decrease
(4)
Sol 1b) (i)
1. Arbitrages are usually instructed by setting up. Zero cost portfolios. This
would imply that at least one of the holdings must be (-)ve. i.e., a short
holding.
2. Not al securities require (-)ve holdings. It my be possible to create an arbitrage
using a (-)ve holding of an equity derivative even if the primary equity market
(cash market for equities) does not allow short holdings
3. The short holding may relate to cash. This simply involves a loan, not a short
holding of a traded security.
4. If other clients have long holdings in an asset, a bank may be ale to use internal
transfers to create a short holding on a particular client's account, without
having a short holding in the bank's overall position.
(3)
Sol 1b)(ii)
1. A risk free rate of return is one that involves no credit-risk. In other words, there
is no possibility of counterparty default.
2. Government bonds [e.g., gilts with UK] are guaranteed by the central
government of the issuing country. It is usually assumed that the possibility of
default on such bonds doesn't exist – at least in respect of the Central
Government Bonds issued by the Central Governments of major economies.
3. The cash flows arising from an investment in the Governmebnt bond are know
with certainty only if the bond is held until redemption. The price obtained on
an earlier sale is subject to market fluctuations.
4. The overall return still depends on the available reinvestment rates.
5. Index-linked bonds are affected by inflation rates. Hence they are not risk free
in the usual sense.
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6. Certain bonds are owned primarily by certain types of institutions (e.g., pension
funds) or may be relatively less marketable. Hence, even though, the rate of
return on these bonds is default risk free, it may not be an appropriate proxy
for the risk free rate that would apply when setting up an arbitrage.
7. The repo rate is the usual benchmark used for the risk free rate because it is a
move flexible measure than the rate based on bonds because it defines an overnight rate for the 24 hour investments.
(4)
[11]
Sol 2a)
1. "Stationary Increments" means that the distribution of Bt-Bs (t>s) depends only
on (t-s)
2. "Independent Increments" means that Bt-Bs is independent of Br whenever
r<s<t
3. "Continuous Sample paths" means that the function t Bt(w) for each
particular realization of w is a continuous function of t
(3)
Sol 2b)(i)
The given SDE can be written as
dxt = (-) dwt + 2dt
Here we are applying the function f(x) = x-1 which has derivatives f1(x) = x-2 and f"(x) = +2x-3 to the above process.
So, using Ito's lemma, we have
df(Xt) = (-1) (-Xt-2) dWt + [2(-Xt-2) + ½ (-1)2 (2Xt-3)]dt
i.e., dRt = Rt2 dWt + (Rt3 – 2Rt2) dt
(2)
Sol 2b)(ii)
The process will have a (+)ve drift when the drift
Coefficient is (+)ve
when
R
t
3
 2Rt )  0
2

3
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0506
i.e., Rt2 (Rt-2) > 0
This will be true whenever (Rt>2)
(1)
Sol 2b) (iii)
A process is mean – reverting if its drift is such that the process is always attracted
towards some fixed value
This process Rt is not mean-reverting
If Rt>2, the drift Is (+)ve and the process tends to move upwards.
If Rt<2, the drift coe is (-)ve [or '0' when rt = 0] and the process tends to move
downwards. Hence this process is repelled away from the value 2.
(2)
Sol 2(b)(iv)
If Rt has a large (+)ve value, both the drift coe and the volatility will have large
(+)ve values. So, the process will tend to increase rapidly, with very large random
fluctuations.
Since Rt = Xt-1 and the original process Xt can take the value zero, it is possible
that Rt will increase so rapidly, that it goes to infinity.
Therefore, the process Rt is unlikely to be an appropriate model for a real world
quantity, which can be expected to take only finite values.
(2)
[10]
Sol 3a)
The formula to be applied to change the measure is:
EQ Xt / Fs  
1
EpMtXt.Fs
Ms
In this formula
(i) P and Q are any two equivalent probability measures
(ii) Fs is the filtration representing the history of the process X up to time s
4
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(iii) Mt is the change of measure process defined by
 dQ

Mt  Ep 
/ Ft 
dP


Where dQ is the Radvn – Nikodym derivative
dP
based on a time horizon T>t
(4)
Sol 3b)(i)
Under p x has the density function
2

x 
exp  1 / 2 
 
2
   

1

The Radin – Nikudym derivative is just the ratio of the two probability
density functions because we are dealing with continuous random variables.
Hence
 1 x    2
 1 x   r  2
dp
 exp   
  exp   
 
dQ

 
 2   
 2
 1  ( x   ) 2  ( x    r ) 2 
 exp   

2

 2
 1  ( 2 x  2   r ) ( r ) 
 exp   

2
 2 

 r
 exp   2
 
r 

( x    ) 
2 

(3)
Sol 3b)(ii)

Ep e kx I (a  x  b)

 dp kx

 EQ 
e I ( a  x  b) 
 dQ

5
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0506


r
r 

 EQ exp  kx  2 )( x    )  I e kx I (a  b)


r 
 r
 EQ exp  2 ( x   )  e kx2 I(a  x  b) 
2 





r
r 
r
 EQ exp  k  2 ) X  2     I (a  b)
2

 


Let r = k2 This will eliminate X in the RHS of the above equation
Therefore,
Ep [ekx I (a<x<b)]
r
 r
 exp  2 (    EQ I (a  z  b)
2

The expectation of an indicator function is just a probability.
Therefore,
Ep [ekx I (a<x<b)]
1


 exp k  k 2 2  PQ a  x  b
2


Under Q, X is N (+r,2) i.e., N (+k2, 2)
Therefore,
PQ [a<X<b] = P[a<N(+k2,2<b)
 a  (   k 2
b  (   k 2 
 P
 N (0, I ) 





b
a  

 P
 k  N (0, I ) 
 k 

 

6
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0506
Therefore,
b  

a  

 
  k    
  k 
 

 

Ep[ekx I(a<x<b)]
1

  b

a  

 exp  K  k 2 2   
  k    
  k 
2




 

(7)
[14]
Sol 4a)
Recall that rho is defined as δf/δr. In this instance, we can estimate is as:

f 33.3  32.9

 0.8% 1 (in paisa )
r
0.5
(2)
Sol 4b)
Theta is negative for a European call option. This is because (all else being equal)
the time value of an option will decrease as the term to expiry reduces. So, if the
intrinsic value is constant (remembering that theta is a partial derivative), then the
total value of the option must likewise decrease as the term to expiry reduced.
Theta may, however, be positive for a deeply in-the-money European put option.
For example, if the current share price is 1 and the strike price 100, then a
European put that can be exercised today may be worth more than an otherwise
identical European put that can be exercised only in a year's time, as the higher
present value of the money received not outweighs the loss of the option.
(1)
7
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Sol 4c)
The correct pairings are:
i)
Call option (strike price 50)
ii) Put option (strike price 50)
iii) Put option (strike price 55)
iv) Put option (strike price 120)
v) Call option (strike price 95)
+0.41
-0.79
-0.70
-0.30
+0.21
Because the payoff on a call option is positively correlated with the price of the
underlying share, call options on a share have positive deltas. In fact, these will be
n the range 0<Δ<1.
Because the payoff on a put option is negatively correlated with the price of the
underlying share, put options on a share have negative deltas. In fact, these will be
in the range -1<Δ<0.
Options that are deeply in-the-money will have deltas with a higher numerical
value, i.e., closer to +1 or -1.
(5)
Sol 4d)
Call option (6-month versus 3-months)
The payoff for a call option is max {S-K,0}. If S<K there is no loss. So the
downside risk is capped, but the upside potential is unlimited.
The longer the life of the option, the more opportunity there is for the underlying
security price to move. Hence the value of the option increases.
In some unusual circumstances other factors might override this effect.
(1)
Sol 4e)
Put option
For similar reasons, the value of a put option is usually also higher for a 6-month
option than for a 3-month option.
8
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In this case the payoff is max {K-S,O}. If S>K there is no loss. So the risk
associated with the underlying security price going up is capped, but the potential
profit associated with the price going down is large.
S would have to fall to 0 before the gain was capped.
(1)
[10]
Que 5a)
Call option with no dividends
0.35 2
 100  
log 
   0.03 
2
 100  
d1 
0.35 2 / 12
 2 
  
  12 
 0.5606
d 2   0.5601 0.35 2 / 12  0.7035
c =100(-0.5606)-110e-0.03x2/12(-0.7035)
=2.39
Que 5b)
Call option with a dividend
Holding a call option is like holding cash whilst waiting to buy a share. If the
share pays a dividend in the meantime, the holder of the option "loses out" on this
dividend. To allow for this, we reduce the share price by the present value of the
dividend:
S = 100-13e-0.03/12 = 87.03
We then recalculate the option price based on the adjusted share price:
0.35 2
 87.03  

log 

0
.
03

 
2
 110  
d1 
0.35 2 / 12
 2 
  
  12   1.5328
d 2   1.5328  0.35 2 / 12  1.6757
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c =100(-0.5328)-110e-0.03x2/12(-1.6757)
=0.32
[5]
Sol 6(i)(a)
Construct the tree and calculate the current derivative price
The current share price is 480. We are given that u = 1.25 and d = 0.8. So the
share price will either increase to 480u = 600 or decrease to 480d = 384. these
share prices are shown inside the boxes in the tree below.
The payoffs, shown above the boxes, are then either (600-500)2 = 10,000 or (384580)2 = 13,456.
10,000
5/9
600
?
480
13,456
4/9
384
The risk-neutral probability corresponding to an up step is given by the formula on
page 45 of the Tables :
e r  d 1.05  0.8 5


u  d 1.25  0.8 9
So, using the risk-neutral pricing formula, the price of the derivative is :
1 5
4

V0  e  rt EQ[ H ] 
 x10,000  x13,456  10,987
1.05 9
9

(2)
Sol 6(i)(b) Estimate Delta
We might be tempted to estimate the derivative's delta directly from the tree as:
H 10,000  13,456

 16
S T
600  384

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However, since the payoff function for this derivative is highly non-linear and the
final share prices of 600 and 384 straddle the critical value of 500, this would not
provide a reliable estimate.
In fact, because of the squaring involved, if we were to increase the initial share
price slightly, this would have a bigger effect at the 600 node than the 384 node.
So the value of delta would actually be positive, not negative!
(4)
Sol 6(ii)(a) Estimate rho
If we change the value of 4, this has no effect on the tree. But it does affect the
calculation of the derivative price, which is now:
V0  e  rt EQ[ H ] 
1 5
4

 X 10,000  X 13,456  10,883
1.06 9
9

We can now calculate an appropriate value for rho:
(2)
Sol 6 (ii)(b) Estimate kappa
If we change the value of , this will change the value of u and d. So we will need
a new tree.
If we increase the value of u to 1.26 (say), the new value of  will be  = log1.26.
The risk-neutral probability is now:
e r  d 1.05 1 / 1.26

 0.549694
u  d 1.26 1 / 1.26
The new tree look like this:
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10,983
0.549694
604,8
480
14,172
380.9524
0.450306
The new price is:
So the approximate value for kappa is:
k
V0 11,828 10,987

 110,000 (to2 SF )
 log 1.26 log 1.25
(2)
Sol 6 (ii)(c) Estimate lambda
If we change the value of q, this will change the final share prices. So again we
will need a new tree. The risk-neutral probabilities will be unchanged.
The risk-neutral probability does not change because the total return on the share
remains the same as before, although part of this total return is now paid as a
dividend rather than as a capital gain.
If we use q = 0.01, the new tree looks like this:
8,842
5/9
594.03
?
480
14,357
4/9
380.1791
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The new price is:
V 10,755  10,987
 0 
 23,000 (to2 SF )
q
0.01  0
So the appropriate value for lambda is:
V0 
1 5
4

 x8,842  x14,357   10,755
1.05  9
9

(2)
Sol 6(iii) Estimate the new derivatives price
We can use our estimates of the Greeks to approximate the change in the price:
V0 
V0
V
V
dr  0 d  0 dS 0
r

S 0
 dr  kd  dS 0
= -11,000(0.0025)+110,000(0.01)+5,500(-2)=9,927.5
This would make the new price 10,987-9,927.5 = 1,059.5
(2)
Sol 6 (iv) Three problems a derivatives manager might experience
 Simple pricing models, such as one-step binomial trees, are not at all
reliable when used with this type of non-linear derivatives.
 It would be difficult to hedge a portfolio containing this type of derivative.
 It is difficult to obtain accurate estimates of the Greeks for a non-linear
derivative, and hence to quantify the levels of risk involved.
 If the manager needs to unwind the portfolio, it might be difficult to find a
buyer or seller for an unusual derivative of this type.
(2)
[15]
13
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Sol 7(a)
Prices of:
Bond A = 40e 0.07x0.5  40e 0.07x1  40e 0.07x1.5  40e 0.07x2  1040 e 0.07x2.5 = Rs. 1019.74
Bond B = 50e 0.069x0.5  50e 0.069x1  50e 0.069x1.5  50e 0.069x2  50e 0.069x2.5  1050 e 0.069x3 =
Rs.1079.35
(2)
Sol 7(b)
Zero rate for maturity of 30 months:
1019.74 = 40e 0.08x0.5  40e 0.075x1  40e 0.072x1.5  40e 0.07x2 1040 e 2.5R
R = 6.98%
Zero Rate for maturity of 36 months:
1079.35 = 50e 0.08x0.5  50e 0.075x1  50e 0.072x1.5  50e 0.07 x2 50e 0.0698x2.5  1050 e 3R
R = 6.86%
(2)
Sol 7(C)
Let annual coupon be Rs. C
6-month par yield
(1000 
C 0.08x 0.5
)e
 1000
2
C = Rs. 81.62; Coupon Rate = 8.16%
12-month Par Yield:
C 0.08x0.5
C
e
 (1000  )e 0.075x1  1000
2
2
C = Rs. 76.52; Coupon Rate = 7.75%
18-month Par Yield
C 0.08x 0.5 C 0.075x1
C
e
 e
 (1000  )e 0.072x1.5  1000
2
2
2
C = Rs. 73.49; Coupon Rate = 7.35%
24-month Par Yield
C 0.08x 0.5 C 0.075x1 C 0.072x1.5
C
e
 e
 e
 (1000  )e 0.07 x 2  1000
2
2
2
2
C = Rs. 71.48: Coupon Rate = 7.15%
30-month Par yield
C 0.08x 0.5 C 0.075x1 C 0.072x1.5 C 0.07 x 2
C
e
 e
 e
 e
 (1000  )e 0.0698x 2.5  1000
2
2
2
2
2
C = Rs. 71.24; Coupon Rate = 7.12%
36-month Par Yield
C 0.08x 0.5 C 0.075x1 C 0.072x1.5 C 0.07 x 2 C 0.0698x 2.5
C
e
 e
 e
 e
 e
 (1000  )e 0.0686x3  1000
2
2
2
2
2
2
Coupon = Rs. 70.07; Coupon Rate = 7.01%
(3)
14
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Sol 7(D)
Forward rates:
7.5 x1  8 x0.5
= 7%
1  0.5
= 7.2 x1.5  7.5 x1 = 6.6%
1.5  1
7.0 x 2  7.2 x1.5
=
= 6.4%
2  1.5
= 6.98 x2.5  7.0 x2 = 6.9%
2.5  2
6.86 x3  6.98 x 2.5
=
= 6.26%
3  2 .5
6 months to 12 months =
12 months to 18 months
18 months to 24 months
24 months to 30 months
30 months to 36 months
(3)
[10]
Sol 8(a)
Company B pays a higher rate of interest than Company A in both fixed and
floating markets because Company B has a worse credit rating than Company A.
Company B pays 1.4% more than Company A in fixed rate and only 0.5% more
than Company A in floating rate market. Company B appears to have a
comparative advantage in floating rate market where as Company A appears to
have a comparative advantage in fixed rate market. Company A borrows fixed rate
fund (at 11.4%) and Company B borrows floating rate fund (at LIBOR + 0.4%).
Then they enter into swap agreement to ensure that Company A ends up with
floating rate funds and Company B ends up with fixed rate fund.
The total gain = 1.4 – 0.5 = 0.9%. This is shared among Company A, Company B
and Bank as : Company A = 0.4%; Company B = 0.4% and Bank = 0.1%.
11.4%
Company
A
LIBOR-0.5%
11.4%
Bank
LIBOR+0.4%
12.4%
Company
B
LIBOR + 0.4%
Company A:
Pays 11.4% to outside lender
Pays LIBOR-0.5% to Bank
Receives 11.4% from Bank
--------------------------------------------15
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Net: Company A Pays LIBOR-0.5%
If Company A had borrowed independently, it would have paid LIBOR-0.1%.
Thus, the net gain to Company A from the swap is 0.4.
Company B:
Pays LIBOR + 0.4% to outside lender
Pays 12.4% to Bank
Receives LIBOR + 0.4% from Bank
--------------------------------------------------Net: Company B Pays 12.4%
If Company B had borrowed independently, it would have paid 12.8%. Thus, the
net gain to Company B from the swap is 0.4.
Bank:
Pays 11.4% to Company A
Pays LIBOR + 0.4% to Company B
Receives LIBOR-0.5% from Company A
Receives 12.4% from Company B
------------------------------------------------Net: Receives 1%
Thus, the net gain to Bank is 1%.
(3)
Sol 8(b)
B fix  10 e
 0.09 x
B fl  208 .6e
4
12
 10 e
 0.09 x
10
12
 210 e
 0.09 x
16
12
= Rs. 205.23 million
4
 0.09
12
=Rs. 202.43 million
Value of Swap to the party paying floating = B fix  B fl = 205.23 – 202.43 = Rs.
2.80 million
Value of Swap to the party paying fixed = B fl  B fix = 202.43 – 205.23 = -Rs. 2.80
million
(3)
Sol 8(c)
Cost of Delivering:
Bond 1: 124.25 – 94.25x1.2232 = 8.96
Bond 2: 139.50 – 94.25x1.3992 = 7.63
Bond 1: 114.00 – 94.25x1.1250 = 7.97
Bond 1: 143.00 – 94.25x1.4127 = 9.85
Bond 2 is cheapest to deliver.
(3)
[9]
16
ST6
0506
Sol 9(a)
The present value of coupon payments:
40e 0.07 x0.25  40e 0.075x0.75  40e 0.07 x1.258 = Rs. 113.40
Forward Price of Bond:
Fo  (980  113 .40 )e
P(0, T )  e
0.08x
16
12
0.08x
16
12
= Rs. 964.15
= 0.8988
If the strike price is the quoted price the would be paid for the bond on exercise,
one month’s accrued interest must be added to X. This produces a value for X of
1000 + 40x0.16667 = Rs. 1006.67
d1 
ln( F0 / X )  ( 2T / 2)
 T
ln(964 .15 / 1006 .67 )  (0.07 2 x

16
x0.5)
12
= -0.4935
16
0.07
12
d 2  d1   T  0.4935  0.0808
= -0.5743
p  P(0, T )[ X(d 2 )  F0 (d1 )]  0.8988 [1006 .67 x(0.5443 )  964 .15(0.4935 )]
p  0.8988[1006.67x0.7171 964.15x0.6892]
= Rs. 51.63
(3)
Sol 9(b)
Fk = 0.08; δk = 0.25; L = Rs. 1,000,000; Rk = 0.09; tk = 1.25; tk+1 = 1.5; Zero rate
with continuous compounding = 4ln(1.02) = 7.921%; P(0, t k 1 )  e 0.07921x1.5 = 0.8880;
 k  0.18
d1 
ln( Fk / Rk )  ( k 2 t k / 2)
 k tk

ln(0.08 / 0.09 )  (0.18 2 x1.25 x0.5)
d 2  d 1   k t k  0.4847  0.2012
0.18 1.25
= -0.4847
= -0.6859
The caplet price is:
L k P(0, t k 1 )[ Fk (d1 )  Rk (d 2 )]  1,000 ,000 x0.25 x0.8880 [0.08 x(0.4847 )  0.09(0.6859 )]
 221,993[0.08x0.3140  0.09x0.2464] =
Rs. 653.09
(3)
[6]
17
ST6
0506
Sol 10(a)
Vasicek Model:
B(t, T )  e a( )b( )r (t ) where  T  t
b( ) 
1  e 


1  e 0.1x10
0.1
a( )  [b( )   ][  
= 6.3212
2
2
]

b( ) 2 =
4
2 2
-0.3343
B(t , T )  e 0.33436.3212x0.10 =0.3805
Price for a zero coupon bond with a principal of Rs. 100 = Rs. 38.05
Cox-Ingersoll-Ross Model:
B(t, T )  e a( )b( )r (t ) where  T  t
    2 2  0.12  2 x0.02 2
b( ) 
a( ) 
2(e

 1)
(   )( e  1)  2
2

2

=0.103923
2(e 0.103923x10  1)
(0.103923  0.1)( e 0.103923x10  1)  2 x0.103923
= 6.2956

 2 x0.1x0.1 
2e (  ) / 2
2 x0.103923 e (0.1039230.1)10 / 2
ln 

ln



0.103923x10
0.02 2
 1)  2 x0.103923
 (   )(e  1)  2 
 (0.103923  0.1)(e



= -0.3671
B(t, T )  e 0.36716.2956x0.10 =
0.3691
Price for a zero coupon bond with a principal of Rs. 100 = Rs. 36.91
(4)
Sol 10(b)
Market risk is the risk that the value of a portfolio will fall due to an adverse
changes in the level or volatility of the market price of interest-rate instruments,
currencies, commodities and equities.
Market risk is measured as the potential level of loss with a specified confidence
interval in a certain time period.
Unlimited. If the share price goes “sky high”, the writer will have to deliver the
shares to the purchaser at the agreed price.
(3)
Sol 10(c)
Credit risk is the risk that a counterparty to a transaction will default, wholly or in
part, on its obligations or that there will be a change in the market’s perception of
the risk of default.
(1)
18
ST6
0506
Sol 10(d)
This is because the purchaser has the choice of whether to exercise the option and
will lose out if he or she elect to exercise but the writer refuses or is unable to
carry out the necessary transaction.
(2)
[10]
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19