Math 225
October 20, 2004
Homework 5 - Answers
1. If you didn’t turn it in already, do Chapter 2, Exercise 1.
P( Z < –0.66 ) = P( Z > +0.66 )
by symmetry
= 1 – P( Z < +0.66 ) <— that’s 1 - (0.66)
= 1 - 0.7454
= 0.2546
P( |Z| < 1.64 ) = P( -1.64 < Z < +1.64 )
= P( Z < +1.64 ) – P( Z < -1.64 )
= P( Z < +1.64 ) – P( Z > +1.64 )
= 2 P( Z < +1.64 ) – 1
<— that’s 2(1.64) -1
= 2 ( 0.9495 ) – 1
= 0.8990
P( |Z| > 2.20 ) = P( Z > +2.20 ) + P( Z < -2.20 )
= 2 P( Z > +2.20 )
= 2 ( 1 – P( Z < +2.20 )) <— that’s 2(1-(2.20))
= 2 ( 1 – 0.9861 )
= 0.0278
2. Assume that X is a normal random variable with mean  = 0.20, standard
deviation  = 2.10. ( It follows that X = 0.20 + 2.10 Z, where Z is a standard normal
random variable. )
a. The condition X < 0.10 is equivalent to what condition on Z ?
X < 0.10 means 0.20 + 2.10Z < 0.10 which is equivalent to
Z < –0.10/2.10, or
Z < –0.0476
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b. What is P ( X < 0.10 ) ?
P( X < 0.10 ) = P( Z < –0.0476 ) = (-0.0476) = 0.4810
(This is from Excel, using “NORMSDIST” for ;
interpolation in the table on page 23 works, too.)
c. What is P ( –2.0 < X < +2.0 ) ?
P( -2.0 < X < +2.0 ) = P( (-2.0-0.20)/2.10 < Z < (+2.00.20)/2.10) )
= P( Z < 0.857 ) – P( Z > -1.048 ) = (0.857)-(-1.048)
= 0.6569
3. Suppose that L(t) is a Brownian Motion process with mean  = 0.10 year –1
and  = 4.0 year –1/2. What is the distribution of X = L(15) – L(10) ? (Give its shape,
mean, and standard deviation.)
It’s the change in L(t) over a time period of length t = 5.
So, it’s normal with mean t = 0.50 yr-1, stdev  t = 8.944
yr-1/2.
4. Do Chapter 3, Exercise 3.1. ( Use the formula at the end of Section 3.1 where you can;
otherwise translate the questions into questions about L(10). )
a. From the formula on p. 33,
E( S(10) ) = s0 exp ( t( + 2/2) )
= 100 exp ( 10( 0.01 + 0.04/2 ) )
= 100 exp ( 0.3 )
= 134.99
b. P( S(10) > 100 )… S(10) is 100*exp(0.1 + 0.2 10 Z) for a
standard normal Z, so S(10) > 100 is equivalent to
Z > (ln 1 – 0.1)/(0.2 10 ) = -0.158.
Now, P( Z > -0.158 ) = 0.563.
c. P( S(10) < 110 )...
equivalent to Z > (ln 1.1 – 0.1)/(0.2 10 ) = -0.007, so
P( Z < 110 ) = 0.497.
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5. Do Chapter 3, Exercise 3.2.
a. E( S(10) ) = s0 exp ( t( + 2/2) )
= 100 exp ( 10( 0.01 + 0.16/2 ) )
= 100 exp ( 0.9 )
= 245.96. That’s a big difference from
exercise 3.1,
considering that the starting price
and
mean growth rate are the same, and
the
volatility just changed from 0.2 to
0.4.
b. P( S(10) > 100 )… S(10) is 100*exp(0.1 + 0.4 10 Z) for a
standard normal Z, so S(10) > 100 is equivalent to
Z > (ln 1 – 0.1)/(0.4 10 ) = -0.079.
Now, P( Z > -0.079 ) = 0.531.
c. P( S(10) < 110 )...
equivalent to Z > (ln 1.1 – 0.1)/(0.4 10 ) = -0.004, so
P( Z < 110 ) = 0.498. Notice that this stock grows almost
twice as fast as the stock in exercise 3.1 in expected
value terms, but it has less chance of finishing above
100 and more chance of finishing below 110. That’s
what a volatility of 0.4 does for you. (And we think
AAPL has a volatility of 0.55.)
6. Assume that the following definite integral is correct:


x 
1
2
exp  
1
 2
x 2  Bx  C  dx

1
 exp  B 2  C
2
.


Use this to derive the formula at the end of Section 3.1, which is the following:

E S (t )  s0e
t  
2
2
.
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Well, S(t) = exp( L(t) ), and L(t) is normal with mean
ln(s0)+t and standard deviation *sqrt(t). So, we
can write L(t) as L(t) = ln(s0)+t +  sqrt(t) Z for
some standard normal variable Z. So:
S(t) = exp ( ln(s0)+t +  sqrt(t) Z ).
Write this as
S(t) = exp ( C + B Z ),
where C = ln(s0)+t and B =  sqrt(t).
Now
E (S (t ))  



z 
z 
exp(C  Bz )
1
exp( 12 z 2 )dz
2
1
exp( 12 z 2  Bz  C )dz.
2
(The second step works because multiplying
exponentials means adding the exponents.)
This is in the required form! So we just fill in the blanks:
E(S(t)) = exp ( (1/2)B2 + C )
= exp ( ln(s0) + t + (1/2)(2t) )
= s0 exp ( t (  + 2/2 ) )
as desired.
(end)
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