Math 225 – Financial Mathematics
November 29, 2004 November 29, 2004 name______
Exam 2 – Answers
You may take up to three hours for this exam. You may take it when and where you like.
You may use any notes prepared entirely by yourself. (This does not include annotated handouts or any published references.) You may also use a calculator, but not a computer or any consultation (except with me, or cell 610-220-4382). You may use this paper, your own paper, or both. It isn’t necessary to turn in scratch paper.
Return this exam in class, to my office, or to the mathematics office by 11:10 am Monday.
This exam has 6 numbered problems on 6 numbered pages.
It is never necessary to reduce your answer to a number. An explicit formula (that is, all numbers with no variable or function names) is always sufficient, if it’s correct and can actually be evaluated. The standard normal cdf, F(x), can be part of an explicit formula. (Of course, you can usually be more confident of your answer if you reduce it to a number.)
1. The current price of a stock is S = $85.00. A call option with expiring exactly T = 1 year from now with a strike price of K = $75.00 is now selling for C=$20.00.
The present value of a guaranteed $100 payment at time T=1 is $98.00.
a. (4 pts) If the stock price is $95.00 at time T, how much will the option be worth at that time?
S(T) – K = $95 - $75 = $20
b. (6 pts) A put option, also with T = 1 and K = 75.00, has current price P. What is P ?
If the present value of $100 is $98, then the present value of K=$75 must be = $73.50. (Or: 100 e-rT = 98 --> K e-rT = 73.50.)
From
S – C + P = present value of K
with S = 85, C = 20, present value of K = 73.50 we get
P = $8.50.
2. A stock price moves according to Geometric Brownian Motion (GBM) with parameters
m = 0.03 per year,
σ = 0.20 per year1/2.
The current price is S(0) = $100.00. ( Note: ln(100) = 4.605170. )
Let S(2) be the random variable representing the price four two years from now.
a. (5 pts) What is the expected value of S(2) ?
b. (5 pts) If L(2) = ln ( S(2) ), what is the expected value of L(2) ?
c. (10 pts) Let R = L(2) – L(0) = ln ( S(2) ) – ln ( S(0) ). Describe the distribution of R.
Mean of R —>
mt = 0.06
Variance of R —> OR Standard deviation of R —>
s2t = 0.08
Say something about the shape of the distribution of R —>
It’s normal
d. (5 pts) What is the probability that S(2) is less than $100 ?
3. A certain stock has a price of $ 110.00 now. Tomorrow, it will be worth either exactly $ 150.00, or exactly $ 50. (Assume r = 0.)
150
110
50
a. (5 pts) What are the “risk-neutral probabilities” associated with the two branches?
They are p, 1-p where p is chosen to satisfy
p(150) + (1-p)(50) = 110. So,
p = 3/5 (for top branch), 1-p = 2/5 (for bottom branch)
b. (2 pts) A certain put option on this stock has a strike price of K = $ 90. What will that option be worth tomorrow if the stock price goes up ?
zero
c. (2 pts) What is the option worth tomorrow if the stock price goes down ?
$40
d. (6 pts) What should the price of the option be today ?
(3/5)(zero) + (2/5)($40) = $16.
e. (10 pts) Design a combination of x shares of stock and y dollars cash that guarantees the same results (tomorrow) as the option.
x and y must satisfy:
150 x + y = 0
50 x + y = 40
The solution is x = -2/5) shares stock, y = +$60.
4. (10 pts) A certain European call option gives its holder the right (but not the obligation) to buy 1 share of ABC stock for $40 on November 22, 2005. Assume that the stock price follows GBM, and assume these parameters:
m (drift parameter for ABC stock) = 0.07 per year
(measured in ordinary dollars)
σ (volatility of ABC stock) = 0.25 per year1/2
today’s date = November 22, 2004 (Yes, I know you’re probably doing the exam five days early, but use November 22 anyway)
r (risk-free interest rate, continuously compounded) = 0.04 per year
current price of ABC stock: $30.
Using any variant of the Black-Scholes formula, compute the value of the option on November 22, 2004.
Here’s the flat-dollar version of the formula:
The inputs are:
s0 = $ 30 (current stock price)
s = 0.25
T = 1 (so also =1) (T = time from Nov. 22 2004 to
Nov. 22 2005; use years to be compatible withs.)
k = Ke-rT where K = strike price of $40, T = 1, and r = 0.04;
therefore k = 38.431578.
So,
ln(k/s0) = 0.247682
s = 0.25
= 0.03125
and
The option is worth about 72 cents.
5. A stock now trades for $40 and has a volatility of σ = 0.40 (for one year). We
want to construct a tree for evaluating options on this stock.
We decide on a time step of D = 0.25 year.
We construct a binomial tree is constructed to represent Geometric Brownian Motion (as well as we can using this time step).
Use flat dollars (or assume r = 0 ) throughout.
Here is the start of the tree:
40
t=0.25 t=0.50 t=0.75
a. (5 pts) What prices should you put at the indicated nodes ?
We want where s = 0.40 and Dt = 0.25. This works out to: u = 1.20, d = 0.80. Now the prices at the nodes are (top to bottom) 40u2=57.60, 40u=48.00, and 40d=32.00, as shown above.
b. (5 pts) What risk-neutral probabilities will you use for the two leftmost branches?
1/2 for each branch.
6. Suppose a stock price moves according to this grid. (All prices are in flat dollars. Yes, I know that this isn’t a good approximation to GBM, but let’s use this model anyway.)
t=0 t=1 t=2 t=3 t=4
For the call option (part a):
140 60
130 50
120 40 120 40
110 30 110 30
100 21.25 100 20 100 20
90 12.50 90 10
80 5 80 0
70 0
60 0
a. (8 pts) A certain European call option allows you to buy the stock for $80 at time 4. That is, the payoff at time 4 is
S(4) – 80, but never less than 0.
What should be the price of the option at time 0? $21.25
b. (8 pts) An American put option allows you to sell the stock for $100 at any time you please, up to time 4. (For simplicity, let’s assume that the strike price is $100 in flat dollars regardless of when you exercise. Or equivalently, assume that r really is zero.)
What should be the price of the put option at time 0? $7.50 (see next page)
c. (4 pts) (continuing part b) If the price of the stock is 90 at time t=1, should you exercise the put option at that time? Yes, exercise for a profit of $10
x No, wait and see
(end of exam)
t=0 t=1 t=2 t=3 t=4
For the put option (part b):
140 0
130 0
120 0 120 0
110 2.50 110 0
100 7.50 100 5 100 0
90 12.50 90 10
80 20 80 20
70 30
60 40
Note: Early exercise values aren’t shown, since they never exceed these wait values. The exercise value at the node (t=1, S=90) is 10, which is why it’s better to wait than exercise at that node (part c).
Reference material:
<— Sorry; s0 omitted in exam.
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