Math 53 – Math of Finance
Monday, March 19, 2007
Homework5---SOLUTIONS
Here’s the Black-Scholes formula:
(1)
or, if you prefer using ,
(2)
Recall that K is the strike price as written in the contract, and k is the same value in flat dollars.
1. Assume these inputs:
S0 = 40.00(dollars, as of now)
= 0.25(yearly volatility)
T= 2.00(years)
K= 45.00(dollars, at time T)
r= 0.05(per year)
so that
k= 40.71768381(flat dollars)
a. Calculate V, the theoretical price of a call option with strike price K, expiration date T.
Do it “by hand” (with a calculator or fresh spreadsheet, but not the
one-click spreadsheet from class) and show steps on paper.
Then verify your answer using the “Black-Scholes.xls” spreadsheet from
the class website or some equivalent tool.
For this and other problems, use either version (1) or (2) of the formula.
$ 5.31124
b. Repeat, but with S0 = $ 41.00. (You don’t need to do part b by hand; just get a
numerical answer by any method that works.)
$ 5.87540
2. Extreme cases of the formula. Assume for this problem that the arguments S0, , T, K, r, k
are always strictly positive. Some of these parts are hard.
It may be helpful to recall that and . Also,
the derivative of is the standard normal density function: .
Also, notice that if you divide V by k, then S0 and k only appear in the remaining
expression as part of the fraction S0/k.
a. Show by any method that V is always strictly positive.
Two approaches. First: We derived the formula for V by calculating the expected value of (S(T)-k)+ given the density function for S(T). The density function is positive everywhere and integrates to 1. The function (S(T)-k)+ is nonnegative everywhere and is positive on a half-infinite interval. When we multiply them together we get a function that is sometimes zero, sometimes positive; so its integral must be positive. General principle: If any random variable X is never negative and has a positive probability of being positive, then its expected value is positive. As a special case, V is positive.
Second, not requiring knowledge of how the formula was derived: When approaches zero, V has limiting value equal to (S0-k)+ (see below, part g). We computed V/ in class, and found it to be always positive. So, for any positive value of , the value of V must always be greater than (S0-k)+, which is nonnegative.
It doesn’t work just to notice that(d1) > (d2). [ Note: d1 and d2 are the conventional names for the arguments of in the formula ] If we also have S0 ≥ k, then it would follow that V = S0(d1) – k(d2) > 0. But when S0 < k, no such inference is possible.
b. Show by any method that V is always strictly greater than S0 – K OR that it is
always strictly greater than S0 – k. (Both are true.)(This is significant because it means that you should never exerciseyour call option early.)
Same two approaches. The first approach is a little more delicate in part b. We derived V as the expected value of (S(T)-k)+, given the density function for S(T). [ At least that’s what we did in flat dollars. If we were working in ordinary dollars, we would be using a different distribution for S(T), we would be getting the expected value of (S(T)-K)+, and we would need to discount the result back to time zero using e-rT at the end. ] In any case, the result is always larger than the expected value of (S(T)-k) (without the “+”). That expected value is the difference between E(S(T)) and E(k)=k. But E(S(T)) = S0 under the flat-dollar, risk-neutral assumptions we have in place, so the difference is always greater than S0 – k.
The second approach is exactly as above: We showed V > (S0-k)+ which implies that V > S0 – k.
c. What is the limit of V as S0 approaches zero? (In this and other parts,
assume that the other arguments remain fixed.)
zero
d. How does V behave as S0 approaches +infinity? Try to give a simple formula
for V that is more precise than “V also goes to infinity.”
V becomes indistinguishable from S0 – k, and so it goes to infinity.
e. What is the limit of V as K (or k) approaches zero?
S0. In most of parts c-i, the main trick is to notice that d1 and d2 each approaches zero or +infinity, and so (d1) and (d2) each approach 0 or 1, with resulting simplifications of the formula for V.
f. What is the limit of V as K (or k) approaches zero +infinity?
zero
g. What is the limit of V as approaches zero?
(S0 – k)+. Here’s why: Suppose 0. Now d1 and d2 approach…well, if S0 < k then they both approach –infinity, so (d1) 0 and (d2) 0, and V 0. But if S0 > k, then d1 and d2 approach +infinity, so (d1) 1 and (d2) 1, and V S0 – k. In either case, the result is the same as (S0 – k)+.
h. What is the limit of V as approaches +infinity?
S0 --- that may seem like a strange result, but it’s true.
i. What is the limit of V as T approaches zero?
(S0 – k)+.
3. Compute . First give a formula for --- it is a function of the same arguments as
V itself --- then calculate its value given the inputs from problem 1. (The answer should be roughly the same as the difference between your answers to #1a and #1b.)
… that is, it’s the same as (d1).
This makes sense, because to first approximation it’s the probability that the option will be exercised. If the option is exercised, an increase of $1 in the initial share price results in an increase of about $1 in S(T) and an increase of $1 in the profit; of course if the option isn’t exercised the increase in S0 doesn’t matter. (But beware, this argument isn’t quite right, and actually is an adjusted probability.)
When you just compute the derivative, you get other terms, but with a lot of work they all cancel.
4. (Whether or not you have a formula for delta…) Show that if I form an investment portfolio
consisting of (a) one call option as described above and (b) shares of the underlying stock, then the (theoretical) value of my portfolio doesn’t change in response to small
changes in the stock price. That is, if W is the value of my portfolio, then .
This practice is called “delta-hedging.” If I do it, I can’t gain or lose money unless changes or I can’t keep up with large changes in the stock price. (Or if anything else
strange happens.)
When S0 chances by a small amount, say h, the value of one call option increases by about h --- that’s what it means for to be equal to V/S0. Also, the value of the - shares changes by -h. So, we break even.
Now we seem to have proved that in practice.
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