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MATH-UA.0250 Mathematics of Finance Spring 2022 A
发布时间：2022-05-16
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MATH-UA.0250
Mathematics of Finance
Spring 2022
TAKE-HOME FINAL
[PROBLEM A]
In this problem we examine two stochastic processes for a stock price:
PROCESS A: “Driftless” geometric Brownian motion (GBM). “Driftless” means no “dt” term. So it’s our familiar process: dS = s S dW with S(0) = 1. s is the volatility.
PROCESS B: dS = a S2 dW for some constant a, with S(0) = 1
As we’ve said in class, for any process the instantaneous return is the random variable:
dS/S = (S(t + dt) - S(t))/S(t)
[1] Explain why, for PROCESS A, the variance of this instantaneous return (VAR[dS/S]) is constant (per unit time). Hint: What’s the variance of dW?
The rest of this problem involves PROCESS B.
[2] For PROCESS B, the statement in [1] is not true. Explain why PROCESS B’s variance of the instantaneous return (per unit time) depends on the value S(t).
Let’s manipulate PROCESS B using a change of variable (and Ito’s Formula) to see what we come up with. Worth a try. Let Y(t) = 1/S(t).
[3] Apply Ito directly and show that we obtain:
dY = (-1/S2)dS + (1/2)(2/S3) (dS)2
[4] Can you reframe this and obtain:
dY = -a dW + a2 S dt (***)
Recall that, for any process S(t), the average value over the interval [0, T] equals the stochastic integral:
A(T) = (1/T) S S(t) dt over [0, T]
(This is a random variable for each future time T. Its value depends on the path that S(t) takes from time 0 to time T.)
[5] Integrate equation (***) over [0,T]. Show that we get:
Y(T) – Y(0) = (1/S(T)) – 1 = S -a dW + a2 T A(T)
[6] Explain why, for each future time T, the above says that:
1/S(T) = U(T) + (constant) * A(T) where U(T) is normally-distributed.
[7] Are U(T) and A(T) uncorrelated?
If we had “solved” PROCESS B’s SDE we’d be able to write down the density function for S(T).
That we have not done.
But we have related the random variables S(T) and A(T) … that’s some information.
[8] Do you think that PROCESS B is an “economically-plausible” process for a stock price? Hint: What does this process say about the variability of the price as the price goes up?
[PROBLEM B]
XYZ stock is currently trading at $50 per share. We ask our favorite trading desk to price a bunch of 6-month options on XYZ: 4 puts and 4 calls. The table above gives the information.
For each option price we’ve run the price through the Black-Scholes formula and solved for the implied volatility.
The table also gives the deltas: the derivative of each option price with respect to the stock price. As you know this means:
n Leave the implied vol, the interest rate, and the dividend rate unchanged;
n Change the current price of the stock by a small amount up and down, say +/- $0.01;
n Apply the call-price or the put-price formulas, C(S,t) or P(S,t), and obtain the “stock-price-up” and “stock-price-down” prices of the options;
n Numerically compute the derivative DOption/DS. This is the option’s delta at that value of the underlying stock and that implied volatility. It’s a measure of how sensitive the option price is to the price of the underlying stock. Note that we could also obtain the delta analytically by the taking the partial derivative (with respect to S) of the functions C(S,t) and P(S,t).
[1] Suppose you buy the $40-strike put at the offer price of $0.77. You come in the next day and see that the stock price has fallen to $44. Will the put price have gone up or down?
[2] Say you bought the $40-strike put (at $0.77); you decide to sell it back after the price of the stock has fallen to $44. Would the trader bid a price corresponding to 32% implied volatility (which is what you “paid” back when the stock was at $50) … or would they bid at a higher or lower implied volatility? Explain.
[3] Why is the delta negative for all the puts and positive for all the calls?
[4] The $55-strike call is priced at $1.20. Suppose the price of the stock moved up from $50 to $52. For this $55-strike call … what would be its (approximate) updated price if it got priced at the same implied volatility (19%) as before?
[5] Suppose you bought the $40-strike put and the $60-strike call. This would cost you
$0.77 + $0.15 = $0.92.
Please draw a graph that shows the value of this combined position at the expiration date. The x-axis will be S(T), the value of the stock at expiration; the y-axis will be the value of the position (put and call).
[6] Recall the general formula (in terms of S(t), R, q, t, T) for the forward price of a stock for delivery date T:
[Forward price for delivery date T] = S(t) * exp((r-q)(T-t))
In our specific example … is the forward price for delivery in 6 months higher or lower than the current price of $50?
[7] Recall the put-call parity relationship:
[PRICE OF K-STRIKE CALL (EXPIRY T)] – [PRICE OF K-STRIKE PUT (EXPIRY T)] =
[VALUE OF FORWARD CONTRACT WITH INVOICE PRICE K (DELIVERY DATE T)]
Look at the table. At what strike do the put and the call have the same value? Is this consistent with what we know from put-call parity?
[8] Let’s revisit the put-call parity relationship:
CK(S,t) - PK(S,t) = VK(S,t)
where the subscript “K” means strike = K (for the two options C and P) and invoice price = K (for the forward contract V).
Assume that the dividend rate is zero (like we have in our specific case). Take the derivative of the put-call parity relationship. Show that: (DELTA of CALL) - (DELTA of PUT) = 1
Do we see this relationship appearing in in the table above?
[9] Let’s focus on the $40-strike put. As we see in the table … the trader is offering this option at $0.77, which is an implied volatility of 32%. At this implied volatility we observe a delta of -0.123. Of course, at 40% implied volatility the price you’d be charged is higher. How about the delta at 40% volatility? Would it be more negative or less negative? Explain.
[PROBLEM C]
This problem references the same set-up as in [PROBLEM B].
The graph shows the price profile (option price vs. strike, in the Black-Scholes model) of 6-month options on XYZ, for two different implied volatilities (IVs): 22% and 40%. Recall that 22% was the implied volatility where the $50-strike put was priced, and 40% was the implied volatility where the $30-strike put was priced.
[1] Can you locate on the graph the price ($2.74) of the $50-strike put (priced at 22% IV) and the price ($0.14) of the $30-strike put (priced at 40% IV)?
[2] At what value of the strike is the price difference (for the two different implied volatilities) the biggest? Any thoughts on this?
[EXTRA CREDIT: JUST FOR FUN]
QUESTION 1:
Can you comment on the statements below?
For European-style options, we know that the put (struck at the forward price) and the call (struck at the forward price) have the same price. This is true in ANY MODEL.
Black-Scholes assumes that log(S(T)) is distributed normally. This means that the resulting distribution for S(T) is not symmetric … it’s fatter to the right than to the left because of the nature of the exponential function.
In Black-Scholes we get, for the two at-the-money-forward options, a call delta >0.5 and a put delta >-0.5. (Look at the deltas for the $50.75 put and call in the PROBLEM B table.)
If we had normality (instead of log-normality) can you suggest what the deltas would be?
QUESTION 2:
Look at the graph in PROBLEM C. For a fixed implied volatility, we see that as we go lower in strike the put price drops … but at a slower and slower rate. Any ideas on why this is happening?
Hint: Recall that the put price can be obtained as the present value of the expected option payoff under the adjusted density function for S(T):
exp(-r(T-t)) S(K – x) pdf(x) dx where the integral is from 0 to K.
What does the model assume about the shape of this pdf?
QUESTION 3:
Let’s look at very-low-strike puts. Suppose there’s a “threshold strike,” call it K1, for which the only way that the stock price could fall below K1 over the life of the option is if the company goes bankrupt over the period. And assume that if that were to happen the stock price would fall to $1.
Show that for any strike K < K1 the put price could be close to:
e-r(T-t) x (K - $1) x (PROBABILITY OF BANKRUPTCY)
