# [PROBLEM B] Current stock price for XYZ $50.00 Interest rate 3% Dividend rate 0% Delta (AOption/As)

[PROBLEM B] Current stock price for XYZ $50.00 Interest rate 3% Dividend rate 0% Delta (AOption/As) Strike Expiration Option Price Implied Vol Option PUT $30.00 6-months $0.14 40% -0.023 $40.00 6-months $0.77 32% -0.123 PUT PUT $50.00 16-months $2.74 22% -0.431 PUT $50.75 6-months $2.97 21% -0.470 CALL $50.00 6-months $3.48 22% 0.569 CALL $50.75 6-months $2.97 21% 0.530 CALL 19% 0.299 $55.00 $60.00 6-months 6-months $1.20 $0.15 CALL 15% 0.065 XYZ stock is currently trading at $50 per share. We ask a trading desk to price 6-month options on stock XYZ: 4 puts and 4 calls. The table above gives the information. Option prices are provided; for each option at its given price we run the price through the Black-Scholes formula and solve for the implied volatility. (Recall that this means answering the question: “What o [volatility] do we need in the Black-Scholes formula to match the given price?'') And now let's use the Black-Scholes formula again to calculate the derivative of the option price with respect to the stock price. This means: [10] 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. [EXTRA CREDIT] Go back to part [8], where we examined the table to find the strike at which the put and call have the same value. As expected, the deltas are opposite signs. Can you explain this statement below? “Black-Scholes assumes that In(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. This explains why … at the $50.75 strike (where the put and call have equal value) … the delta of the call is >0.5 and the delta of the put is >-0.5. If we had normality (instead of log-normality) the deltas would be exactly 0.5 and -0.5.” [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). Note that a graph of the profit/loss would be the same except that it would be “shifted down” by $0.92. [6] Write the general formula (in terms of S, R, q, t, T) for the forward price of a stock for delivery date T. In our specific example … is the forward price for delivery in 6 months higher or lower than the current price of $50? [7] Write the general formula for V(St), the value of a forward contract (in terms of S, R, q, t, T, and L, where L is the delivery price). [8] Recall the put-call parity relationship: [PRICE OF K-STRIKE CALL (EXPIRY T)] – [PRICE OF K-STRIKE PUT (EXPIRY T)] = [PRICE OF FORWARD CONTRACT WITH DELIVERY 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? [9] Let's revisit the put-call parity relationship: CK(St) – PK(St) = V«(St) where the subscript “K” means strike = K (for the two options C and P) and delivery 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?