Homework Set 6 Solutions

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1 MATH Introduction to Mathematical Finance Prof. D. A. Edwards Due: Apr. 11, 018 P Homework Set 6 Solutions K z K + z S 1. The payoff diagram shown is for a strangle. Denote its option value by V z. (a) (5 points) Construct a portfolio of puts and/or calls that replicates this payoff. Draw the payoff diagrams of the individual options. Write V z in terms of the values of the various options (you needn t write out the full Black-Scholes formulas). P K z K + z S Decomposition of strangle into put and call options. Solution. See the figure above. The dotted line represents buying a put with strike K z [which we shall denote P (S, t; K z)], and the dashed line represents buying a call Copyright 018 D. A. Edwards All Rights Reserved

2 with strike K + z (which we shall denote C(S, t; K + z)]. Therefore, we have that M667S18Sol6. V z (S, t) = P (S, t; K z) + C(S, t; K + z). (A) (b) (7 points) Show that the sensitivity of V z to changes in z is given by e r(t t) [N( d ) + N(d+ )], d± = log(s/(k ± z)) + (r σ /)(T t) σ. T t Solution. Differentiating (A) with respect to z, we have V z z = P C (S, t; K z) + (S, t; K + z). K K Calculating the derivative with respect to strike, we have C K = [SN(d 1) Ke r(t t) N(d )] K Continuing to simplify, we have = Sn(d 1 ) d 1 K e r(t t) [ N(d ) + Kn(d ) d ]. K C K = d [ ] 1 Sn(d 1 ) Ke r(t t) n(d ) e r(t t) N(d ) = e r(t t) N(d ), K where we have used the fact that d 1 /K = d /K, and notes in class that the bracketed quantity is 0. Then using put-call parity, we have P K = C K + e r(t t) = e r(t T ) [1 N(d )] = e r(t T ) N( d ), where we have used the property that N(x) + N( x) = 1. Continuing to simplify, we have as required. V z z = e r(t t) N( d ) e r(t t) N(d + ) = e r(t t) [N( d ) + N(d+ )], d ± = log(s/(k ± z)) + (r σ /)(T t) σ, T t (c) (3 points) Use your answer to (b) to show that a strangle is always cheaper than a straddle. Solution. Note that a straddle corresponds to a strangle with z = 0. We also note that V z /z < 0, so V z is a strictly decreasing function of z. Hence a strangle s value decreases with increasing z, and is always cheaper than a straddle, which has z = 0. (d) ( points) Explain your answer to (c) financially. Solution. A straddle has a positive payout for any S K, so financially it should be worth more than a strangle which has a zero payout for a finite range of S.

3 M667S18Sol6.3 (e) ( points) Note that the strangle can still be expensive because of the possibility of large payouts if S becomes large. Explain how one could introduce another option into the portfolio to make the portfolio cheaper. Do NOT work through all the details. Solution. To make the portfolio cheaper, we could cap the maximum payout in a similar way to that on a spread. In particular, we could sell a call option with some strike greater than K > K + z. This would then maximize the payout at a fixed value for large S.. Let P (t; K, T ) be the value of a put option at time t with strike K and expiration T. Let C P (t; K 1, T 1 ) be a compound call on the underlying put option (call-on-put) with strike K 1 and expiration T 1 < T, and let P P (t; K 1, T 1 ) be a compound put on the underlying put option (put-on-put), also with strike K 1 and expiration T 1. (a) ( points) What is the payoff function for P P? Solution. The payoff is just the difference between the strike price and the underlying put price, as long as it is positive: P P (T 1 ; K 1, T 1 ) = [K 1 P (T 1 ; K, T )] +. (b) (3 points) Construct a put-call parity relationship for C P and P P. Solution. We construct a portfolio where we buy C P and sell P P. Then the payoff for this portfolio is Π(T 1 ; K 1, T 1 ) = [P (T 1 ; K, T ) K 1 ] + [K 1 P (T 1 ; K, T )] + = P (T 1 ; K, T ) K 1, where we have used notes in class to write down the payoff function for C P. But then the value of this portfolio for any time t < T 1 is given by Π(t; K 1, T 1 ) = P (t; K, T ) K 1 e r(t 1 t), so we have C P (t; K, T ) P P (t; K, T ) = P (t; K, T ) K 1 e r(t 1 t). 3. (5 points) Suppose that in order to minimize fluctuations in the option price near expiry, we limit the average to some fixed time frame before expiry: I = t 0 T0 0 f(s(τ), τ) dτ, t T 0 < T, f(s(τ), τ) dτ, T 0 t T. Note that this is not consistent with the definition given in class, since the integrand depends on t. Derive a single PDE for V (S, I, t) in this case similar to the one we gave in class.

4 M667S18Sol6.4 Solution. Note that we have not changed I at all for the time t t 0. Hence in that case the result is the same as before: V t = σ S V V x S + f(s, t)v I rv, t T 0. But for t > T 0, I does not vary at all, so di = 0. Hence when we expand V, the V/I term will vanish, and we have V t = σ S Putting these two equations together, we have V t = σ S where H( ) is the Heaviside step function. V V x S rv, t T 0. V V x S + f(s, t)h(t 0 t) V I rv, 4. Consider the case of the lookback floating strike put, and suppose that we assume that V (S, M, t) = MF (y, t), y = S M. (6.1) (a) (6 points) Show that with this assumption, the domain in y is fixed, the final condition for F can be written in terms of y alone, and the boundary condition is given by F (1, t) = F (1, t). y Solution. The domain for the original problem is 0 S M, which then becomes 0 S/M 1, or 0 y 1. So the domain in y is fixed. The final condition becomes V (S, M, T ) = (M S) + MF (y, T ) = M(1 S/M) + F (y, T ) = (1 y) +. To do the boundary condition, we note that M = y M y = S M y = y M y. (B) Using this result, the boundary condition at S = M becomes V M ( (MF ) (M, M, t) = M F ) ( M M + F (y = 1, t) = y F y + F (M, M, t) = 0 ) (1, t) = 0 F (1, t) = F (1, t), y

5 Powered by TCPDF ( M667S18Sol6.5 where we have substituted in y = 1. (b) (5 points) Show that if one assumes (6.1), the governing equation for F is given by the Black-Scholes equation with S replaced by y. Solution. We just have S-derivatives in the equation from class, so we compute S = y S y = 1 M S = 1 M y. y Using this result in the Black-Scholes equation, we have (MF ) t (MF ) t = σ S = σ y F t = σ y 1 (MF ) M y 1 (MF ) r(mf ) M y (MF ) y + ry (MF ) r(mf ) y F F + ry y y rf, which is just the Black-Scholes equation with V replaced by F and S replaced by y.