Lookback options Simone Calogero February 12, 2020 Lookback options are non-standard European style derivatives whose pay-off depends on the minimum or maximum of the stock price within a given time period until maturity. There exists four main types of lookback options. • A lookback call option with floating strike and maturity T > 0 gives to the owner the right to buy the underlying stock at maturity for the minimum price of the stock in the interval [0;T ]. Thus the pay-off for this lookback option is float YLC = S(T ) − minfS(t); t 2 [0;T ]g: • A lookback put option with floating strike and maturity T > 0 gives to the owner the right to sell the underlying stock at maturity for the maximum price of the stock in the interval [0;T ]. Thus the pay-off for this lookback option is float YLP = maxfS(t); t 2 [0;T ]g − S(T ): • A lookback call option with fixed strike K > 0 and maturity T > 0 pays the buyer the difference between the maximum of the stock price in the interval [0;T ] and the strike K, provided this difference is positive. Hence the pay-off for this lookback option is fixed YLC = (maxfS(t); t 2 [0;T ]g − K)+: • A lookback put option with fixed strike K > 0 and maturity T > 0 pays the buyer the difference between the strike price K and the minimum of the stock price in the interval [0;T ], provided this difference is positive. Hence the pay-off for this lookback option is fixed YLP = (K − minfS(t); t 2 [0;T ]g)+: In the following section we focus on the lookback call option with floating strike in a Black- Scholes market. In particular the stock price is given by (r− 1 σ2)t+σW (t) S(t) = S(0)e 2 f and the risk-neutral price for the floating strike lookback call option with maturity T > 0 is float −r(T −t) ΠLC (t) = e Ee[S(T ) − minfS(τ); τ 2 [0;T ]gjFW (t)]: 1 1 Pricing PDE for the lookback call option with float- ing strike The main purpose of this section is to derive the PDE satisfied by the pricing function of lookback call options with floating strike. Theorem 1. Let v : (0;T ) × (0; 1) × (0; 1), v = v(t; x; y), satisfy 1 @ v + rx@ v + σ2x2@2v = rv; t 2 (0;T ); x > 0; 0 < y < x; (1a) t x 2 x @yv(t; x; x) = 0; t 2 [0;T ]; x > 0; (1b) v(T; x; y) = x − y; 0 ≤ y ≤ x: (1c) float Then ΠLC (t) = v(t; S(t); min0≤τ≤t S(τ)). Proof. Let Y (t) = min0≤τ≤t S(τ); note that fY (t)gt≥0 is a non-increasing process in the 0 space C [FW (t)]. However fY (t)gt≥0 is not a diffusion process. We now show that dY (t)dY (t) = 0: Recall that this means that the quadratic variation of fY (t)gt≥0 is zero in any interval [0;T ] along any sequence of partitions fΠngn2N of this interval such that kΠnk ! 0, as n ! 1. (n) n Letting Πn = ft0 = 0; t1 ; : : : ; tm(n) = T g, we have to prove that m(n) X (n) (n) 2 2 lim (Y (tj ) − Y (tj−1)) = 0 in L (Ω): n!1 j=1 But m(n) X (n) (n) 2 (n) (n) X (n) (n) (Y (tj ) − Y (tj−1)) ≤ max jY (tj ) − Y (tj−1)j jY (tj ) − Y (tj−1)j j j=1 j (n) (n) X (n) (n) = max jY (tj−1) − Y (tj )j (Y (tj−1) − Y (tj )) j j (n) (n) = max jY (tj−1) − Y (tj )j(Y (0) − Y (T )); j where in the sum we used that Y (t) is non-increasing to write jY (t) − Y (s)j = Y (s) − Y (t), (n) (n) for t ≥ s. As Y is continous in time, then maxj jY (tj−1)) − Y (tj )j ! 0, pointwise in ! 2 Ω. (n 2 As Y (tj ) ≤ Y (0), then, by the dominated convergence theorem, the limit is zero also in L , which completes the proof of dY (t)dY (t) = 0. Similarly one can prove that dS(t)dY (t) = 0 (see Exercise 1). Hence applying It^o'sformula we obtain 1 d(e−rtv(t; S(t);Y (t))) = e−rt(@ v + rx@ v + σ2x2@2v − rv)(t; S(t);Y (t)) dt t x 2 x −rt −rt + e σS(t)@xv(t; S(t);Y (t))dWf(t) + e @yv(t; S(t);Y (t))dY (t): 2 The drift term (::: ) dt is zero by the PDE (1a). We now show that the term (::: )dY (t) is also −rt zero, thereby concluding that fe v(t; S(t)Y (t)gt≥0 is a Pe-martingale relative to fFW (t)gt≥0. Since Y (t) is non-increasing, then it has bounded first variation and therefore the integral Z t @yv(τ; S(τ);Y (τ))dY (τ) 0 can be understood in the Riemann{Stieltjes sense. We divide this integral as Z t Z @yv(τ; S(τ);Y (τ))dY (τ) = @yv(τ; S(τ);Y (τ))dY (τ) 0 S(τ)>Y (τ) Z + @yv(τ; S(τ);Y (τ))dY (τ): S(τ)=Y (τ) The second piece is zero by the boundary condition (1b). The second piece is also zero, because fS(τ) > Y (τ)g is an open set (as S; Y are time-continuous) and Y (τ) is constant −rt in this set (that is \dY (τ) = 0"). It follows that the process fe v(t; S(t)Y (t)gt≥0 is a Pe-martingale relative to fFW (t)gt≥0; in particular −rT −rt Ee[e v(T;S(T );Y (T )]jFW (t)] = e v(t; S(t);Y (t)): Hence, by the terminal condition (1c), −r(T −t) v(t; S(t);Y (t)) = e Ee[S(T ) − Y (T )]jFW (t)]; which is the claim. Exercise 1. Prove the property dS(t)dY (t) = 0 used in the previous theorem. To study the problem (1) one needs a complete set of boundary consitions for strong solutions. Assume first that y ! 0. This means that the stock price has reached the value zero at some time 0 ≤ τ ≤ t, in which case of course the minimum stock price in the interval [0;T ] will be zero with probability 1. Hence for y ! 0+, the lookback call price converges to its highest possible value, that is v(t; x; 0) = x: t 2 [0;T ]: (2) The boundary value as x ! 1 is not so obvious. In the next theorem we show that the 1+2 dimensional problem (1) can be reduced to a 1+1 dimensional problem with explicit boundary conditions. Theorem 2. Let u : (1; 1) ! R satisfy 1 @ u + rz@ u + σ2z2@ u = ru; t 2 (0;T ); z > 1 (3a) t z 2 zz with terminal condition u(T; z) = z − 1 (3b) 3 and boundary conditions lim (u(t; z) − z) = 0; u(t; 1) − @zu(t; 1) = 0: (3c) z!1 Then x v(t; x; y) = yu t; y solves (1) as well as (2). Exercise 2. Prove the previous theorem. A project which aims to price floating strike lookback call options by solving numerically the problem (3) is described in the next section. 2 A project on lookback call options with floating strike The main purpose of this project is to derive numerically some qualitative properties of lookback call options with floating strike and to compare the finite differences method with the Monte Carlo method. Part 1 • Write a short introduction on the Lookback option, where you should discuss in par- ticular why this option has been introduced (you can find plenty of info on the web). Outline the content of the rest of the report. • Solve Exercise 2. • Write a finite difference scheme that solves the problem (3). Use the Crank-Nicolson method. HINTS: (1) recall to first transform the terminal value problem into an initial value problem by doing the change of variables t ! T − t. (2) Think carefully on how to impose the boundary condition u(t; 1) = @zu(t; 1) numerically Part 2 • Write a Matlab function that implements the finite difference scheme derived in Part 1. The parameters S0; r; σ; T , must appear as input variables of your function. • Plot the initial price of the lookback call as a function of S0; r; σ; T and discuss the results. • Compare the price obtained by the finite difference method and by the Monte Carlo method for different value of σ and compare the efficiency of the two methods, e.g., by performing speed tests. Present your results using tables. Include your matlab codes in an appendix. 4.