MSc Financial Engineering CHRISTMAS ASSIGNMENT: MERTON S JUMP-DIFFUSION MODEL. To be handed in by monday January 28, 2013

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1 MSc Financial Engineering CHRISTMAS ASSIGNMENT: MERTON S JUMP-DIFFUSION MODEL To be handed in by monday January 28, 2013 Department EMS, Birkbeck Introduction The assignment consists of Reading and understanding the introductory text on Ito calculus for jump-diffusions in Part I below. Answering the three exercise of Part II, in which you are asked to develop part of the Merton jump-diffusion model for option pricing. Literature may of course be consulted, but the answers will have to be phrased in your own words. 1. Part I: Ito calculus for jump diffusions Stock prices can exhibit sudden large jumps, for example on arrival of unexpected good or bad news. These jumps can t be modelled very well by Brownian motion, and have to be added separatedly. The archetypical stochastic jump process is the so-called Poisson process, and we will model our stock return by the sum of a Brownian diffusion and a Poisson process with random jumps. Such processes are known as jump diffusions. This section introduces the Poisson process, and extends the Ito-lemma to jump-diffusions. In part 2 you will be asked to apply this to option pricing Poisson process. A stochastic process (N t ) t 0 is called a Poisson process if 1. N t only takes on values 0, 1, 2,... and N 0 = If s t, then N t N s is Poisson-distributed with mean λ(t s): this means that (λ(t s))n (1) P(N t N s = n) = e λ(t s), n = 0, 1, 2,... n! 3. If u s < t, then N u and N t N s are independent. Here λ > 0 is a constant which is called the intensity of the Poisson process. The definition of a Poisson process is quite similar to that of a Brownian motion, except that we have replaced the normal distribution by the 1

2 2 Poisson distribution. Note N t is an increasing (in the sense of nondecreasing) function of t: s t implies that N s N t, since the event that N t N s < 0 has probability 0, by 2. A Poisson process will jump from 0 to 1 at some time T 1, then from 1 to 2 at some time T 2, etc. In general, it will jump to i at (2) T i := the smallest t such that N t i. We artificially put T 0 = 0. The jump times T 1, T 2,... are random times, and it can be shown that the waiting times between T i T i 1 between jumps are independent and exponentially distributed. Instead of just jumping by an amount of 1, we can generalize the process by admitting stochastic jumps of J i at the successive jump times T i of a Poisson process. We will do this below, but first have a look at Ito calculus for a Poisson process Stochastic Calculus with Poisson processes. We start by considering the stochastic differential of N t, and then study the differential of functions on N t. By definition, (3) dn t := N t+dt N t ; Note that dn t can only take on values 0, 1, 2, etc. It will do so with probabilities 1 P(dN t = k) = λk (dt) k e λt = 0 if k 2, k! since dt is an infinitesimal: (dt) 2 = (dt) 3 = = 0. It follows that dn t is either 0 or 1, with probabilities and e λdt = 1 λdt + (λdt)2 2 = 1 λdt, (λdt)e λdt = λdt(1 λdt + (λdt)2 ) = λdt, 2 respectively, using the Taylor expansion of the exponential and the fact that dt is an infinitesimal. Summarizing: 1, probability λdt (4) dn t = 0, probability 1 λdt. If we now consider a function f(n t ) of N t, then df(n t ) := f(n t+dt ) f(n t ) = f(n t + dn t ) f(n t ), and we therefore have f(nt + 1) f(n df(n t ) = t ), probability λdt 0, probability 1 λdt. 1 take t s = dt in (1)

3 Since df(n t ), like dn t, takes on only two different values, one of which is 0, and with the same probabilities as dn t does, this can be summarized in a single equation: (5) df(n t ) = (f(n t + 1) f(n t )) dn t Stochastic calculus: Poisson processes with random jumps. More generally, let us consider a process (X t ) t 0 which jumps at the same times as a Poisson process, but whose jumps are given by random variables J t (if the jump takes place at time t). Using stochastic differentials, such a process can be denoted by (6) dx t = J t dn t. By similar arguments as before we now have f(xt + J df(x t ) = t ) f(x t ), probability λdt 0, probability 1 λdt hich e can write more succintly as (7) df(x t ) = (f(x t + J t ) f(x t )) dn t. We will always assume that the jumps are independent of the Poisson process. Later on we will also assume that they are mutually independant and identically distributed (that is, have the same probability distributions) Stochastic calculus with jump-diffusions. A jump-diffusion process is a stochastic process (X t ) t 0 whose stochastic differential is given by (8) dx t = A t dt + B t dw t + J t dn t, where (W t ) t 0 and (N t ) t 0 are, mutually independent, Brownian motion and Poisson processes, respectively. We assume that A t and B t are adapted to the filtration generated by (W s, N s ) : s < t}, that is, A t and B t are functions of the Brownian and Poisson trajectories up till, but not including 2 time t; we also say in this situation that A t and B t are previsible. We furthermore assume that J t is conditionally independent of dn t, given this filtration (that is, given such trajectories). We then have the Theorem 1.1. (Ito s lemma for jump-diffusion processes). Let f(x, t) be a C 2 -fundtion of x and t. Then ( f df(x t, t) = t + A f t x dt + 1 ) 2 f f (9) 2 B2 t dt + B x 2 t x dw t + (f(x t + J t, t) f(x t, t)) dn t, } 3 2 for Brownian motion it does not matter whether we include t or not: by continuity, knowing W s for all s < t implies knowing W t; for N t it is important not to include t, since a jump might happen precisely at t

4 4 were all derivatives of f are to be evaluated at (X t, t), that is, just before any (potential) jump in (t, t + dt]. Proof. The proof is a combination of the Ito-lemma for Brownian motion and the previous arguments for pure jump processes. To simplify notations, we put dy t = A t dt + B t dw t, the diffusion part of dx t. Also to simplify the exposition, we limit ourselves to functions f(x) of x only, leaving the more general case to the reader. Then df(x t, t) = f(x t+dt ) f(x t ) f(xt + dy = t ) f(x t ), probability 1 λdt f(x t + J t + dy t ) f(x t ), probability λdt. The usual Ito-lemma applied to the first line gives We can re-write the second line as f (X t )dy t f (X t )(dy t ) 2. f(x t + J t + dy t ) f(x t + J t ) + f(x t + J t ) f(x t ), and again apply the usual Ito-lemma to the first term. Hence f df(x t, t) = (X t )dy t + 1f (X 2 t )(dy t ) 2, f (X t + J t )dy t + 1f (X 2 t + J t )(dy t ) 2 + f(x t + J t ) f(x t ), with probabilities 1 λdt and λdt, respectively. In terms of dn t, df(x t ) = f (X t )dy t f (X t )(dy t ) 2 + (f (X t + J t ) f (X t )) dy t + 12 (f (X t + J t ) f (X t )) (dy t ) 2 } dn t + f(x t + J t ) f(x t )} dn t. We now argue that all terms in the second line are equal to 0. Indeed, dy t dn t and (dy t ) 2 dn t are all random variables with mean and variance 0: we have for example that E(dW t dn t ) = E(dW t )E(dN t ) = 0, and 3 var(dw t dn t ) = E(dW 2 t )E(dN t ) 2 ) = dt (λdt + λ 2 (dt) 2 ) = 0. Hence dw t dn t = 0. Similarly, dtdn t = 0. Using the adaptedness of the coefficient processes A t, B t, the same conclusion follows for dy t dn t and, a forteriori, for (dy t ) 2 dn t. Hence df(x t ) = f (X t )dy t f (X t )(dy t ) 2 + ( f(x t + J t ) f(x t ) ) dn t, which is the same as (9) for the f s considered. 3 since one easily computes that the variance of Nt N s is λ(t s), t > s

5 Now that we have extended the Ito-calculus to jump-diffusion processes, we can do derivative pricing à la Black and Scholes, using such processes as the fundamental drivers of asset prices Merton s jump-diffusion model. Merton s model assumes that the underlying asset price evolves according to ds t (10) = µdt + σdw t + (J t 1)dN t, S t where S t stands for the value of S just before a jump at t, if there is one. We assume that jump-sizes J t are identically distributed and mutually independent. We also assume that the three processes (W t ) t 0, (N t ) t 0 and (J t ) t 0 are independent. To explain the J t 1, consider what happens with the asset price when a jump occurs at T i. Letting S Ti be the asset price just before the jump, we have that ds Ti S Ti = S T i S Ti S Ti = (J Ti 1), or S Ti = J Ti S Ti. Since prices should not become negative, we will assume wlog that J Ti 0. With the previous observation, the SDE (10) becomes easy to solve: S t evolves like a Geometric Brownian motion between jumps, and gets multiplied by J Ti at the jump-times T 1 < T 2 < of the Poisson process. To simplify notations, we will write J i := J Ti. Then, S 0 e (µ 1 2 σ2 )t+σw t, 0 t < T 1, S (11) S t = 0 J 1 e (µ 1 2 σ2 )t+σw t, T 1 t < T 2, S 0 J 1 J 2 e (µ 1 2 σ2 )t+σw t, T 2 t < T 3, etc. Equivalently, the usual GBM gets multiplied by J 1 J k if there are k jumps in [0, t]: (12) S t = S 0 J 1 J 2 J k e (µ 1 2 σ2 )t+σw t if N t = k, which sometimes is summarized in the single formula: ( Nt ) (13) S t = S 0 J ν e (µ 1 2 σ2 )t+σw t, ν=0 where we artificially put J 0 := 1. 5

6 6 2. Part II: Derivative pricing in jump-diffusion models Suppose the asset price S t follow a jump-diffusion process (10), and consider a derivative contract written on the asset, with price a function V (S t, t) of S t and of t. Exercise 2.1. Set up a portfolio Π t at time t by going long the derivative, and shorting t of the underlying asset. The amount t can only depend on information avialable prior to a (potential) jump at t. (a) By using Ito s lemma for jump diffusions, show that ( ( ) V V dπ t = t + µs t S t + 1 ) 2 σ2 St 2 2 V dt S ( ) 2 V + σs t S t dw t + (V (JS t, t) V (S t, t)) t S t (J t 1)) dn t, with derivatives evaluated in (S t, t) (that is, pre-jump). (b) There are now two sources of risk: the familiar diffusion risk coming from the dw t -term, but also the jump risk. Explain why it is not possible anymore to completely eliminate the risk, and why it is not even possible to totally eliminate the jump-risk on its own. (What would t have to be? Can it, given that it has to depend on pre-jump information only?). (c) Choose t = V/ S(S t, t) to eliminate the diffusion risk. Merton assumed that financial markets would not reward jump-risk, since it is asset specific, and can in principle be diversified away. The expected return of the resulting portfolio would therefore be the risk-free return. Under this assumption, show that V (S, t) satisfies (14) V t σ2 S 2 2 V V + (r κ)s + λe (V (JS, t) V (S, t)) = rv, S2 S where J is a random variable which is identically distributed with J t, and where κ is given by κ = λe(j 1), λ being the intensity of the Poisson process. For a European-style derivative this has to be solved for t < T, with boundary condition V (S, T ) = F (S), F (S) representing the pay-off. (d) Equation (14) is sometimes called a partial integral-differential equation or PIDE. To understand why, re-write the equation in terms of the pdf f J (x) of the jump J.

7 Exercise 2.2. (Feynman-Kac for jump diffusions) There are several ways of solving (14). This exercise explains the Feynman-Kac approach. Introduce a new process (Ŝt) t 0 by (15) dŝt = (r κ)ŝt dt + σŝt dw t + (J 1)Ŝt dn t. 7 (a) Suppose that V (S, t) solves (14). Show that e rt V (Ŝt, t) is a martingale. ( ) Hint: Compute d e rt V (Ŝt, t) = A t dt + B t dw t + C t dn t ; this is a martingale precisely when 0 = E t (dx t ) = A t dt + E t (C t dn t ), E t being expectation conditional on S t. To compute the E t (C t dn t )- term, use the independence of the jumps. (b) Suppose that V (S, T ) = F (S). Show that ( ) V (S, t) = e r(t t) E F (ŜT ) Ŝt = S, which is the Feynman-Kac solution for (14) with final condition V (S, T ) = F (S). (c) By using the explicit solution of the SDE (15), show that ( ( ( Nτ ) ) ) V (S, t) = e rτ E F S J ν e (r κ 1 2 σ2 )τ+σ(w T W t) (16) = k=0 ν=0 (λτ) k ( ( ) ) e (r+λ)τ E F SJ 1 J k e (r κ 1 2 σ2 )τ+σ(w T W t). k! where τ := T t and N τ = N T N t. Hint: for the second line, condition on the different values N τ can take.

8 8 Exercise 2.3. (Merton s formula for European call prices under jump diffusion) Suppose now that the jumps are log-normally distributed: Show that C(S, t; K, T ) = where k=0 J i = e Y i, Y i N(α, β 2 ). (λτ) k ) e (Se λτ κτ+kα+ 1 2 kβ2 Φ(d 1,k ) Ke rτ Φ(d 2,k ), k! d 1,k = log(s/k) + (r σ2 κ)τ + k(α + β 2 ) σ2 τ + kβ 2, and d 2,k = log(s/k) + (r 1 2 σ2 κ)τ + kα σ2 τ + kβ 2, with τ = T t, and κ = λe(j 1) = λ(e α+ 1 2 β2 1). In practice, λτ would usually be small, and one only would take the first few terms of this series (e.g. first two). Hint: apply (16); exploit that SJ 1 J k e (r κ 1 2 σ2 )τ+σ(w T W t) = e (r κ 1 2 σ2 )τ+σ(w T W t)+y 1 + +Y k =: Se R k, with R k a normal random variable whose mean and variance are easily computed as sum of independent normals. To work out the expectations with the call-type pay-off, use that if S, K are positive constants, and if R N(a, b 2 ), then E ( max ( Se R K, 0 )) = Se a+ 1 2 b2 Φ(d 1 ) K Φ(d 2 ), where log(s/k) + a + b2 d 1 =, d 2 = log(s/k) + a = d + b, b b and Φ is the standard normal cumulative distribution function. (This can be shown by similar computations as the ones we used for establishing the Black and Scholes formula.)