# Trouble understanding jump part in Kou double exponential jump diffusion model

I am trying to work with Kou's double exponential Jump-diffusion model and simulate a price path in a programming language.

So the dynamics of the asset price in Kou's model follow: $$‎\frac{dS(t)}{S(t-)}=\mu‎‏ ‎dt+\sigma ‎dW(‎t)+d(\sum_{i=1}^{N(t)}(V_i-1))‎$$

where W(t) is a standard Brownian motion, N(t) is a Poisson process with rate ‎‎λ , and {Vi} is a sequence of independent identically distributed (i.i.d.) non negative random variables such that Y=log(V) has an asymmetric double exponential distribution with the density: $$f_Y(y)=p.‎\eta_1 e^{-‎\eta_{1}y‎‎}‎\upharpoonleft_{y‎\geq 0‎}+q.‎\eta_2 e^{‎\eta_2 y‎} \upharpoonleft_{y<0},\eta_{1}>1,\eta_{2}>0 ‎‎‎ ‎$$

Solving this SDE gives: $$S(t)=S(0)\exp\{(\mu- \frac{1}{2}\sigma^2)t+\sigma W(t)\} \prod_{i=1}^{N(t)}V_i$$

I generate the Yi-s in a simulation program via the asymmetric double exponential distribution. So let's say I have generated the following four jumps: $$\{12.8277,-14.4736,7.287,-10.1267\}$$

EDIT: I simulate these values with the following Matlab code:

y=binornd(1,p,N,1); %1 = upwards jump, 0  = downwards jump
Y=y.*exprnd(e1,N,1)-(1-y).*exprnd(e2,N,1);


Now the part which I do not get is the following. Because Y = log(V), the Vi-s in the price equation are: $$V_i = e^{Y_i}$$ right?

So when the first jump occurs at time t1, I am adding the jump part in the price equation (the multiplication with Vi). To do so, I take the exponential of 12.8277, but then the stock price explodes (because exp(12.8277)>372).

I think I am mixing things up with the exponential in the equation, because multiplying with the exponential of the generated Yi-s leads to incorrect stock prices.

Could someone explain to me the part which I am interpreting wrong?

• If I understand your question correctly, then the four values are realizations of $Y$? What parameters ($p$, $\eta_1$, $\eta_2$) did you use for generating them? It looks like either you used non-sense parameters or your sampling has a problem. – LocalVolatility Jun 30 '17 at 12:32
• p = 0.4, eta1 = 10 and eta2 = 5 – Peter Lawrence Jun 30 '17 at 13:00
• That means the mean down-jump (in log returns) is -20% and the mean up-jump is +10%. Thus, you'd expect numbers of roughly that magnitude from your sampling. So I suppose your sampling procedure has a problem. – LocalVolatility Jun 30 '17 at 13:03
• I have edited my question with the Matlab code I am using to sample the values. With these parameters, the sample path should still be within reasonable range right? – Peter Lawrence Jun 30 '17 at 13:05
• Yes - it should. See my below answer. – LocalVolatility Jun 30 '17 at 13:07

Your problem is that $\eta$ in your density for $Y$ is the rate parameter of the exponential distribution such that its mean is $1 / \eta$. MATLAB however requires you to provide the mean as mu in the exprnd function. I.e. instead of passing e1 and e2 you should pass their inverses.