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Choosing Exponential Rate Parameters in Kou Jump Simulations

Article Quant Q&A · Author: Peter Lawrence

Summary

The document clarifies a parameterization issue when simulating jumps in Kou’s double exponential jump diffusion model. The log jump size is drawn from an asymmetric mixture of positive and negative exponential distributions, then exponentiated to obtain the multiplicative price jump. The questioner worries that large simulated log jumps cause implausibly large price moves and suspects the exponential transformation is being applied incorrectly.

The accepted explanation identifies the mismatch: the model’s eta parameters are exponential rates, whose corresponding means are their reciprocals, while MATLAB’s exponential random-number function expects a mean parameter. The proposed correction is therefore to pass the inverse eta values when generating the positive and negative exponential components. The note addresses this implementation detail, not the choice or calibration of jump parameters, simulation stability, or risk-neutral drift adjustments.

Key ideas

  • Kou’s model represents log jump sizes with a mixture of positive and negative exponential distributions.
  • Exponentiating a log jump converts it to the multiplicative jump factor in the price process.
  • The model’s eta values are exponential rate parameters, not means.
  • MATLAB’s exponential generator expects a mean, so the input should use the reciprocal of each eta.
  • Correct parameterization resolves this simulation issue but does not determine calibrated model parameters.

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


# 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: \begin{equation} ‎\frac{dS(t)}{S(t-)}=\mu‎‏ ‎dt+\sigma ‎dW(‎t)+d(\sum_{i=1}^{N(t)}(V_i-1))‎ \end{equation}

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: \begin{equation} 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 ‎‎‎ ‎\end{equation}

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

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: \begin{equation} \{12.8277,-14.4736,7.287,-10.1267\} \end{equation}

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: \begin{equation} V_i = e^{Y_i} \end{equation} 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?

## Answer by LocalVolatility (score 0, accepted)

https://quant.stackexchange.com/a/34931

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.

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