Can someone help me finding the expected value of the solution to Merton's jump diffusion model:

\begin{align} S_t &= S_0 \exp \left( \left(r - \frac{\sigma^2}{2} - \lambda k \right) t + \sigma W_t \right) \prod_{j=1}^{N_t} (1+\epsilon_i) \end{align}

where $W_t$ is a BM and $N_t$ is a Poisson process with intensity $\lambda$ and $k$ is the expectation of $\epsilon_i$. The Brownian Motion and Poisson Process are independent.

I know that

\begin{align} E \left[ \exp \left( \left(r - \frac{\sigma^2}{2} \right) t + \sigma W_t \right) \right] = \exp(rt) \end{align}

but what is

\begin{align} E \left[ \prod_{j=1}^{N_t} (1+\epsilon_i) \right] = ? \end{align}

  • $\begingroup$ The product of $N_t$ factors. $\endgroup$
    – Gordon
    Oct 8, 2017 at 19:02
  • $\begingroup$ Can you elaborate a little bit? $\endgroup$
    – Andri
    Oct 8, 2017 at 19:48
  • $\begingroup$ For a given sample $\omega$, $N_t(\omega)$ is an integer. $\endgroup$
    – Gordon
    Oct 8, 2017 at 20:22
  • $\begingroup$ Hint: you can leverage the law of iterated expectations (or total expectations) to deal with the random variable $N_t$. $\endgroup$ Oct 8, 2017 at 21:14
  • $\begingroup$ So if I condition on the number of jumps to be n then there will be n terms in the product and since each jump is independent the expectation will become $(1+k)^n$ ? I know that I am supposed to get $\exp(\lambda k t)$ but how can I get that from $(1+k)^n$ ? $\endgroup$
    – Andri
    Oct 8, 2017 at 21:42

1 Answer 1


Given for all $i$ the mean of $\epsilon_i$ is $k$ and that the $\{\epsilon_i\}_i$ are i.i.d., we have$^{\text{(1)}}$:

$$\begin{align} E\left[\prod_{i=1}^{N_t}(1+\epsilon_i)\right] &=E\left[E\left[\prod_{i=1}^{N_t}(1+\epsilon_i)|N_t\right]\right] \\[6pt] &=E\left[\prod_{i=1}^{N_t}E\left[(1+\epsilon_i)|N_t\right]\right] \\[6pt] &=E\left[\prod_{i=1}^{N_t}(1+k)\right] \\[13pt] &= E\left[(1+k)^{N_t}\right] \end{align}$$

By definition of the expectation and the distributional properties of $N_t$:

$$\begin{align} E\left[(1+k)^{N_t}\right]&=\sum_{n=0}^{\infty}(1+k)^{n}\frac{(\lambda t)^n}{n!}e^{-\lambda t} \\[6pt] &= e^{k\lambda t} \end{align}$$

$\text{(1)}$ Note a subtlety here that got me momentarily confused: the product of $n$ i.d.d. random variables $\epsilon_1,\cdots,\epsilon_n$ with same distribution as $\epsilon$ is not the same as the $\text{n}^{\text{th}}$ power of variable $\epsilon$. Hence, you cannot directly collapse the product $\prod_{1\leq i \leq n}(1+\epsilon_i)$ to the power $(1+\epsilon)^n$, you first need to "inject" the expectation into the product.


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