I would appreciate a hint regarding the following question (taken from Durret, Essentials of Stochastic Processes, questions 2.38

"Let $S_t$ be the price of stock at time t and suppose that at times of a Poisson process with rate  the price is multiplied by a random variable $X_i$ > 0 with mean $\mu$ and variance $\sigma^2$. That is,

$S_t = S_0\prod_{i=1}^{N(t)}X_i$

where the product is 1 if $N(t) = 0$. Find ES(t)/ and varS(t).

Is this an example of thinning?

  • $\begingroup$ No. This has nothing to do with thinning. You need to take conditional expectation with respect to N(t) to get the mean and conditional variance formula to get the variance. $\endgroup$ – Lost1 Feb 27 '14 at 0:37

You start by the conditional expectation with respect to $N_t$.

$E[S_t|N_t] = S_0\prod_{i=1}^{N(t)} EX_i$ with the additional hypothesis that the process $N$ and the $X_i$s are independent, so

$$E[S_t|N_t] = S_0 \mu^{N(t)} $$

Let $r$ be the rate of the Poisson process, we have $$E[S_t] = EE[S_t|N_t] = S_0 \exp(-rt)\sum_{k=0}^\infty \mu^{k}\frac{(rt)^k}{k!} = S_0 \exp(rt(\mu-1))$$

We also have via the same computation: $$E[S_t^2] = E[S_0^2 (\sigma^2 + \mu^2)^N_t] = S_0^2 \exp(rt(\sigma^2 + \mu^2 - 1))$$

so $Var[S_t] = S_0^2 (\exp(rt(\sigma^2 + \mu^2 - 1)) - \exp(2rt(\mu-1)))$

One can actually compute the characteristic function of such a process (in terms of the characteristic function of the $X_i$s), this is a simple case of the Lévy–Khintchin representation.

  • $\begingroup$ Hello, I downvoted your answer. This is not because I think it is a bad answer, but on this site, it is the norm that people do not give answers to questions which shows no attempt. This, imo, can easily be a homework question and the OP showed no attempt at it. This is why I left a comment below the question on something to look at, but I did not post answer myself. I hope you can understand. $\endgroup$ – Lost1 Feb 27 '14 at 0:49
  • $\begingroup$ Sure. Thank you for your comment. $\endgroup$ – mookid Feb 27 '14 at 1:25
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    $\begingroup$ @Lost1 shouldn't you actually downvote the question, then? The answer is useful to others who will eventually reach this page when searching for help in similar problems. $\endgroup$ – Andre Terra Jan 28 '16 at 23:34
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    $\begingroup$ @AndréTerra that goes without saying... $\endgroup$ – Lost1 Jan 29 '16 at 9:26
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    $\begingroup$ Why do you have that $E[S_{t}^{2}] = E[S_{0}^{2}(\sigma^{2} + \mu^{2})^{N(t)}$ and not $E[S_{t}^{2}] = E[(S_{0}\mu^{N(t)})^{2}] = E[(S_{0}^{2}\mu^{2N(t)})]$? $\endgroup$ – sir_thursday Mar 14 '16 at 21:12

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