# What is the expectation of exponential of the product of two random variables?

I am looking for examples of probability distributions that would allow me to characterize the distribution (at least approximately) and to compute the first two moments exactly of:

$$e^{aXY}$$

where $a$ is a positive real and $X$ and $Y$ are scalar random variables defined on the positive real line and jointly distributed with mean/variance $(\mu_{X},\sigma^2_{X})$ and $(\mu_{Y},\sigma^2_{Y})$ and covariance $\sigma_{XY}$.

Any suggestions very welcome. I'm looking for examples, so hopefully there should be more than one approach to this.

-
I'm not tied to the lognormal distribution, it's just an example that came to mind... Is the exponential-exponential of a normal a base-e-squared lognormal? (or is this last sentence total nonsense) – PatrickT Aug 20 '12 at 23:23
Did you try using en.wikipedia.org/wiki/Moment-generating_function ? – Stat-R Aug 21 '12 at 2:03
Thanks Stat-R. I don't know an awful lot about stats. I have looked at the mgf page. If I understand, it's about expressions of the form $e^{tX}$, where $X$ is a random vector but $t$ is deterministic, so in effect it's an exponential of a sum of rvs, as opposed to an exponential of products. So not sure how to proceed from there... – PatrickT Aug 21 '12 at 17:27
Thanks Macro. I'm only looking for examples where the distribution of $e^{aXY}$ can be characterized, at least approximately, and the first two moments computed explicitly. I have no assumption about $X$ and $Y$, except that they are positive and continuous (not discrete, I forgot to insist on that). I wrote down the first moment when $X$ and $Y$ are LogNormal, but it's messy. But that doesn't mean it can't be solved: writing out the expected value of the Normal is very messy too, until you know what changes of variable to apply. – PatrickT Aug 22 '12 at 18:53
@PatrickT: The mgf of a lognormal does not exist for any positive $t$. – cardinal Aug 23 '12 at 0:33
show 2 more comments

## 1 Answer

I'm entering this as an "answer" to my own question even though it's probably not much of an answer. But hopefully this will help in showing what I'm after. I will edit it or delete it if asked. Edit: This is still work in progress. I should probably mention that Stat-R, cardinal, StasK, and Macro have guided me every step of the way to ... where I have stopped now. Please do feel free to edit/correct this draft.

I'm looking for examples of a joint distributions of the dependent rvs $X$ and $Y$ such that (1) the distribution can be characterized, at least approximately, (2) the following can be computed exactly: $$E[e^{aXY}]$$

Example 1. Let: $$X = x + \epsilon$$ $$Y = y + \epsilon$$ $$\epsilon \sim \mathcal{N}(0,\sigma^{2})$$ where $x$ and $y$ are fixed real numbers (the known means of $X$ and $Y$).

Remark: $$e^{\epsilon} \sim \mathcal{LN}(0,\sigma^{2})$$ $$E[e^{\epsilon}]=e^{\sigma^{2}/2}$$ It follows from a substitution of $x+\epsilon$ and $y+\epsilon$ into $XY$ that: $$E[e^{aXY}] = e^{axy} \times E[e^{a(x+y)\epsilon} e^{a\epsilon^{2}}]$$

Edit: The $\epsilon^{2}$ is not negligeable, as explained in the comments. Thus,

$$E[e^{aXY}] = e^{axy} \times E[e^{a(x+y)\epsilon}] \times E[e^{a\epsilon^{2}}] + e^{2axy} \times COV[e^{a(x+y)\epsilon},e^{a\epsilon^{2}}]$$

The first expectation on the rhs: $$E[e^{a(x+y)\epsilon}] = e^{a^{2}(x+y)^{2}\sigma^{2}/2}$$

The second expectation on the rhs features the square of a Normal, which is a Chi-squared. Edit: I have been shown, in the comments, how to compute the expectation by exploiting the fact that it's an evaluation of the MGF of a chi-squared, since $(\epsilon/\sigma)^{2}\sim\chi_{1}^{2}$. Therefore, $$E[e^{a\epsilon^{2}}] = (1-2a/\sigma^{2})^{-k/2} = 1/\sqrt{1-2a/\sigma^{2}}, \ a < \sigma^{2}/2$$ where $k$ denotes the degrees of freedom, which in this case is $k=1$, as the chi-squared is obtained from $1$ squared Normal. If $a>\sigma^{2}/2$, then the expectation diverges.

Lastly, the covariance term. I think here I need to compute this from the integral, hopefully some known results can be used along the way. If someone knows how to finish this it would be great, but I'm stuck.

-
I have neglected the covariance between $e^{a(x+y)\epsilon}$ and $e^{a\epsilon^{2}}$, which may not be a smart move... – PatrickT Aug 23 '12 at 17:43
${\rm Prob}[\epsilon^2 > \epsilon] = {\rm Prob}[\epsilon^2 > 1] =$ 32%, which is not negligible. – StasK Aug 23 '12 at 19:12
Thanks StasK, yes that was a badly inspired simplification... – PatrickT Aug 23 '12 at 20:36
You should have edited your question with the information rather than put it out as an answer. – Michael Chernick Aug 23 '12 at 20:44
Re: "Not sure how difficult it would be to compute the expectation of the exponential of a chi-square", you could use the moment generating function for that. From wikipedia, if $X$ has a $\chi^2$ distribution with $k$ degrees of freedom, then $m_{X}(a) = E (e^{aX}) = (1 − 2a)^{−k/2}$ for $a < 1/2$. In general, it may help you to look at how various moment generating functions are derived to get a handle on some tricks for calculating $E(e^{aXY})$. Most theoretical statistics texts should have some examples. – Macro Aug 23 '12 at 21:40
show 10 more comments