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.

  • $\begingroup$ 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) $\endgroup$
    – PatrickT
    Aug 20, 2012 at 23:23
  • $\begingroup$ Did you try using en.wikipedia.org/wiki/Moment-generating_function ? $\endgroup$
    – Stat-R
    Aug 21, 2012 at 2:03
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    $\begingroup$ 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... $\endgroup$
    – PatrickT
    Aug 21, 2012 at 17:27
  • $\begingroup$ 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. $\endgroup$
    – PatrickT
    Aug 22, 2012 at 18:53
  • $\begingroup$ @PatrickT: The mgf of a lognormal does not exist for any positive $t$. $\endgroup$
    – cardinal
    Aug 23, 2012 at 0:33

1 Answer 1


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.

  • $\begingroup$ I have neglected the covariance between $e^{a(x+y)\epsilon}$ and $e^{a\epsilon^{2}}$, which may not be a smart move... $\endgroup$
    – PatrickT
    Aug 23, 2012 at 17:43
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    $\begingroup$ ${\rm Prob}[\epsilon^2 > \epsilon] = {\rm Prob}[\epsilon^2 > 1] =$ 32%, which is not negligible. $\endgroup$
    – StasK
    Aug 23, 2012 at 19:12
  • $\begingroup$ Thanks StasK, yes that was a badly inspired simplification... $\endgroup$
    – PatrickT
    Aug 23, 2012 at 20:36
  • $\begingroup$ You should have edited your question with the information rather than put it out as an answer. $\endgroup$ Aug 23, 2012 at 20:44
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    $\begingroup$ Be careful, this answer contains a mistake. It states that $$ E(e^{a\epsilon^{2}}) = (1-2a/\sigma^{2})^{-k/2} = 1/\sqrt{1-2a/\sigma^{2}}, \ a < \sigma^{2}/2 $$ Let's start with the first comment from Macro. As he points out, it's true that if $X \sim \chi_{k}^{2}$ then $$ E (e^{aX}) = (1 − 2a)^{−k/2} $$ Which for $z \sim \mathcal{N}(0,1)$ means $$ E (e^{az^{2}}) = (1 − 2a)^{−1/2} $$ And for $\epsilon \sim \mathcal{N}(0,\sigma^{2})$ this means $$ E (e^{a\epsilon^{2}}) = E (e^{a\sigma^{2}z^{2}}) = (1 − 2a\sigma^{2})^{−1/2} = 1/\sqrt{1-2a\sigma^{2}}, \ a < \frac{1}{2\sigma^{2}} $$ $\endgroup$
    – Javier
    Mar 6, 2017 at 18:13

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