# Expression for $\mathbb{E}(\exp( ab X Y))$ when $X \sim \text{N}(0, \sigma_X^2)$ and $Y \sim \text{N}(0, \sigma_Y^2)$

Let $$X \sim \text{N}(0, \sigma_X^2)$$ and $$Y \sim \text{N}(0, \sigma_Y^2)$$ be independent normal random variables with zero mean, but (possibly) different variances. Given some constants $$a$$ and $$b$$, I would like to obtain an expression for:

$$\mathbb{E}(\exp( ab X Y)).$$

I am able to solve the univariate case where I only have one of these random variables in the exponential, but I cannot solve this bivariate case.

• It's not sufficient to know the marginal distributions; you need information about their joint distribution. Are they jointly normal? Are they independent? Nov 28 '19 at 23:30

Without loss of generality, the factor $$ab$$ can be omitted as including this factor has the same effect as changing $$\sigma_X^2\sigma_Y^2$$ by a factor of $$a^2b^2$$.

Using the law of total expectation, the definition of the moment generating function, the mgfs of the normal and chi-square distributions, \begin{align} Ee^{XY} &=E(Ee^{XY}|Y) \\&=EM_X(Y) \\&=Ee^{\frac12\sigma_X^2Y^2} \\&=Ee^{\frac12\sigma_X^2\sigma_Y^2\chi_1^2} \\&=M_{\chi_1^2}(\frac12\sigma_X^2\sigma_Y^2) \\&=(1 - \sigma_X^2\sigma_Y^2)^{-1/2}. \end{align} This only works for $$\sigma_X\sigma_Y<1$$ which suggest that the expectation is perhaps not finite when $$\sigma_X\sigma_Y\ge 1$$.

Verifying the result against simulations:

> X <- rnorm(1e+6,sd=.9)
> Y <- rnorm(1e+6,sd=.7)
> mean(exp(X*Y))
 1.284138
> (1-.9^2*.7^2)^(-.5)
 1.287672

• so if I use the parameters a,b it should be like this?: =(1−a^2*b^2*σX^2*σY^2)^(−1/2) where the parameters are squared (sorry I don't know how to edit equation yet) Nov 29 '19 at 3:26
• @Miguel Yes, that's right. Nov 29 '19 at 11:29
• one more question. The result above change if some of the parameters is negative? example: E(exp(-abXY) or does not matter because a is squared in the result? Nov 29 '19 at 16:43
• sorry for ask again. @Jarlee, do yo know how change the result above when the problem takes the form: E(exp(X/Y)), when X is divided by Y? Thanks very much Dec 3 '19 at 2:48
• @MiguelAlegre Neither $X/Y$ nor $\exp(X/Y)$ will have finite expectations, see stats.stackexchange.com/questions/299722/…. Dec 3 '19 at 8:55