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.

  • $\begingroup$ It's not sufficient to know the marginal distributions; you need information about their joint distribution. Are they jointly normal? Are they independent? $\endgroup$
    – Glen_b
    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] 1.284138
> (1-.9^2*.7^2)^(-.5)
[1] 1.287672
  • $\begingroup$ 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) $\endgroup$ Nov 29 '19 at 3:26
  • $\begingroup$ @Miguel Yes, that's right. $\endgroup$ Nov 29 '19 at 11:29
  • $\begingroup$ 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? $\endgroup$ Nov 29 '19 at 16:43
  • $\begingroup$ 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 $\endgroup$ Dec 3 '19 at 2:48
  • $\begingroup$ @MiguelAlegre Neither $X/Y$ nor $\exp(X/Y)$ will have finite expectations, see stats.stackexchange.com/questions/299722/…. $\endgroup$ Dec 3 '19 at 8:55

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