Show that $\Bbb{E}(e^{sX-s^2/2+tY-t^2/2})=e^{st\Bbb{E}(XY)}$ for standard jointly normal random variables $(X,Y)$

Question

Let $X,Y$ be two normal random variables such that $\Bbb{E}(X)=0,\;\Bbb{E}(Y)=0,\;V(X)=1,\;V(Y)=1.$

I would like to prove that for all $s,t\in\Bbb{R}$ we have $$\Bbb{E}\bigl(\exp(sX-s^2/2)\exp(tY-t^2/2)\bigr)=\exp(st\Bbb{E}(XY)).$$

As $X,Y$ are not independent I am not sure how can I tackle this problem. For exemple, I don't know the density $f_{XY}.$

EDIT: I have badly translated the problem (english to french). Indeed the assumption is 'with joint Gaussian distribution'.

Answer 1

Assuming furthermore that $(X,Y)$ is jointly normal, one knows that there exists some standard normal random variable $Z$ independent of $X$ such that $Y=uX+vZ$ with $u=E(XY)$ and $u^2+v^2=1$.

Then, by independence of $X$ and $Z$, the desired expectation $$(\ast):=E(e^{sX-s^2/2+tY-t^2/2})$$ is $$(\ast)=E(e^{sX-s^2/2+t(uX+vZ)-t^2/2})=E(e^{(s+tu)X})E(e^{tvZ})e^{-s^2/2-t^2/2}$$ Using the identity $$E(e^{aX})=E(e^{aZ})=e^{a^2/2}$$ for every $a$, one sees that $$(\ast)=e^{(s+ut)^2/2}e^{(tv)^2/2}e^{-s^2/2-t^2/2}=e^{ust}$$ that is, the desired result.

Answer 2

This is false. Without some assumption on joint distribution we cannot prove this. In fact, the stated equation gives a formula for $Ee^{sX+tY}$ and using this we can easily see that $(X,Y)$ has a 2-dimensional normal distribution. This is absurd since the fact that marginals are normal does not imply that the joint distribution is normal.