Vtyjkyuk

Let $X,Y$ be two normal random variables such that $BbbE(X)=0,;BbbE(Y)=0,;V(X)=1,;V(Y)=1.$

I would like to prove that for all $s,tinBbbR$ we have $$BbbEbigl(exp(sX-s^2/2)exp(tY-t^2/2)bigr)=exp(stBbbE(XY)).$$

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'.

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 $$(ast)=e^(s+ut)^2/2e^(tv)^2/2e^-s^2/2-t^2/2=e^ust$$ that is, the desired result.

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.