without computing it, show that $I = \mathbb{E}[e^{XY} |X] \geq 1$

Question

$X, Y$ are two independent $\mathcal{N}(0,1)$ random variables

this question was a follow up question of this one

I honestly don't know if that series of equivalences could be of any help.

my reasoning was as follows :

$ \mathbb{E}[I] = \mathbb{E}[e^{\frac{X^{2}}{2}}] = +\infty$

now let's assume that $I<1$ because $e^{XY} \geq0$ then $0\leq I<1 $ this gives $0\leq \mathbb{E}[I]<1 $

contradiction !

is this it ?

Answer 1

Since $Y$ is standard normal, $E(e^{xY})=e^{x^2/2}$ for every $x$, hence, by independence of $X$ and $Y$, $E(e^{XY}\mid X)=e^{X^2/2}$ almost surely. Thus, $E(e^{XY}\mid X)\geqslant1$ almost surely.

The proof does not use the distribution of $X$, only the distribution of $Y$ and the independence of $X$ and $Y$.

Your approach does not yield a contradiction: you are asked to show that some random variable $Z$ is such that $Z\geqslant1$ almost surely. The negation of this property is not that $Z<1$ almost surely, but that $P(Z<1)\ne0$. And the hypothesis that $P(Z<1)\ne0$ does not imply that $E(Z)<1$.

Edit: If one is forbidden to compute $E(e^{XY}\mid X)$, one can note that, by Jensen convexity inequality, for every $x$, $E(e^{xY})\geqslant e^{xE(Y)}=1$, hence, again by independence, $E(e^{XY}\mid X)\geqslant1$ almost surely.

This uses only that $E(Y)=0$ and the independence of $X$ and $Y$.

Answer 2

$1+E[XY|X]\leq E[e^{XY}|X]$ a.s and $E[XY|X]=XE[Y]=0$ by independence!! So we have the desired inequality without any computing

Answer 3

Invoking the convexity of the exponential function appears to be the simplest proof here. Since the exponential function is convex we have that

$$E[\exp\{XY\} \mid X] \geq \exp\{E(XY \mid X)\} = \exp\{XE(Y)\} = \exp\{X\cdot 0\}=1.$$

We have used the zero mean property of $Y$, and the independence of $X,Y$.