Two i.i.d random variables inequality If $X$ and $Y$ are i.i.d positive random variables,


*

*Prove that $\Bbb E(X/Y) \ge 1$: I use Jensen's inequality $\Bbb E[\exp(\log(X/Y))]$ and get the answer. One can also use the A-G inequality to prove this.

*(Difficult) if ${\rm Var}(\log(X))\ge 2$, prove that $\Bbb E(X/Y) \ge 2$.


While the first inequality is not that difficult, the second one is rather hard...
 A: Here is how I worked this when I made my comment.

Since $X$ and $Y$ are identically distributed:
$$
\newcommand{\Var}{\operatorname{Var}}\newcommand{\E}{\operatorname{E}}
\begin{align}
\E(\log(X/Y))
&=\E(\log(X))-\E(\log(Y))\\
&=0
\end{align}
$$
Therefore, since $X$ and $Y$ are independent
$$
\begin{align}
\E\left(\log(X/Y)^2\right)
&=\E\left(\log(X/Y)^2\right)-\E(\log(X/Y))^2\\
&=\Var(\log(X/Y))\\
&=\Var(\log(X)-\log(Y))\\
&=\Var(\log(X))+\Var(\log(Y))\\
&=2\Var(\log(X))
\end{align}
$$
Jensen gives that for $k\ge1$,
$$
\begin{align}
\E\left(\log(X/Y)^{2k}\right)
&\ge\E\!\left(\log(X/Y)^2\right)^k\\
&=\sqrt{2\Var(\log(X))}^{\,2k}
\end{align}
$$
Thus, because $X$ and $Y$ are identically distributed
$$
\begin{align}
\E(X/Y)
&=\tfrac12\left(\E(X/Y)+\E(Y/X)\right)\\[3pt]
&=\E\left(\sum_{k=0}^\infty\frac1{2k!}\log(X/Y)^{2k}\right)\\
&\ge\sum_{k=0}^\infty\frac1{2k!}\sqrt{2\Var(\log(X))}^{\,2k}\\[9pt]
&=\cosh\left(\sqrt{2\Var(\log(X))}\,\right)
\end{align}
$$
A: The second one can be proved in an analogous way. Let us write $U = \log X$ and $V = \log Y$. Then we may write
$$ \Bbb{E}[X/Y] = \Bbb{E}[e^{U-V}] = \frac{1}{2}\Bbb{E}[e^{U-V}] + \frac{1}{2}\Bbb{E}[e^{V-U}] = \Bbb{E}\cosh(U-V). $$
Now let $f(z) = \sum_{n=0}^{\infty} \frac{1}{(2n)!}z^n = \cosh\sqrt{z}$. This function is convex on $[0, \infty)$. Thus by the Jensen's inequality,
$$ \Bbb{E}\cosh(U-V) = \Bbb{E}f((U-V)^2) \geq f( \Bbb{E}[(U-V)^2] ). $$
Simplifying the last expression gives $f(2\operatorname{var}(U))$ and therefore
$$ \Bbb{E}[X/Y] \geq \cosh\sqrt{2\smash[b]{\operatorname{var}(\log X)}} $$
as @robjohn commented.
