TITLE: Moments of a random variable and of its conditional expectation
QUESTION [7 upvotes]: Let $X$ be a bounded random variable with $\mathbb{E}X=0$. Since $X$ is bounded, all its moments exist. Let $\mathcal{G}$ be any $\sigma$-field and let $Y:=\mathbb{E}[X|\mathcal{G}].$ I am interested in proving the following inequality relating even moments of $X$ with even moments of $Y:$ 
For non-negative integers $i<j,$ we have 
$$\mathbb{E}X^{2i}\cdot\mathbb{E}Y^{2j}\leq \mathbb{E}X^{2j}\cdot\mathbb{E}Y^{2i}.$$
Some special cases:
a) If $i=0,$ then this follows from Jensen’s inequality.
b) If $X$ is equiprobable on $\{-1,+1\},$ then this follows from the fact that $|Y|\leq 1$ almost surely.
Intuitively, I expect this to be true because $Y$ is a smoothed version of $X.$
Does the inequality follow easily from some known results? If not, what tools might help me prove it?

REPLY [4 votes]: So I think this is false. I want to consider the case $i=1$, $j=\infty$ first. Let $Y$ take values in $\{\pm 1,\pm\frac 12\}$ with equal probability of $\frac 14$ each. Now let $X$ take values $\pm 1$ with probability $\frac 14$ each and $\pm \frac 34,\pm\frac 14$ with probability $\frac 18$ each. The $\sigma$-algebra is $\mathcal G=\{\{X=1\},\{X=-1\},\{X\in \{\frac 14,\frac 34\},\{X\in -\frac 14,-\frac 34\}\}$ so that $\mathbb E(X|\mathcal G)=Y$.
Now we have $\mathbb EX^{2i}\mathbb EY^{2j}=\frac 12\mathbb EX^2$ while the right side is $\frac 12\mathbb EY^2$ so that the inequality goes in the opposite direction. However $\mathbb EX^{2j}$ and $\mathbb EY^{2j}$ converge to $\frac 12$ as $j\to\infty$, so that the wrong inequality persists for finite $j$.