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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?

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    $\begingroup$ Just a comment: I believe your inequality should hold if and only if the following discrete version holds. Let $(b_k)_{1\le k\le N}$ and $(c_{k,l})_{1\le k,l\le N}$ satisfy $\sum b_k=0$ and $\sum_l c_{k,l}=0$ for each $k$. Then the discrete version is $\sum_{k,l} (b_k+c_{k,l})^{2i}\sum_k b_k^{2j}\le \sum_{k,l}(b_k+c_{k,l})^{2j}\sum_k b_k^{2i}$. $\endgroup$
    – Anthony Quas
    Jul 7, 2015 at 7:15
  • $\begingroup$ One more comment: there is no need to assume that $X$ is of expectation 0: if $X$ is a random variable with arbitrary expectation, then let $Z$ be a random variable that is $\pm 1$ with equal probabilities (independent of $X$) and let $\tilde X=ZX$ and $\tilde Y=ZY$; and $\tilde {\mathcal G}=\mathcal G\vee \mathcal P_Z$. Then $\mathbb E(\tilde X|\tilde G)=\tilde Y$ and $\mathbb E(\tilde X)=0$. Now the inequality for $\tilde X$ and $\tilde Y$ implies the inequality for $X$ and $Y$. $\endgroup$
    – Anthony Quas
    Jul 7, 2015 at 14:30

1 Answer 1

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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 $j$.

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  • $\begingroup$ Thanks Anthony. You're right. This is a bummer for me, I will try to post another conjecture which will help me prove what I want to. $\endgroup$
    – Hedonist
    Jul 7, 2015 at 16:02

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