Inequality involving moments of a distribution

Question

Let X be a real random variable. Under what conditions on the distribution do we have that

$$\mathbb{E}( X^{2n + 2}) \geq \mathbb{E}( X^{2n}) \mathbb{E}( X^{2})$$

for all integer $n$?

I tried using the Cauchy Shwartz inequality but it seems I would need to have the expectation values squared. Is there some general inequality you would recommend I start from?

Answer 1

We have, for all positive integer $n$, $$\mathbb{E}(X^{2n + 2}) - \mathbb{E}(X^{2n}) \mathbb{E}(X^{2}) = \frac{1}{2}\mathbb{E}[(X^{2n} - Y^{2n})(X^2 - Y^2)] \ge 0$$ where $Y$ is independent of $X$ and $Y \sim X$, and we have used $$(a^n - b^n)(a - b) = (a - b)^2(a^{n-1} + a^{n-2}b + \cdots + a b^{n - 2} + b^{n-1}) \ge 0$$ for all $a, b \ge 0$.

Answer 2

It is always true for non-negative integers $n$. $EX^{2}\leq (EX^{2n+2})^{2/(2n+2)}$ and $EX^{2n}\leq (EX^{2n+2})^{2n/(2n+2)}$ by Hoder's inequality. Multiply these two to get $EX^{2}EX^{2n} \leq EX^{2n+2}$.