Is it true that $\mathbb{E}[X^n] > \mathbb{E}[X^{n-1}]\mathbb{E}[X]$ for all $n \geq 2$?

Question

Let $X$ denote a random variable with a smooth distribution over $[0, 1]$. Is it true that

$$\mathbb{E}[X^n] > \mathbb{E}[X^{n-1}]\mathbb{E}[X]$$

for all integers $n = 2, 3, ...$? This seems to be a simple application of Jensen's inequality: however, the exact proof eludes me!

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My thoughts so far: In the case of $n = 2$, we know that

$$\mathbb{E}[X^2] > \mathbb{E}[X]\mathbb{E}[X] = \mathbb{E}[X]^2$$

by Jensen's inequality (since the function $f(x) = x^2$ is strictly convex on $[0, 1]$).

Similarly, Jensen's inequality tells us that

$$\mathbb{E}[X^n] > \mathbb{E}[X]^{n} = \mathbb{E}[X]^{n-1}\mathbb{E}[X]$$

but that is, unfortunately, not quite the inequality I am after.

Finally, in the case where $X$ is uniformly distributed, one can calculate that

$$ \mathbb{E}[X^n] = \frac{1}{n+1}$$

and similarly

$$ \mathbb{E}[X^{n-1}] = \frac{1}{n}$$

$$ \mathbb{E}[X] = \frac{1}{2}$$

which allows one to verify that

$$\mathbb{E}[X^n] = \frac{1}{n+1} > \mathbb{E}[X^{n-1}]\mathbb{E}[X] = \frac{1}{2n}$$

for all $n > 1$. So this inequality holds in the uniform case, at least.

Answer 1

By Hölder's inequality, for any random variable $X$ for which $\mathbb E[|X|^n]$ exists we have

$$ \mathbb E[|X|^{n-1}] \le (\mathbb E[|X|^n])^{(n-1)/n} $$ and $$ \mathbb E[|X|] \le (\mathbb E[|X|^n])^{1/n} $$ The product of these is $$ \mathbb E[|X|^{n-1}]\; \mathbb E[|X|] \le \mathbb E[|X|^n]$$

Moreover, equality holds iff $|X|$ is almost surely equal to a constant.