# Prove that E[x^n] >= (EX)^n for n = 2k [duplicate]

Prove that $$E[x^n] \geq (Ex)^n$$ for $$n = 2k$$

I only have the formula of E(x) but I don't know how to prove it.

• Welcome to Cross Validated! Can you use Jensen’s inequality?
– Dave
Feb 7, 2022 at 10:57
• Maybe you can try mathematical induction? Feb 7, 2022 at 11:09
• Please add the self-study tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. Feb 7, 2022 at 13:39
• 1. Prove that all power functions of the type $f(x) = x^n ,n \epsilon N$ are convex functions. 2. Use Jensen’s inequality as pointed out by Dave Feb 7, 2022 at 14:18

First recall that $$\text{Var}(X) = \mathbb{E}[X^2] - \mathbb{E}[X]^2$$ and that $$\text{Var}(X) \geq 0$$.
So $$\mathbb{E}[X^{2k}] = \mathbb{E}[(X^k)^2] = \text{Var}(X^k) + \mathbb{E}[X^k]^2.$$
Since $$\text{Var}(X^k) \geq 0$$, $$\, \, \, \, \mathbb{E}[X^{2k}] \geq \mathbb{E}[X^k]^2$$
Edit: Just realised this doesn't answer your question. This only works when $$n$$ is a power of 2 (by repeating the argument multiple times). It doesn't work for all even $$n$$
• Why does this not work for all even (positive) $n$? Perhaps you are contemplating an inductive argument, but there are other approaches. For instance, $n$ needn't be integral at all: you should be analyzing $|X|$ rather than $X.$ The only reason in the question to write $2k$ was to make it so that $X^{2k}=|X|^{2k}.$