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.
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2$\begingroup$ Welcome to Cross Validated! Can you use Jensen’s inequality? $\endgroup$– DaveCommented Feb 7, 2022 at 10:57
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$\begingroup$ Maybe you can try mathematical induction? $\endgroup$– toni057Commented Feb 7, 2022 at 11:09
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1$\begingroup$ 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. $\endgroup$– kjetil b halvorsen ♦Commented Feb 7, 2022 at 13:39
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$\begingroup$ 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 $\endgroup$– RomainCommented Feb 7, 2022 at 14:18
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1 Answer
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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 $n$
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1$\begingroup$ 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 $X^{2k}=|X|^{2k}.$ $\endgroup$– whuber ♦Commented Feb 7, 2022 at 16:16