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If $X$ is a real-valued random variable, then the $k$th raw moment is $\mathbb{E}[X^k]$, while the $k$th central moment is $\mathbb{E}[(X-\mathbb{E}[X])^k]$. If $k$ is even, is the $k$th central moment always bounded above the $k$th raw moment?

When $k = 2$, then $\mathbb{E}[(X-\mathbb{E}[X])^2] = \mathbb{E}[X^2]-\mathbb{E}[X]^2$, and because $\mathbb{E}[X]^2$ is always positive, it follows that this is less than or equal to $\mathbb{E}[X^2]$. But I'm having trouble extending this to larger moments.

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To complement the accepted answer, we can show that $E[(X - E[X])^m] \leq E[X^m]$ when $X \geq 0$. Assume without loss of generality that $E[X] = 1$ (rescaling $X$ if necessary). Then, omitting the dependence on $\omega$ for notational convenience, \begin{align} & \int_{\Omega} X^m \, d \Omega - \int_{\Omega} (X - 1)^m \, dP(\omega) \\ & \quad = \int_{X \leq 1/2} \left[ X^m - (X - 1)^m \right] \, dP(\omega) + \int_{1/2 < X \leq 1} \left[ X^m - (X - 1)^m \right] \, dP(\omega) \\ & \quad + \int_{X > 1} \left[ X^m - (X - 1)^m \right] \, dP(\omega). \\ & \quad \geq \int_{X \leq 1/2} \left[ X^m - (X - 1)^m \right] \, dP(\omega) + \int_{X > 1} \left[ X^m - (X - 1)^m \right] \, dP(\omega) =: I_1 + I_2. \end{align} For $m$ odd, both $I_1$ and $I_2$ are positive and the statement is proved. We assume from now on that $m$ is even. It is easy to see that: \begin{align} |a^m - (a - 1)^m| \leq (1 - a) \quad & \text{ if } \quad 0 \leq a \leq 1/2; \\ a^m - (a - 1)^m \geq (a - 1) \quad & \text{ if } \quad a > 1. \end{align} Using these inequalities, we deduce that $$ I_1 + I_2 \geq \int_{X > 1} (X - 1) \, dP(\omega) - \int_{X \leq 1/2} (1 - X) \, dP(\omega). $$ To conclude, we note that, by definition of the expected value, \begin{align} & \int_{X \leq 1/2} (X(\omega) - 1) \, dP(\omega) + \int_{1/2 < X \leq 1} (X(\omega) - 1) \, dP(\omega) + \int_{X > 1} (X(\omega) - 1) \, dP(\omega) = 0 \\ & \rightarrow \int_{X > 1} (X(\omega) - 1) \, dP(\omega) \geq \int_{X \leq 1/2} (1 - X(\omega)) \, dP(\omega). \end{align}

EDIT: Ixob Nil's proof in a different answer is a nice simplification of this proof.

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The statement is not true in general and it is easy to construct a counterexample. For example, we want $\mathbb{E}(X^4)\leq \mathbb{E}((X-\mathbb{E}(X))^4)$, we can let $X$ has a small probability of taking a large positive value, while keeping $\mathbb{E}(X)$ negative. Let $X$ be a random variable with $P(X=-2)=1-\epsilon$ and $P(x=M)=\epsilon$. We choose $\epsilon=1/(M+2)$ such that $\mathbb{E}(X)=-1$.

Then we have $$ \mathbb{E}(X^2)=M+2,\quad \mathbb{E}((X-\mathbb{E}(X))^2)=M+1 $$ and $$ \mathbb{E}(X^4)=M^3-2M^2+4M+8,\quad \mathbb{E}((X-\mathbb{E}(X))^4)=M^3+2M^2+2M+1. $$ Obviously, if $M$ is large enough, we have $\mathbb{E}(X^4)\leq \mathbb{E}((X-\mathbb{E}(X))^4)$.

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  • $\begingroup$ Thanks! Is it also easy to construct counterexamples in the case when $X$ is always positive? $\endgroup$
    – user13491
    Commented Dec 28, 2015 at 15:36
  • $\begingroup$ If $X$ is always positive, so is $\mathbb{E}(X)$. As a result $|X-\mathbb{E}(X)|\leq |X|$, and of course $\mathbb{E}\big((X-\mathbb{E}(X))^k\big)\leq \mathbb{E}(X^k)$. $\endgroup$
    – yhhuang
    Commented Dec 28, 2015 at 16:51
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    $\begingroup$ $X-\mathbb{E}[X]$ might be negative, so I don't think it's necessarily true that $|X-\mathbb{E}[X]| \leq |X|$. For example, if $X = 1$ or $X = 9$, with equal probability. $\endgroup$
    – user13491
    Commented Dec 28, 2015 at 23:04
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Roberto Rastapopoulos' proof could be simplified by firstly proving the inequality holds when $a\ge 0$: $$a^m-(a-1)^m\ge a-1;$$

Then similarly for $X\ge0$ with $E(X)=1$, $$E(X^r)-E[(X-1)^r] = \int_{X\ge 0}[x^r-(x-1)^r]dP(\omega)\ge \int_{X\ge 0}(x-1)dP(\omega)=0.$$

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  • $\begingroup$ Much simpler indeed! :) $\endgroup$ Commented Feb 8, 2021 at 19:14

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