Is the kth central moment less than the kth raw moment for even k? 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.
 A: 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 $\mathbb{E}(X^4)\leq \mathbb{E}((X-\mathbb{E}(X))^4)$.
A: 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.
A: 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 $E(X)=1$,
$$E(X^r)-E[(X-1)^r] = \int_{X\ge 0}[x^r-(x-1)^r]dP(\omega)\ge \int_{X\ge 0}(x-1)dP(\omega)=0.$$
