Can we show that $\operatorname E[(X-\operatorname E[X])^k]\le\operatorname E[X^k]$ if $X\ge 0$? If $X$ is a integrable nonnegative random variable and $k\in\mathbb N_0$, are we able to show that $\operatorname E[(X-\operatorname E[X])^k]\le\operatorname E[X^k]$?
That's clearly not true if $X$ might be negative.
 A: The claim is trivial if $k$ is odd. So let us assume that $k \in \{2, 4, 6, \cdots\}$. In such case, define $p(x)$ by
$$ p(x) = x^k - (x-1)^k + 1 - x. $$
We claim that

Claim. $p(x) \geq 0$ for $x \geq 0$.

Before proving this, let us see how this is related to our problem. Write $Y = X/\mathbf{E}[X]$. Then $Y \geq 0$ and $\mathbf{E}[Y] = 1$, and so, the claim tells that $ 0 \leq \mathbf{E}[p(Y)] = \mathbf{E}[Y^k - (Y-1)^k] $, which reduces to  the desired inequality.
Now we turn to proving the claim itself. Assume first that $0 \leq x \leq \frac{1}{2}$. Then


*

*Since $p''(x) = k(k-1)(x^{k-2} - (x-1)^{k-2})$, we easily find that $p$ is concave on $(-\infty, \frac{1}{2}]$.

*By concavity, the graph of $p$ lies above the line joining $(0, p(0)) = (0, 0)$ and $(\frac{1}{2}, p(\frac{1}{2})) = (\frac{1}{2}, \frac{1}{2})$. This line segment is part of the line $y = x$, and so, $p(x) \geq x \geq 0$.

*Again by concavity, $p(x) \leq p(\frac{1}{2}) + p'(\frac{1}{2})(x - \frac{1}{2}) $. We easily check that $p'(\frac{1}{2}) = n2^{2-n} - 1 > -1$, and so, $p(x) \leq \frac{1}{2} - (x - \frac{1}{2}) = 1 - x$.
Combining altogether, we find that $0 \leq p(x) \leq 1$ for $x \in [0, \frac{1}{2}]$. But since $p(x) + p(1-x) = 1$, it follows that $p(x) \geq 0$ for $x \in [0, 1]$. Finally, $p(x) \geq 0$ is easy to check for $x \geq 1$. Therefore the claim follows. $\square$

The following figure is the graph of $p(x)$ for $k = 4, 6, \cdots, 16$ on $[0, 1]$, which numerically confirms that $x \leq p(x) \leq 1-x$ for $[0, \frac{1}{2}]$ and the graph of $p$ is symmetric around $(\frac{1}{2}, \frac{1}{2})$.

