TITLE: $E[|X-\mu|^n] \le 2 E[|X-\mu|^{n+1}]$ for Integer Random Variables
QUESTION [11 upvotes]: I would like to prove the following moment inequality for all integer random variables ($X\in\mathbb Z$) for which the $n+1$th moment is defined:
$$\mathrm{E}[|X-\mu|^n] \le 2 \mathrm E[|X-\mu|^{n+1}].$$
($n\ge1$, but probably doesn't have to be an integer.)
The constant, 2, can't be improved, since it is sharp for the Bernoulli distribution uniform over $\{0,1\}$.
We can assume $\mu=\mathrm E[X]\in[0,1/2]$ by shifting.
I know how to show that at least $\mathrm{E}[|X-\mu|^n] \le C \mathrm E[|X-\mu|^{n+1}]$ for some constant $C>0$, but I wonder how I might get the exact value 2.

REPLY [5 votes]: Proof by induction for integer $n\geq 1.$
Without loss of generality, assume $\mu \in [0, 1)$.
Proof for $n=1$. (basis for induction)
First notice that
\begin{align}
0 = E[X-\mu] &= - \sum_{i = 0}^\infty  p_{-i} (\mu + i) + \sum_{j=1}^\infty p_j (j- \mu)
\end{align}
Thus,
\begin{align}
E[|X-\mu|] &= \sum_{i = 0}^\infty  p_{-i} (\mu + i) + \sum_{j=1}^\infty p_j (j- \mu)\\
& = 2\sum_{i = 0}^\infty  p_{-i} (\mu + i) = 2\sum_{j=1}^\infty p_j (j- \mu).
\end{align}
We have
\begin{align}
E[|X - \mu|^2] &= \sum_{i = 0}^\infty  p_{-i} (\mu + i)^{2} + \sum_{j=1}^\infty  p_j (j-\mu)^{2} \\
& \geq \mu \sum_{i = 0}^\infty  p_{-i} (\mu + i) + (1-\mu) \sum_{j=1}^\infty  p_j (j-\mu)\\
& = \frac{\mu}{2} E[|X-\mu|] + \frac{1-\mu}{2} E[|X-\mu|]\\
& = \frac{1}{2} E[|X-\mu|],
\end{align}
which completes the proof for $n=1$. The inequality is achieved with equality if and only if $X \sim \mathcal{B}(q)$ for some $q \in [0,1)$.
Proof for integer $n\geq 1$. (proof due to @user8675309 in the comments).
The proof proceeds by induction. The base case for $n=1$ is already provided.
Induction hypothesis. Assume that the inequality holds for some integer $n$.
Then,
\begin{equation}
E[|X - \mu|^{n+1}]^2 \leq  E[|X - \mu|^{n}] E[|X - \mu|^{n+2}] \leq 2E[|X - \mu|^{n+1}] E[|X - \mu|^{n+2}],
\end{equation}
completing the proof. The first inequality is due to Cauchy-Schwarz, and the second inequality follows from the induction hypothesis. The first inequality is tight if and only if $|X-\mu|$ is a constant, and the second inequality is achieved if and only if $X \sim \mathcal{B}(q)$ for some $q \in [0,1)$. Putting the two together, for $n>1$, the inequality is strict unless $X \sim \mathcal{B}\left(\frac{1}{2}\right)$ or $X \sim \mathcal{B}\left( 0 \right)$ (i.e., $X=0$). For $n=1$, inequality is strict unless $X \sim \mathcal{B}(q)$ for some $q \in [0,1)$.