$E[|X-\mu|^n] \le 2 E[|X-\mu|^{n+1}]$ for Integer Random Variables

Question

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.

Answer 1

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 $q \in [0,1)$.

Answer 2

Here is a proof: unfortunately, I had to do a distinction of cases...

Let us assume wlog, as stated in the OP, that $\mu \in [0,1/2]$. First, observe that the desired inequality is equivalent to $$ \mathbb{E}[ f_n(X-\mu) ] \geq 0, \qquad n \geq 1 \tag{1} $$ where $f_n(x) = |x|^n(|x|-1/2)$. Now, by our assumption on $X$ (integer-valued) and $\mu\in[0,1/2]$, $f_n(X-\mu)$ is only negative for $X=0$.

• Case 1: Suppose $1-p_0 \leq \mu$. Then, we can rewrite $$\begin{align*} \mathbb{E}[ f_n(X-\mu) ] &= p_0\cdot f_n(\mu) + (1-p_0) \sum_{k\neq 0} \frac{p_k}{1-p_0} f_n(k-\mu)\\ &= p_0\cdot f_n(\mu) + (1-p_0) \sum_{k\neq 0} \frac{p_k}{1-p_0} g_n(k-\mu) \end{align*}$$ where we define $g_n(x)= |x|^n(|x|-1/2)_+\geq 0$ as the "corrected" version of $f_n$, such that $g_n$ is convex and $$ g_n(x) = \begin{cases} 0 & \text{ if } |x| \leq 1/2\\ f_n(x) & \text{ otherwise.} \end{cases} $$ By Jensen's inequality, we then have, letting $\nu := \sum_{k\neq 0} \frac{p_k}{1-p_0}\cdot k$, $$ \mathbb{E}[ f_n(X-\mu) ] \geq p_0\cdot f_n(\mu) + (1-p_0) \cdot g_n(\nu -\mu) \geq p_0 f_n(\mu) + (1-p_0) f_n(\nu -\mu) \tag{2} $$ the second inequality since $g_n\geq f_n$ by construction. Note that $(1-p_0)\nu = \mu$; thus, we just showed that the extremal case is for the two-point setting, i.e., letting $Y\sim\mathrm{Bern}(1-p_0)$ $$ \mathbb{E}[ f_n(X-\mu) ] \geq \mathbb{E}[ f_n(\nu Y-\mu) ] \tag{3} $$ for which is is relatively easy to check that the inequality holds, using that from our assumption we have $\nu \geq 1$. (The idea here is that this is necessary, as for $\nu < 1$ we would drop in in our reduction the initial assumption that $X$ cannot take values in $(0,1)$, which is crucial.) To see why, recalling that $\nu-\mu = \frac{\mu p_0}{1-p_0} \geq p_0 \geq 1-\mu \geq \frac12$, $$ \begin{align} \mathbb{E}[ f_n(\nu Y-\mu) ] &= p_0 \mu^n\left(\mu-\frac12\right) + (1-p_0) \mu^n\left( \frac{p_0}{1-p_0} \right)^n\left(\frac{\mu p_0}{1-p_0}-\frac12\right) \\ &\geq p_0 \mu^n\left(\mu-\frac12\right) + (1-p_0) \mu^n\left( \frac{p_0}{1-p_0} \right)^n\left(\frac12 - \mu\right) \\ &= \mu^n p_0 \left(\frac12 - \mu\right) \left( \left( \frac{p_0}{1-p_0} \right)^{n-1} - 1\right) \geq 0 \end{align} $$ the last inequality recalling that $p_0\geq 1/2$ from our assumption.

• Case 2: Suppose $1-p_0 > \mu$. Then, since $f_n$ is even, and increasing on $[1/2, \infty)$, $f_n(k-\mu) \geq f_n(1-\mu)$ for all $k\neq 0$, so that we have $$\begin{align*} \mathbb{E}[ f_n(X-\mu) ] &= p_0\cdot f_n(\mu) + \sum_{k\neq 0} p_k\cdot f_n(k-\mu)\\ &\geq p_0\cdot f_n(\mu) + (1-p_0) \cdot f_n(1-\mu) \\ &= p_0 \mu^n (\mu-\tfrac12) + (1-p_0) (1-\mu)^n (\tfrac12-\mu) \\ &= (\tfrac12-\mu)( (1-p_0) (1-\mu)^n - p_0\mu^n )\\ &\geq (\tfrac12-\mu)\mu (1-\mu)( (1-\mu)^{n-1} - \mu^{n-1} ) \tag{as $1-p_0 > \mu$}\\ &\geq 0 \end{align*}$$ the last inequality since $\mu \leq 1/2$. $\;\;\square$