How can I show the expected inequality

Question

For any $x, \mu = E[x]$,

it has been mentioned that

$$E\left[ \frac{x^m - \mu^m}{x-\mu}\right] \gt m \mu^{m-1}.$$

How can we say this for all $x$ is somewhat confusing to me. I can understand it to be true for $x > \mu$.

Answer 1

Let $m=4$ and $X:\left(\genfrac{}{}{0pt}{}{-1}{0.5}\,\genfrac{}{}{0pt}{}{0}{0.5}\right)$, i.e. $X$ is $-1$ with probability $\frac12$.

Then $\mu=EX=-\frac12$. So we have: \begin{align} E\left[\frac{X^4-\mu^4}{X-\mu}\right]&=E\left[X^3+\mu X^2+ \mu^2X+\mu^3\right]\\ &=-\frac12+\left(-\frac12\cdot\frac12\right)+\left(\frac14\cdot-\frac12\right)+\left(-\frac18\right)\\ &=-1 < -\frac48 = 4\mu^{3},\end{align}

so the inequality doesn't hold for all $x$ and all $m$, as you suspected.

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However, let $X$ be a positive random variable, and $m\in\mathbb{N}$.

Then by Jensen's inequality, because $f(x)=x^k$ is convex for $x>0$, we would have $E[X^k]>(EX)^k=\mu^k$ for all $k\in\mathbb{N}$. Then, after using the difference of powers formula as before, we have: $$E\left[X^{m-1}+\mu X^{m-2}+\dots+\mu^{m-2}X+\mu^{m-1}\right] >\mu^{m-1}+\dots+\mu^{m-1}=m\mu^{m-1},$$ so the inequality is correct for all positive $x$ and $m\in\mathbb{N}$.

Answer 2

It's certainly not always true. For example, if $m=2$, $$ \frac{x^2-\mu^2}{x-\mu} = x+\mu > 2 \mu^1\ \text{iff}\ x > \mu$$