Let $X$ be a nonnegative random variable with distribution $F$ and mean $\mu=E(X)>0$. Let $A_{\mu}=[\mu, \infty)$. Is it true that
$$
\int_{A_\mu} x dF(x) \geq \mu/2
$$
must then hold?
I'm trying to find counterexamples, both for continuous or discrete random variables, but I'm not finding any, so I've started suspecting the inequality is actually true. Also, I'm wondering whether there's a quick proof which establishes it, so that I'm maybe missing something stupid. I've tried a proof by contradiction, but I did not get that far.
Edit: The inequality trivially holds if the mean $\mu$ is strictly smaller than the median. Indeed,
$$
\mu = E[X|X\geq \mu] P(X\geq \mu)+E[X|X < \mu] P(X < \mu)
$$
thus
$$
\int_{A_\mu} x dF(x) =E[X|X\geq \mu] P(X\geq \mu)\\
=\mu-E[X|X < \mu] P(X < \mu)\\
\geq \mu\{1-P(X < \mu)\}\\
\geq \mu/2.
$$
But what if the opposite is true, i.e. if the mean is larger than the median?
Best Answer
In general, the inequality does not hold. Consider the following counterexample: $X$ takes value $6/4$ with probability $3/4$ and value $7/2$ with probability $1/4$. In this case, $\mu=2$.
Saying that the inequality does not hold is the same as saying that $$ \int_{A_\mu^c}xdF(x)=E[X|X<2]P(X<2)> \mu/2=1. $$ Clearly, $E[X|X<2]=6/4$, thus the above inequality rewrites $$ P(X<2)> 1/(6/4)=4/6=2/3 $$ which holds true since $P(X<2)=P(X=6/4)=3/4$.