Skewness implications for conditional expected value Suppose the random variable $X$ is continuously distributed over the interval $[0,1]$ with full support. Further, suppose the expected value of $X$ is $E[X]=\mu<1/2$. 
Does this specification already imply that $E[X|x \in [1/2-\epsilon,1/2+\epsilon]]\geq \mu$ for all $\epsilon \in [0,1/2]$? If yes, how would a proof look like?
My intuition is the following: The distribution has to have more mass at some values smaller than $1/2$. Otherwise, the mean $\mu$ would not be smaller than $1/2$. Thus, the distribution is positively skewed. Given this intuition, the implication should always hold as for $\epsilon=0$, $E[X|x =1/2]=1/2> \mu$. Also, for $\epsilon=1/2$, $E[X|x \in [0,1]]=\mu$.
Thank you!
 A: The proposed inequality is false, unless more restrictions are in place. 
To construct a counterexample, one can follow this guideline: to have $\mu < 1/2$, place some heavy mass near-and-below $x = 1/2$. Then let the density dies out towards $x=0$ on the left tail, while having relatively larger right tail. With that, upon conditioning, the right tail sacrifices more than the left, and the mean will shift left. 
As a toy example, consider this piecewise-uniform density (two nonzero columns):
$$f(x) = \begin{cases}
0 & ,~0< x \leq \frac14 \\
3/4 & ,~ \frac14 < x \leq \frac12 \\
1/4 & ,~ \frac12 < x \leq 1
\end{cases}$$ 
The overall mean is conveniently computed as a combination of "column center location times column mass" $$\mu = \frac38 \frac34 + \frac34\frac14 = \frac{15}{32} = 0.46875 < \frac12 \qquad \text{as requested.}$$
Now, for span of $\epsilon = \frac14$, excluding half of the rightmost column, the conditional mass is
$$P_{\epsilon} \equiv \Pr\left\{\frac12-\epsilon < X < \frac12+\epsilon \right\} = \frac34 + \frac12\frac14 = \frac78$$
so that the conditional mean is
\begin{align}
\mu_{\epsilon} \equiv E\left[X \,\middle|\, x \in \Bigl[\frac12-\epsilon,\frac12+\epsilon \Bigr]\right] &= \frac1{P_\epsilon}\left(\frac38 \frac34 + \frac58 \Bigl(\frac12\frac14\Bigr) \right)\\
&= \frac38 \frac67 + \frac58\frac17 = \frac{23}{56} \approx 0.41 < \mu
\end{align}
where $6/7$ and $1/7$ are the heights of the conditional density (also piecewise-uniform).
With some minor efforts one can make the density "smooth" while still achieving the same goal of having $\mu_{\epsilon} < \mu < 1/2$ for at least some $\epsilon > 0$.
