Consider a random variable $X:\Omega\to[0,1]$ and the transformation to a Bernoulli variable $$ T(X) = \begin{cases}1 &\text{if }X\geq m\\ 0 &\text{otherwise}\end{cases} $$ where $m$ is the median of $X$. Show that ${\it Var}(T(X))\geq {\it Var}(X)$.
Remarks:
Intuitively, the variance should increase when weight is "squeezed to the sides". But it's been tricky to prove.
If $P(X=m)=0$, then ${\it Var}(T(X)) = 1/4 \geq {\it Var}(X)$. But I need the general case where $P(X=m)\not=0$.
Edit: It turns out that this conjecture is false. I posted a more general version here Transformation that increases variance of bounded random variable
The result does not hold.
For a simple counterexample, assume that $P(X=0)=x$, $P(X=x)=\frac12$ and $P(X=1)=\frac12-x$, for some fixed $x$ in $(0,\frac12)$.
The median of $X$ is (unique and) equal to $x$, thus $T(X)$ is Bernoulli with parameter $P(X\geqslant x)=1-x$ and its variance is $x(1-x)$.
On the other hand, $E(X)=\frac12(1-x)$ and $E(X^2)=\frac12(1-x)^2$ hence the variance of $X$ is $\frac14(1-x)^2$.
One sees that $\mathrm{Var}(X)>\mathrm{Var}(T(X))$ for every $x<\frac15$... and that $\mathrm{Var}(X)<\mathrm{Var}(T(X))$ for every $x>\frac15$.