Is the following property for positive random variables fulfilled in general? Suppose we have a continuous random variable $X$, defined on the interval $[0, \infty)$, which has density $f(x)$ and a finite expectation and variance.
I am wondering whether the following is true in general:
Take any value $\alpha > 0$, and let
$x_1 := \mathbb{E}(X|X \leq \alpha)$, while
$x_2 := \mathbb{E}(X|X > \alpha)$.
Will a new, discrete random variable $\tilde{X}$, where
$\tilde{X} = x_1$ with probability $F(\alpha)$, and $\tilde{X} = x_2$ with probability $1-F(\alpha)$, always have a lower variance than $X$?
I have checked that this property is true for the uniform distribution. Formally, it is straightforward to see that the property is fulfilled if for all $\alpha > 0$,
$\int_{0}^{\infty} x^2 f(x)dx \geq \frac{\left(\int_0^\alpha x f(x) dx\right)^2}{F(\alpha)} + \frac{\left(\int_\alpha^\infty f(x) dx\right)^2}{1-F(\alpha)}$.
Using Jensen's inequality, a sufficient condition for the above is that
$\left(\int_{0}^{\infty} x f(x)dx\right)^2 \geq \frac{\left(\int_0^\alpha x f(x) dx\right)^2}{F(\alpha)} + \frac{\left(\int_\alpha^\infty f(x) dx\right)^2}{1-F(\alpha)}$.
However, I am not really sure where to go from there. The integral in the LHS can still be split up into the range $0$ to $\alpha$ and $\alpha$ to $\infty$, but doesn't seem to help me much.
Maybe the property is not true in general, after all?
Any input is appreciated. Many thanks in advance!
 A: 
The property does hold, in great generality.

Let $X$ denote any (integrable) real valued random variable and $a$ some real number such that $A=[X\leqslant a]$ and $A^c=[X\gt a]$ both have positive probability. Let $X_a$ denote any random variable such that $X_a=E[X\mid A]$ with probability $P[A]$ and $X_a=E[X\mid A^c]$ with probability $P[A^c]=1-P[A]$. Then $E[X_a]=E[X]$ and 
$$
E[X_a^2]=E[X:A]^2P[A]^{-1}+E[X:A^c]^2P[A^c]^{-1}.
$$
For every event $B$, Cauchy-Schwarz inequality applied to $(X\mathbf 1_B)\cdot\mathbf 1_B$ yields
$$
E[X:B]^2=E[(X\mathbf 1_B)\cdot\mathbf 1_B]^2\leqslant E[(X\mathbf 1_B)^2]E[\mathbf 1_B^2]=E[X^2:B]P[B],
$$
hence, for every event $B$ with positive probability,
$$
E[X:B]^2P[B]^{-1}\leqslant E[X^2:B].
$$
Applying this to $B=A$ and to $B=A^c$ and summing the resulting upper bounds yields
$$
E[X_a^2]\leqslant E[X^2:A]+E[X^2:A^c]=E[X^2].
$$
Finally, since $E[X_a]=E[X]$, 

$$\mathrm{var}(X_a)\leqslant\mathrm{var}(X).$$

Although the above fully proves the desired result, it may help to put it in a more general context. Let $\mathcal A=\{\varnothing,A,A^c,\Omega\}$ denote the sigma-algebra generated by $A$, then $Y=E[X\mid\mathcal A]$ is an example of a random variable $X_a$, since $Y=E[X\mid A]\mathbf 1_A+E[X\mid A^c]\mathbf 1_{A^c}$. It happens that, for every sigma-algebra $\mathcal G$, $Z=E[X\mid\mathcal G]$ is such that $E[Z]=E[X]$ and $Z^2\leqslant E[X^2\mid\mathcal G]$. Integrating the last inequality (called conditional Cauchy-Schwarz inequality), one gets $E[Z^2]\leqslant E[E[X^2\mid\mathcal G]]=E[X^2]$. Thus, $E[Z]=E[X]$ and $E[Z^2]\leqslant E[X^2]$, which proves the general result that 
$$
\mathrm{var}(E[X\mid\mathcal G])\leqslant\mathrm{var}(X).
$$
