Is $Var(x | x \le \tau)$ weakly increasing in $\tau$? I am interested in $Var(x | x\le \tau)$ is increasing in $\tau$, where $x$ is some random variable with differentiable cdf. I can show that $E[x | x\le \tau]$ is increasing in $\tau$, which is intuitively obvious, by the following computation:
$$
\frac{\partial}{\partial\tau}E[x\mid x\le\tau]=\frac{\partial}{\partial\tau}\left(\int_{-\infty}^{\tau}\frac{xf(x)}{F(\tau)}dx\right)=\frac{f(\tau)}{F(\tau)}(\tau-E[x\mid x\le\tau]).
$$
In a similar vein, I tried the following:
$$
\begin{align*}
\frac{\partial}{\partial\tau}Var[x\mid x\le\tau] & =\frac{\partial}{\partial\tau}\left(E[x^{2}\mid x\le\tau]-E[x\mid x\le\tau]^{2}\right)\\
 & =\frac{\partial}{\partial\tau}\left(\int_{-\infty}^{\tau}\frac{x^{2}f(x)}{F(\tau)}dx-(\int_{-\infty}^{\tau}\frac{x f(x)}{F(\tau)}dx)^{2}\right)\\
 & =\frac{f(\tau)}{F(\tau)}\left[\tau^{2}-E[x^{2}\mid x\le\tau]-2(E[x\mid x\le\tau]-\tau)\right]
\end{align*}
$$
Is this correct? I have a doubt because (i) a simulation result does not match with the analytical formula I have here (although the derivative of $E[x\mid x\le \tau]$ is verified by a simulation) and (ii) it is not clear if $Var[x\mid x\le \tau]$ is increasing in $\tau$ from the result, although a bunch of simulation suggests it is increasing.
If $Var(x\mid x\le \tau)$ is not increasing in general, under what conditions are they increasing? For example, what if $x$ is supported on positive values?
 A: I doubt it is possible for the conditional variance $\text{Var}(X \mid X \le \tau) $ to be strictly decreasing for all $\tau$ but it is certainly possible to be weakly decreasing for all $\tau$ and strictly decreasing for some $\tau$.
An alternative formulation is to take $Y=-X$ and $t=-\tau$ and ask about  $\text{Var}(Y \mid Y \ge t)$ as a function of $t$.
If $Y$ has an exponential distribution with rate $\lambda$ then $\text{Var}(Y \mid Y \ge t)$ is constant $\frac{1}{\lambda^2}$ for all $t$: this is a direct consequence of the memoryless property of the exponential distribution.  So some adjustment to $Y$ should get the desired property.
One such is $Z=Y^2$, for example setting $\lambda=1$ and with CDF $F_Z(z) = 1-e^{-\sqrt{z}}$ and density $f_Z(z) = \frac{e^{-\sqrt{z}}}{2\sqrt{z}}$ when $z \ge 0$. You then get

*

*$\mathbb E[Z \mid Z \ge t] = t+2 \sqrt{t}+2$

*$\text{Var}(Z \mid Z \ge t) = 4t+16 \sqrt{t}+20$
when $t \ge 0$, while $\mathbb E[Z \mid Z \ge t] = \mathbb E[Z]=2$ and  $\text{Var}(Z \mid Z \ge t) = \text{Var}(Z)=20$ when $t \le 0$.  These are increasing functions of $t$.
We just reverse the signs to answer your question. If $X$ has CDF $F_X(x) = e^{-\sqrt{-x}}$ and density $f_X(x) = \frac{e^{-\sqrt{-x}}}{2\sqrt{-x}}$ for $x \le 0$ then $\text{Var}(X \mid X \le \tau) = -4\tau+16 \sqrt{-\tau}+20$ for $\tau\le 0$, which is a decreasing function of $\tau$.
If you want want an example to be strictly decreasing for some positive $\tau$, just shift $X$, for example with CDF $F_X(x) = e^{-\sqrt{5-x}}$ when $x \le 5$ the conditional variance will still be decreasing, and strictly decreasing for $\tau \le 5$.
