Truncated variance less than variance? Let the random variable X be distributed (not necessarily normally) with some mean $\mu$ and some variance $\sigma^2$. Under what conditions is it true that
$$ var(X|X<a) \leq var(X)$$ for some $a$ in the support of $X$? I can show it to be true for a normal random variable, and I believe it to be true for the Gumbel and logistic distributions (I just compute the variances in R for some different parameters). It is not true for the gamma distribution. 
We can make assumptions on the support of $X$, or on, say, the log-concavity. But how do I guarantee the statement?
 A: Without loss of generality, and simplicity of calculations, assume $E[X]=\mu=0$. Subtracting a constant from a random variable does not change its variance. Define Y to be the truncated random variable:
$$f_y(y)= b f_x(y)$$ for $y<a$ and zero otherwise, where $b^{-1}=\int_{-\infty}^a f_x(x)dx\le1$. Then,
$$E[Y^2] = \int_{-\infty}^a y^2 b f_x(y)\, dy = b\left(E[X^2]-\int_a^{\infty} y^2 f_x(y)\, dy\right)$$
$$E[Y] = \int_{-\infty}^a y b f_x(y)\, dy = b\left(E[X]-\int_a^{\infty} y f_x(y)\, dy\right)= -b\int_a^{\infty} y f_x(y)\, dy$$
so,
$$V[Y]=E[Y^2]-E[Y]^2=bV[X] - b\int_a^{\infty} y^2 f_x(y)\, dy -b^2\left(\int_a^{\infty} y f_x(y)\, dy\right)^2$$
Define Z to be the random variable:
$$f_z(z)= c f_x(z)$$ for $z>a$ and zero otherwise, where $c^{-1}=\int_a^{\infty} f_x(x)dx = 1-b^{-1}$.
$$V[Y]=bV[X] - {b\over c}E[Z^2] -\left({b\over c}E[Z]\right)^2=bV[X] - (b-1)E[Z^2] -\left((b-1)E[Z]\right)^2$$
$$V[Y]=bV[X] - (b-1)V[Z] -b(b-1)E[Z]^2$$
$$\Delta=V[Y]-V[X]=(b-1)\left(V[X]-V[Z]-bE[Z]^2\right)$$
If $b=1$, then $V[Y]=V[X]$, otherwise $b>1$ in which case, $V[Y] \le V[X]$ iff:
$$V[Z]+bE[Z]^2 \ge V[X]$$
where $Z$ is the random variable with a distribution corresponding to the removed part of $X$ for the case $\mu=0$. For the general case, the condition is:
$$V[Z]+bE[Z-\mu]^2 \ge V[X]$$
A: Here's a much quicker answer than the accepted one:
As in @Dean's solution, without loss of generality assume $E[X]= 0$.
Let

*

*$X^+ = \begin{cases}X : X > a \\ 0 : \text{otherwise} \end{cases}$

*$X^-=\begin{cases}X : X \le a \\ 0 : \text{otherwise} \end{cases}$
Clearly, $X = X^+ + X^-$. Also note that $X^+X^-=0$ (with probability $1$) because they cannot both be nonzero simultaneously.
\begin{align*}
Var[X] &= E[X^2] - \underbrace{E[X]^2}_{=0} \\
       &= E[(X^+ + X^-)^2]\\
       &= E[(X^+)^2) + (X^-)^2 + 2 X^+X^-]\\
       &= E[(X^+)^2)] + E[(X^-)^2] + 2 E[\underbrace{X^+X^-}_{=0}]\\
       &= E[(X^+)^2)] + E[(X^-)^2].
\end{align*}
Now, note
\begin{align*}
Var[X^+] &= E[(X^+)^2] - E[X^+]^2\\
         &\le E[(X^+)^2] \qquad \qquad \hspace{2em} \qquad \qquad \text{(since $E[X^+]^2\ge 0$)}\\
         &\le E[(X^+)^2]+E[(X^-)^2]\qquad \qquad \qquad \text{(since $E[(X^-)^2]\ge 0$)}\\
         &= Var[X].
\end{align*}
