Do we have the following inequality $\operatorname{Var}(X\mid X\in A) \le \operatorname{Var}(X)$? $\newcommand{\Var}{\operatorname{Var}}$Let $X$ be some random veriable. Is the following inequality true
$$
\Var(X\mid X\in A) \le \Var(X)
$$
where $A \subseteq \mathbb{R}$ is some measurable set.  The conditional variance is defined as
$$
E[(X-E[X\mid X\in A])^2\mid X\in A]  = \Var(X\mid X\in A). 
$$
For expected value it is easy to find a counter example. For example, take $X = \pm 1$ equally likely. Then,
\begin{align}
& E[X]=0 \\
& E[X\mid X \in (0,2)]=1
\end{align}
If we do this for variance, we get that
\begin{align}
E[(X-E[X\mid X\in A])^2\mid X\in A]= 0
\end{align}
This inequality appears to be true for the case when $X$ is normal with variance $\sigma$ and $A$ is an interval.
We have that
$$
\Var(X\mid X\in [-t,t])= \sigma^2 \rho(t)
$$
where $\rho(t) \le 1$. See [wiki][1] for more details
I tried to show this but only got to the following inequality
\begin{align}
\Var(X\mid X\in A)&= E[(X-E[X\mid X\in A])^2\mid X\in A]\\
& \le E[(X-E[X])^2\mid X\in A] \text{ use $E[X\mid X\in A]$ is minimizer of quadratic term} \\
& = \frac{E[(X-E[X])^2 1_{A}(X)]}{P(X\in A)}\\
& \le \frac{E[(X-E[X])^2 ]}{P(X\in A)}\\
& = \frac{\Var(X)}{P(X\in A)}\\
\end{align}
[1]: https://en.wikipedia.org/wiki/Truncated_normal_distribution
 A: Counterexample:  $$X \sim \operatorname{Binomial}(n = 3, p = 1/2),$$ then $$\operatorname{Var}[X] = 3/4,$$ but $$Y = (X \mid X \ne \{1, 2\})$$ is a scaled Bernoulli distribution with scaling constant $3$ and $p = 1/2$, thus $$\operatorname{Var}[Y] = 9(1/2)(1-1/2) = \frac{9}{4}.$$
A: I entered some commands into R:
y <- qnorm(seq(10000)/10001)

This sets $y$ equal to the tuple $\left( \Phi^{-1} \left(\tfrac i {10001}\right) : i = 1,\ldots,1000 \right),$ where $\Phi$ is the cumulative probability distribution of the standard normal distribution.
Then:
a <- y[abs(y) > 1]
b <- y[abs(y) < 1]
var(y); var(a); var(b)

The output was this:
[1] 0.9984691
[1] 2.521473
[1] 0.2912116

The first is close to $1,$ as one expects since the variance of the standard normal distribution is $1.$ The second is much bigger.
If one wishes not to conclude that $\operatorname E(Z\mid |Z|>1)> \operatorname E(Z)=1$ on the grounds that what I did is only an approximation, one can still say that a certain discrete distribution that has $10\,000$ possible values, each with probability $1/10\,000,$ satisfies that inequality.
So the proposition is false.
