Do we have the following inequality $\operatorname{Var}(X\mid X\in A) \le \operatorname{Var}(X)$?

Question

$\newcommand{\Var}{\operatorname{Var}}$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

Answer 1

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 $$\operatorname{Var}[Y] = 9(1/2)(1-1/2) = \frac{9}{4}.$$

Answer 2

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.