Is the truncated squared expected value less than the variance? Let the continuous random variable $X$ be distributed with mean $\mu_x$ and variance $\sigma^2_x$ with support $[0, \infty]$.
Let the random variable $Y$ be the right truncation of $X$, with truncation point $a$.
One can find conditons for when $\mathtt{Var}[Y] \le \mathtt{Var}[X]$.
Is it true that $\mathbf{E}[(Y-\mu_x)^2] \le \mathbf{E}[(X-\mu_x)^2]= \mathtt{Var}[X] =\sigma^2_x$? How would one show this?
Expressed differently, if $f_X(x)$ is the probability distribution of $X$, then the probability distribution of $Y$ is $f_Y(x)= b f_X(x)$ where
$$
b=\frac{1}{\int_0^{a}f_X(x) dx} \ge 1.
$$
Can one find conditions for which
$$
 \int_{0}^{a} (x-\mu_x)^2 b f_{X}(x) dx  \le 
         \int_{0}^{a} (x-\mu_x)^2 f_{X}(x) dx + \int_{a}^{\infty} (x-\mu_x)^2 f_{X}(x) dx?
$$

It is not true in general.
Let it be that $\mu_x>a$.
Then $Y(\omega)\leq a$ assures that $(Y(\omega)-\mu_x)^2\geq(\mu_x-a)^2$ for every $\omega\in\Omega$.
Consequently $\mathbb E(Y-\mu_x)^2\geq(\mu_x-a)^2$.
However, nothing prevents us to choose $X$ is such a way that $\mathsf{Var}(X)<(\mu_x-a)^2$.
