Variance of the minimum between a constant and a random variable Let $X$ be a random variable, and $c$ a constant.
How can we prove that:
$var[min(X,c)]\le var[X]$?
 A: There are two trivial cases: if $Pr(X \le c)=1$ then you have equality, while if $Pr(X \ge c)=1$ you have $)$ less than or equal to a non-negative variance.  Otherwise $c$ is within the interval that supports $X$.
I am going to let $Y=c-X$ and then prove the equivalent statement $var[\max(Y,0)]\le var[Y]$.  It is not necessary, but it makes what is non-negative or negative more obvious and reduces the the number of terms.  Let:


*

*$p_1=\Pr(Y \ge 0)$, which is non-negative

*$p_2=\Pr(Y \lt 0)$, which is non-negative and $p_2=1-p_1$

*$\mu_1 = E[Y | Y \ge 0]$, which is non-negative

*$\mu_2 = E[Y | Y \lt 0]$, which is non-positive

*$s_1^2 = E[Y^2 | Y \ge 0]$, which is non-negative and $s_1^2 \ge \mu_1^2$ from the conditional variance

*$s_1^2 = E[Y^2 | Y \lt 0]$, which is non-negative and $s_2^2 \ge \mu_2^2$ from the conditional variance

*$Z=\max(Y,0)$, which is non-negative


Then:


*

*$E[Y]=p_1\mu_1+p_2\mu_2$

*$E[Z]=p_1\mu_1$

*$E[Y^2]=p_1 s_1^2+p_2 s_2^2$ 

*$E[Z^2]=p_1 s_1^2$ 

*$var[Y]=p_1 s_1^2+p_2 s_2^2 - p_1^2\mu_1^2-2p_1 p_2\mu_1\mu_2-p_2^2\mu_2^2$

*$var[Z]=p_1 s_1^2- p_1^2\mu_1^2$

*$var[Y]-var[Z] = p_2 s_2^2 -2p_1 p_2\mu_1\mu_2-p_2^2\mu_2^2 = p_2(s_2^2-\mu_2^2)-2p_1 p_2\mu_1\mu_2 + p_1 p_2\mu_2^2$


and we know that $p_2(s_2^2-\mu_2^2)\ge 0$, that $-2p_1 p_2\mu_1\mu_2 \ge 0$, and that $p_1 p_2\mu_2^2 \ge 0$, 
so $var[Y]-var[Z] \ge 0$, i.e. $var[\max(Y,0)]\le var[Y]$ and thus $var[\min(X,c)]\le var[X]$.
