Inequality with conditional expectation of positive random variables

Question

Assume $X$ and $Y$ are a.s. positive random variables. Moreover, $X \le Y$ a.s.

It seems to me that this inequality is not always true: $$ \mathbb{E}[X ~|~ X > a] \mathbb{P}(X > a) \le \mathbb{E}[Y ~|~ Y > a) \mathbb{P}(Y > a) $$ and I look for the conditions on $X$ and $Y$ under which the above inequality holds.

Answer 1

Your original inequality holds all the time, but your reasoning in the main comment thread is wrong.

Claim: $E[X \mid X > a] P(X > a) \le E[Y \mid Y > a] P(Y > a)$

Proof: For any event $A$, let $1_{A}$ denote its ($0,1$-valued) indicator random variable. Then

• $LHS = E[X \cdot 1_{X>a}]$

• $RHS = E[Y \cdot 1_{Y>a}]$

• $Y \ge X$ a.s. implies whenever $X>a$ then $Y>a$ a.s., therefore $1_{Y>a} \ge 1_{X>a}$ a.s.

• $Y \ge X > 0$ a.s. and $1_{Y>a} \ge 1_{X>a}\ge 0$ a.s. $\implies Y \cdot 1_{Y>a} \ge X \cdot 1_{X>a}$ a.s. $\implies RHS \ge LHS$

Claim: $E[X \mid X > a] \le E[Y \mid Y > a] $ can be false

Counter-example: $X$ takes value $1$ or $100$ with equal prob $1/2$ each, and let $Y=X+2$ (surely). Thus, $Y$ takes value $3$ or $102$ with equal prob $1/2$ each. Conditioned on each variable being $> a = 2$, we have:

• $E[X \mid X > 2] = 100$ since $X=1$ is ruled out

• $E[Y \mid Y > 2] = {3 + 102 \over 2} < 100$