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The question has been edited.

Let $X,Y$ be two positive real-valued random variables with continuous densities. Assume that $\mathbb{P}[X>Y]>0$. Does it always hold that: \begin{equation} \mathbb{E}[X|X<Y] <\mathbb{E}[X] \end{equation} If so, how could I formally prove this result?

andreasvr
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2 Answers2

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No, this is not so. And the reason is: it is possible that $$\operatorname{E}[X|X < Y]>\operatorname{E}[X]$$ As an example, consider random variables $X$ and $Y$ with joint PMF given by

\begin{equation} p_{X,Y}= \begin{cases} 0.5 &\text{if }X=2, Y=3\\ 0.5 &\text{if }X=1, Y=0\\ 0 &\text{otherwise} \end{cases} \end{equation}

Evidently \begin{align} \operatorname{E}[X]&=1.5\\ \operatorname{E}[X|X < Y]&=2 \end{align}

kludg
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  • If I add the assumption that the two densities of X,Y are continuos functions, and that $\mathbb{P}[X>Y]>0$? Does it holds in this case? – andreasvr Aug 13 '20 at 15:34
  • I edited the question based on the above comment – andreasvr Aug 16 '20 at 09:38
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    This example still applies if we make the densities continuous. Instead of saying $P(X=2, Y=3)=.5$, the example could be that $X \sim N(2, .001)$ and $Y \sim N(3, .001)$ with probability $.5$ and still get the same $E[X|X < Y] \approx 2$. Independence and overlapping supports would give you the result you want that $E[X|X<Y] <E[X]$. – Sherwin Lott Aug 16 '20 at 13:51
  • @SherwinLott. Assuming independence and overlapping support, how can you prove the desire conclusion? Could you write this proof down as an answer so that I can flag it and assign to bounty? – andreasvr Aug 17 '20 at 06:21
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For this proof assume $X$ and $Y$ are independent continuous real-valued random variables with overlapping support. Let $f$ and $F$ denote the probability and cumulative density functions respectively, and let $c = \mathbb{P}(X<Y)^{-1}$. $$\begin{align} F_{X|X<Y}(x) &\equiv \mathbb{P}(X \leq x | X < Y) \\ &= c \int_{-\infty}^{x} \int_{x}^{\infty} f_{X}(\hat{x})f_{Y}(\hat{y})d\hat{y} d\hat{x} \\ &= c \int_{-\infty}^{x} f_{X}(\hat{x})(1-F_{Y}(\hat{x})) d\hat{x} \end{align}$$ $$\Rightarrow f_{X|X<Y}(x) = c f_{X}(x)(1-F_{Y}(x))$$

This rescales $f_{X}(x)$ by $c(1-F_{Y}(x))$ which is a decreasing function of $x$ (strictly where the supports overlap). Thus, $X$ strictly stochastically dominates $X|X<Y$ and we are done!

(The following makes rigorous the strict stochastic dominance.)

Claim: $F_{X}(x) \leq F_{X|X<Y}(x)$ for all $x$.

Proof: Suppose not, then there exists a $\hat{x}$ such that $F_{X}(\hat{x}) > F_{X|X<Y}(\hat{x})$, which implies $c(1-F_{Y}(\hat{x})) < 1$ (otherwise $f_{X}(x) \leq f_{X|X<Y}(x)$ for all $x \leq \hat{x}$ contradicting the assumption). Thus, $f_{X}(x) \geq f_{X|X<Y}(x)$ for all $x \geq \hat{x}$ since $c(1-F_{Y}(\hat{x})) < 1$ by monotonicity of $F_{Y}$. However, this implies $1 = F_{X}(\infty)>F_{X|X<Y}(\infty) = 1$, which proves the claim by contradiction.

Claim: $c(1-F_{Y}(x))>1$ for some interval of positive support with respect to $X$.

Proof: Suppose not, then $F_{X}(x) \geq F_{X|X<Y}(x)$ for all $x$ in the support of $X$, so $F_{X}(x) = F_{X|X<Y}(x)$ for all $x$ in the support of $X$. This implies $c(1-F_{Y}(x))=1$ for all $x$ in the support of $X$, which contradicts it strictly decreasing on the overlapping support.

Corollary: $F_{X}(x) < F_{X|X<Y}(x)$ over some interval of positive support with respect to $X$.

Sherwin Lott
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