Let $X$, $Y$ be independent bounded integrable random variables and let $a$ be any constant such that $$P(X > Y + a) > 0$$

Is $$E[X | X > Y + a]$$ weakly increasing in $a$?

This thread shows this is true if $X$ and $Y$ are normal, and this thread shows its true if $Y$ is deterministic. I have a counterexample that this is not true for general correlated $X$ and $Y$. If anyone could provide a proof or counterexample I'd be very grateful!

  • $\begingroup$ The word 'bounded' was probably unintentional. Normal random variables, for example, are not bounded. Besides, 'bounded integrable' doesn't make much sense. $\endgroup$ Feb 5, 2019 at 23:40
  • $\begingroup$ Thanks Kavi. What I intended was flexibility in regularity conditions. For example, if you can prove it subject to being able to apply dominated convergence, that would be just fine. Likewise, if you can prove it but only if X,Y have unbounded support, that's also great. $\endgroup$
    – SA28
    Feb 5, 2019 at 23:46

2 Answers 2


I believe I have found a numerical counterexample. Let $X$ be a Beta random variable with parameters (2.3414, .4885) and $Y$ an independent Beta random variable with parameters (2.1760, .5057). Then based on a simulation with 90 million draws, it appears $E[X | X > Y] > E[X | X > Y + 0.1]$.

That said, I figured out that for arbitrary independent $X,Y$ we have

$$E[X | X > Y] \geq E[X].$$

To see this, write

$$E[X | X > Y] = \frac{E[X \textbf{1}(X > Y)]}{P(X > Y)} = \frac{E[X E[\textbf{1}(X > Y)|X]]}{P(X > Y)},$$

and by independence this equals

$$\frac{E[X F_Y(X)]}{P(X > Y)}$$

where $F_Y$ is the marginal CDF of $Y$. Rearranging and noting $E[F_Y(X)] = P(X > Y)$,

$$\frac{E[X F_Y(X)]}{P(X > Y)} =\frac{E[X] E[F_Y(X)]}{P(X > Y)} + \frac{Cov(X , F_Y(X))}{P(X > Y)} = E[X] + \frac{Cov(X , F_Y(X))}{P(X > Y)}$$

Now, since $F_Y(X)$ is an increasing function, it follows that $X$ and $F_Y(X)$ have positive covariance, so the result follows.


If I understand your question correctly, the conjecture is false. Here is a simple counterexample:

  • $X = 1, 2$ with equal probability $1/2$.

  • $Y = 0, 2$ with probabilities $\epsilon, 1-\epsilon$; where $\epsilon$ is a very small positive number (think $\epsilon = 10^{-9}$).

So, conditioned on $X > Y - 0.5$, this allows the sample points $(X,Y) = (2,0), (2,2), (1,0)$. Of these $3$ sample points, $(2,2)$ dominates because of $Y=2$ is much more likely than $Y=0$. So $E[X | X > Y - 0.5] \approx 2$.

Meanwhile, conditioned on $X > Y + 0.5$, this allows the sample points $(X,Y) = (2,0), (1,0)$, and so $E[X | X > Y + 0.5] = (2+1)/2 = 1.5$.


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