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!
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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 $E[X | X > Y + 0.5] = (2+1)/2 = 1.5$.