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Suppose $X$ is a pdf over $[0,m]$ and $Y$ is a binary experiment on $X$ such that $P(Y=1|X)$ is continuous, and we have that $\mathbb{E}[X|Y=1] = \mu_y$ and $\mathbb{E}[X] < \mu_y$. Is it always the case that if $\mathbb{E}[X|X>k] = \mu_y$, $\mathbb{E}[X|X\le k] \le \mathbb{E}[X|Y=0]$? This seems intuitive, but is turning out to be deceptively difficult (for me) to prove.

Any ideas for how to prove this simple result, or intuition/counter examples explaining why it is not true, would be very helpful!

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$E[X] - E[X | Y=0] = \frac{P(Y=1)}{P(Y=0)} (E[X | Y=1] - E[X]) = \frac{P(Y=0)}{1-P(Y=0)}(\mu_y -E[X])$.

Similarly, $E[X] - E[X | X \le k] = \frac{P(X\gt k)}{1-P(X\gt k)} (\mu_y - E[X]).$

So, you need to show that $P(Y=1) \le P(X\gt k)$.

$E[X | X\gt k]$ is the highest possible conditional expected value of $X$ over any event with probability $P(X \gt k)$, and any larger event must have a lower conditional expected value of $X$. If you condition on $Y=1$ and get the same expected value, then $P(Y=1) \le P(X \gt k)$.

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  • $\begingroup$ Thanks for the response. Agreed up to reducing the condition to $P(Y=1) \le P(X > k)$. This is the same strategy I've been trying. Intuitively I agree with your argument in the next paragraph, however, I'm having trouble rigorously proving that. Can you elaborate a bit? Thanks! $\endgroup$
    – armostfalous
    Commented Apr 23, 2013 at 12:31
  • $\begingroup$ Make a conditional quantiles function for each event. The conditional expectation is the average of this quantiles function. Each quantile for $X\gt k$ is at least as large as the corresponding quantile of any other event of the same probability. The difference between the conditional expectations is the average of a nonpositive function. $\endgroup$
    – Douglas Zare
    Commented Apr 23, 2013 at 20:15

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