Is $\mathbb{E}(X|X \geq Y) \geq \mathbb{E}(X|X = Y)$ necessarily? $X$ and $Y$ are independent random variables, each having the same, finite, support. In particular, $Y$ is the max of $n$ i.i.d. random variables, each independent of $X$ and following the same distribution as $X$. That is, if $f(x):=Pr(X=x)$, the CDF of $Y$ is given by $Pr(Y \leq y)= F(y)^n$.
My question is, is it always true that $\mathbb{E}(X|X \geq Y) \geq \mathbb{E}(X|X = Y)?$
At the first glance it seems like this should be obviously true, not only for this particular case but for any two independent RVs $X$ and $Y$?
But when I tried to prove it I found that it is clearly not "obvious". And in this particular case - which is what I have to prove - I have played around with the formulas a lot, without being able to prove this!
Any help is most appreciated.
 A: Let $n=1$ and
$$X,Y\sim 
\begin{cases}
1000, \text{ with probability } p=0.98 \\
0, \text{ with probability } q=0.01\\
-1, \text{ with probability } r=0.01
\end{cases}.
$$
Then
$$\mathbb{E}(X|X\geq Y)= \frac{1000p-r^2}{p+q(1-p)+r^2}=\frac{1000(0.98)-0.01^2}{0.98+0.01(1-0.98)+0.01^2}\approx 999.69,$$
and
$$\mathbb{E}(X|X= Y) = \frac{1000p^2-r^2}{p^2+q^2 + r^2}=\frac{1000(0.98)^2-(0.01)^2}{0.98^2+0.01^2 + 0.01^2}\approx 999.79.$$
A: For completeness, I'll add the proof that a support of size $3$ is minimal for the counterexample, respectively that the strong claim holds for binary supports.
Claim: For $X,Y$ independent with support $\mathcal X\subseteq\mathbb R$, $|\mathcal X|\le 2$, we have $\mathbb E[X|X\ge Y]\ge\mathbb E[X|X=Y]$.
For $\mathcal X=\{x\}$ we have $\mathbb E[X|X\ge Y]=x=\mathbb E[X|X=Y]$. For $\mathcal X=\{x_-,x_+\}$ with $x_-<x_+$ we have
\begin{align*}
\mathbb E[X|X\ge Y]=\mathbb P(X=Y|X\ge Y)\mathbb E[X|X=Y]+\mathbb P(X>Y|X\ge Y)x_+\ge\mathbb E[X|X=Y].
\end{align*}
A: I'm not sure if this is correct, as it looks very easy, and I might very well be misunderstanding the underlying notation that you are using. But here it is anyway:
Let $y \in Y(\Omega)$,
$$\mathbb{E}(X \mid X \geq Y)(y) \space = \space \mathbb{E}(X \space \mid X \geq y) = \mathbb{E}(X \space \mid X = y) + \space  \mathbb{E}(X \space \mid X > y)$$
$$\mathbb{E}(X \mid X \geq Y)(y) \space = \mathbb{E}(X \space \mid X = Y)(y) + \space  \mathbb{E}(X \space \mid X > y)  $$
Since $\mathbb{E}(X \space \mid X > y) \geq 0$, it follows that:
$$\mathbb{E}(X \mid X \geq Y)(y) \space \geq \space \mathbb{E}(X \space \mid X = Y)(y)$$
Thus:  $$\mathbb{E}(X \mid X \geq Y) \space \geq \space \mathbb{E}(X \space \mid X = Y)$$
