I'm doing research and for a proof I need the smaller result that for X, Y random, independent (but not identical) variables we have
$$\mathbb{E}\left[X|X>Y\right] \geq \mathbb{E} \left[X\right]$$
This seems very intuitive, but I've had no luck proving it so far. I've managed to derive the pdf of X|X>Y and use it to calculate the expectation:
$$\mathbb{E}\left[X|X>Y\right] = \int_{-\infty}^\infty\frac{f_X(x)F_Y(x)}{1-\int_{-\infty}^\infty F_X(y)f_Y(y)dy}xdx$$
but I find it very hard to bound this. Does someone have any ideas on how to proceed?
Thanks!
I suppose that $\mathbb P(X>Y)>0$, otherwise the conditional expected value does not make sense. We have that $\mathbb E[X\vert X>Y]=\frac{\mathbb E[X1_{\{X>Y\}}]}{\mathbb P(X>Y)}$, so the problem amounts to show that $$ \mathbb E[X1_{\{X>Y\}}]\ge\mathbb P(X>Y)\mathbb E[X]. $$
Since $X$ and $Y$ are independent, the left-hand side is $\mathbb E[\varphi(Y)]$, where $\varphi:y\mapsto\mathbb E[X1_{\{X>y\}}]$, and the right-hand side is $\mathbb E[\psi(Y)]\mathbb E[X]$, where $\psi:y\mapsto\mathbb P(X>y)$. So it suffices to prove that for all $y\in\mathbb R$, $$ \mathbb E[X1_{\{X>y\}}]\ge\mathbb P(X>y)\mathbb E[X] $$ (or in other words, without loss of generality we can assume that $Y$ is a deterministic constant).
If $\mathbb P(X>y)\in\{0,1\}$ the latter inequality is trivial, so let us assume that $\mathbb P(X>y)\in(0,1)$. As $\mathbb E[X]=\mathbb E[X1_{\{X>y\}}]+\mathbb E[X1_{\{X\le y\}}]$, the latter inequality is equivalent to $$ \frac{\mathbb E[X1_{\{X>y\}}]}{\mathbb P(X>y)}\ge\frac{\mathbb E[X1_{\{X\le y\}}]}{\mathbb P(X\le y)}\cdot $$
Since $X1_{\{X>y\}}\ge y1_{\{X>y\}}$ and $y1_{\{X\le y\}}\ge X1_{\{X\le y\}}$, we have that $$ \frac{\mathbb E[X1_{\{X>y\}}]}{\mathbb P(X>y)}\ge y\ge\frac{\mathbb E[X1_{\{X\le y\}}]}{\mathbb P(X\le y)}, $$ which shows the desired inequality and concludes the proof.