Can anyone give me some hints on the following proof?
$\mathbb{E}${g(X)*$\mathbb{E}${X|Y}|Y}=$\mathbb{E}${g(X)*X|Y}
I only know g(.) is a convex function. I am not good at math. Thank you in advance!
If there is no solution, can I prove that $\ge$ holds?
Your statement is not true. Suppose $\xi_1, \xi_2$ are i.i.d. standard normal distributed and define $Y := \xi_1$ and $X := \xi_1 + \xi_2$. Consider $g(x) := x$. Then $$ \mathbb{E}\left[g(X) X \, \middle\vert \, Y \right] = \mathbb{E}\left[X^2 \, \middle\vert \, Y \right] = \mathbb{E}\left[\xi_1^2 + 2\xi_1 \xi_2 + \xi_2^2 \, \middle\vert \, \xi_1 \right] = \xi_1^2 + 2 \mathbb{E}[\xi_1]\xi_2 + \mathbb{E}[\xi_2^2]\\ = \xi_2^2 + 1, $$ but $$ \mathbb{E}\left[X \, \middle\vert \, Y \right] = \mathbb{E}\left[\xi_1 + \xi_2 \, \middle\vert \, \xi_1 \right] = \xi_1 + \mathbb{E}[\xi_2] = \xi_1, $$ so $$ \mathbb{E}\left[ g(X) \mathbb{E}[X \vert Y] \, \middle\vert \, Y\right] = \mathbb{E}\left[(\xi_1 + \xi_2)\xi_1 \, \middle\vert \, \xi_1 \right] = \xi_1^2. $$.