Prove $\mathsf E^Y \mathsf E^Y [X] = \mathsf E^Y [X]$

Question

Prove that$$\mathsf E^Y \mathsf E^Y [X] = \mathsf E^Y [X]$$ by using $$\mathsf E^{Y=y} [X] = \int x\,p(x\mid y)\operatorname d x$$ or $$E^{Y=y} [X] =\sum_x x\,p(x\mid y)$$

I am having trouble on where to start with this, I know that $\mathsf E X=\mathsf E\,\mathsf E^Y[X]$, but I am not sure how I can use what is being given to assist me.

Answer 1

In more conventional notation we wish to show: $\mathsf E(\mathsf E(X\mid Y)\mid Y)= \mathsf E(X\mid Y)$

$$\begin{align}\mathsf E(\mathsf E(X\mid Y)\mid Y) & = \mathsf E(\int x p(x\mid Y)\operatorname d x)\mid Y) \\[1ex] & \vdots & \textsf{what and why?} \\[1ex] & = \mathsf E(X\mid Y)\end{align}$$

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$$\begin{align}\mathsf E^Y \mathsf E^Y X & = \mathsf E^Y \int x p(x\mid Y)\operatorname d x) \\[1ex] & \vdots & \textsf{what and why?} \\[1ex] & = \mathsf E^Y X\end{align}$$

Fill in the blanks.