Law of iterated expectation - $\mathbb{E}[\mathbb{E}[X\mid Y] \mid Z ] = \mathbb{E}[X]$

Question

I am stuck doing this problem:

$X,Y,Z$ be random variables, then: $$\mathbb{E}[\mathbb{E}[X\mid Y] \mid Z ] = \mathbb{E}[X]$$

Surely this is from LIE and conditioning, but I can't seem to set it up correctly for it to work out nicely, can anyone point me in the right direction, or correct my error below?

Below is the way I was attacking this, but quickly got stuck and it's making me think I simply didn't set up my summation correctly. The expectation inside the conditional is really throwing me for a loop.

My take (likely incorrect): \begin{align} \mathbb{E}[\mathbb{E}[X\mid Y]\mid Z] &= \sum_zP(Z=z) (\mathbb{E}[X\mid Y])\\ &=\sum_zP(Z=z)\sum_x xP(X=x\mid Y=y)\\ &= \sum_zP(Z=z)\sum_x x \frac{P(X=x, Y=y)}{P(Y=y)}\end{align}

and I don't see where to go from here. Any help is greatly appreciated.

Edit: Thank you to everyone who commented and pointed out this is generally false. After asking my professor the question should really be

$$\mathbb{E}[\mathbb{E}[\mathbb{E}[X\mid Y] \mid Z ]] = \mathbb{E}[X]$$

I haven't done the proof yet, but I assume this follows cleanly from LIE.

Answer 1

[We are presuming $X,Y,Z$ are discrete random variables.]

Since $\Bbb E(X\mid Y)$ is a function of random variable $Y$, then we have:

$\qquad\begin{align}\Bbb E(\Bbb E(X\mid Y)\mid Z)&=\left[\sum_y\mathsf E(X\mid Y\,{=}\,y)\,\mathsf P(Y\,{=}\,y\mid Z\,{=}\,z)\right]_{z=Z} \\ &=\left[\sum\limits_y\left(\sum\limits_x x\,\mathsf P(X\,{=}\,x\mid Y\,{=}\,y)\right)\mathsf P(Y\,{=}\,y\mid Z\,{=}\,z)\right]_{z=Z}\\ &=\left[\sum\limits_{x} x\sum\limits_{y}\mathsf P(X\,{=}\,x\mid Y\,{=}\,y)\,\mathsf P(Y\,{=}\,y\mid Z\,{=}\,z)\right]_{z=Z}\end{align}$

Thus it shall not generally be so that $\Bbb E(\Bbb E(X\mid Y)\mid Z)=\Bbb E(X)$ unless, for example, $Y$ is independent of $Z$.

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Update

The above expectation is a function of $Z$. The OP has updated the question to be about the expectation of this. So let us do that, distribute the series and apply the law of total probability...

$\qquad\begin{align}\Bbb E(\Bbb E(\Bbb E(X\mid Y)\mid Z))&=\sum\limits_z\sum\limits_{x} x\sum\limits_{y}\mathsf P(X\,{=}\,x\mid Y\,{=}\,y)\,\mathsf P(Y\,{=}\,y\mid Z\,{=}\,z)\,\mathsf P(Z\,{=}\,z)\\&=\sum\limits_{x} x\sum\limits_{y}\mathsf P(X\,{=}\,x\mid Y\,{=}\,y)\sum\limits_z\mathsf P(Y\,{=}\,y\mid Z\,{=}\,z)\,\mathsf P(Z\,{=}\,z)\\&=\sum\limits_{x} x\sum\limits_{y}\mathsf P(X\,{=}\,x\mid Y\,{=}\,y)\,\mathsf P(Y\,{=}\,y)\\&=\sum\limits_x x\,\mathsf P(X\,{=}\,x)\\&=\Bbb E(X)\end{align}$