If $Y, X, Z$ are random variables, is it true that $E[Y|X=x] = \sum_{z} E[Y|X=x, Z=z]P(Z=z|X=x)$ by law of total expectation? If $Y, X, Z$ are random variables. In particular, $Z$ is a positive discrete random variable.
Is it true that $E[Y|X=x] = \sum_{z} E[Y|X=x, Z=z]P(Z=z|X)$ by the law of total expectation/probability?
Here's my attempt:
\begin{align*}
E[Y|X=x] &= E[E[Y|X=x, Z]|X]\\
&= \sum_{z} E[Y|X=x, Z=z]P(Z=z|X=x)
\end{align*}
 A: I am really having trouble to understand your notation. In the following I will try to give a meaning to this equation and provide a counterexample.
From the Doob Dynkin lemma we know that there exist a function $f$ such that
\begin{align}
\mathbb E\big[Y\big|X,Z\big]=f(X,Z)
\end{align}
holds. It is common to write
$$
f(x,z)=\mathbb E\big[Y\big|X=x,Z=z\big]\,.
$$
It looks like by $\mathbb E\big[Y\big|X,Z=z\big]$ you mean
\begin{align}
\mathbb E\big[Y\big|X,Z=z\big]=f(X,z)\,.
\end{align}
Likewise, we also know
$$\tag{1}
\mathbb E\big[Y\big|X\big]=g(X)\quad\text{ where }\quad g(x)=\mathbb E\big[Y\big|X=x\big]\,.
$$
Now to the question: It looks like you are asking if
$$\tag{2}
g(x)=\sum_{z}f(x,z)\frac{\mathbb P(\{Z=z\}\cap\{X=x\})}{\mathbb P\{X=x\}}
$$
holds. (This is my only way to give your equation a well defined meaning).
Counter Example
Let $X$ be a discrete random variables taking values in $\{-2,-1,+1,+2\}$ with probability $1/4$ for each. Let
$$
Z:=X^2\quad\text{ and }\quad Y:=X^+:=\max(X,0)\,.
$$
($Z$ takes values in $\{1,4\}$ with probability $1/2$ for each. $Y$
takes values in $\{0,1,2\}$ with probability $1/2$ for $0$ and $1/4$ for the other two.)
Further,
\begin{align}
\mathbb E[Y|X]=Y=X^+\,,\quad g(x)=x^+=1_{\{x=1\}}+2\cdot 1_{\{x=2\}}\,.
\end{align}
Since $\sigma(Z)\subset\sigma(X)\,,$
$$
\mathbb E\big[Y\big|X,Z\big]=E\big[Y\big|Z\big]\,,
$$
that is, $f(x,z)$ does not depend on $x\,,$ only on $z=x^2\,.$
From $\{X=x\}\subset\{Z=x^2\}$ it follows that
$$
\frac{\mathbb P(\{Z=z\}\cap\{X=x\})}{\mathbb P\{X=x\}}
=\left\{\begin{array}~ \mathbb P\{Z=z\}&\text{ for }z=x^2\,,\\0&\text{ else. }\end{array}\right.
$$
Therefore the RHS of (2) contains only two terms, one for
$z=1$ and one for $z=4\,:$
$$\tag{3}
f(1)\frac{1}{2}+f(4)\frac{1}{2}\,.
$$
In particular this does not depend on $x$ while
the RHS of (2) is
$$\tag{4}
g(x)=\left\{\begin{array}~ 1&\text{ for }x=1\,,\\2&\text{ for }x=2\,,\\0&\text{ else }\,.\end{array}\right.
$$
