Let $\mathbb{X,Y,Z}$ be random variables such that $\mathbb{X}$ has finite expected value and $\mathbb{Y}$ is bounded. Prove that: $$a)\quad \mathbb{E[Y|Z]} \quad is \quad bounded$$ $$b)\quad \mathbb{E[YE[X|Z]]=\mathbb{E[XE[Y|Z]]}}$$ I have tried to proof both subsections but I think that I'm just closer to the first one:
Let's suposse that $\mathbb{Y,Z}$ are continuous, so: $$\mathbb{E[Y|Z]}=\int_{-\infty}^{\infty}y\mathbb{P[Y=y|Z]}dy\le k\int_{-\infty}^{\infty}\mathbb{P[Y=y|Z]}dy \quad because\quad \mathbb{Y}\quad is\quad bounded$$ $$k\int_{-\infty}^{\infty}\mathbb{P[Y=y|Z]}dy = k$$ $$\therefore \mathbb{E[Y|Z]\le k}$$ If $\mathbb{Y,Z}$ are discrete, so: $$\mathbb{E[Y|Z]}=\sum_{Y}y\mathbb{P[Y=y|Z]}\le k\sum_Y\mathbb{P[Y=y|Z]}=k$$
(a) $Y$ is bounded , we can use Monotonicity property of conditional expectation,so
\begin{align} a<&Y<b \\ \mathbb E(a|Z)<&\mathbb E(Y|Z)<\mathbb E(b|Z) \\ a<&\mathbb E(Y|Z)<b \end{align}
(b) By conditioning on $Z$ and use Low of total expectation
\begin{align} \mathbb E(Y\mathbb E(X|Z))&=\mathbb E \color{blue}{(}\mathbb E\color{red}{(}Y\mathbb E(X|Z)\mid Z\color{red}{)}\color{blue}{)} \overset{(1)}{=} \mathbb E \color{blue}{(} \mathbb E(X|Z) \mathbb E\color{red}{(}Y\mid Z\color{red}{)}\color{blue}{)}\\ \mathbb E(X\mathbb E(Y|Z))&=\mathbb E \color{blue}{(}\mathbb E\color{red}{(}X\mathbb E(Y|Z)\mid Z\color{red}{)}\color{blue}{)} \overset{(2)}{=} \mathbb E \color{blue}{(} \mathbb E(Y|Z) \mathbb E\color{red}{(}X\mid Z\color{red}{)}\color{blue}{)} \end{align} So both are same.
In (1) and (2) we used the Pulling out known factors property of conditional expectation for example for (2)
$$\mathbb E \color{blue}{(}\mathbb E\color{red}{(}X\mathbb E(Y|Z)\mid Z\color{red}{)}\color{blue}{)}=\mathbb E \color{blue}{(}\mathbb E\color{red}{(}X\mathbb g(Z)\mid Z\color{red}{)}\color{blue}{)}=\mathbb E \color{blue}{(}g(Z) \mathbb E\color{red}{(}X \mid Z\color{red}{)}\color{blue}{)}=\mathbb E \color{blue}{(}\mathbb E(Y|Z) \mathbb E\color{red}{(}X \mid Z\color{red}{)}\color{blue}{)}$$