If $Z$ is independent of $(X, Y)$, then $E(X|(Y, Z)) = E(X|Y)$ almost surely.

Question

Suppose $X, Y, Z$ are continuous random variables, with $Z$ independent of $(X, Y)$. I wish to show that $E(X|(Y, Z)) = E(X|Y)$ almost surely. This question has already been asked here, but I remain uncertain as to how to fully proceed with its answer.

By definition, \begin{align} E[(X-E(X|Y, Z))H(Y, Z)] = 0 \tag{1} \end{align} for any $H(Y, Z)$, and \begin{align} E[(X-E(X|Y))h(Y)] = 0 \tag{2} \end{align} for any $h(Y).$ If one is able to show that $E(X|Y)$ satisfies \begin{align} E[(X-E(X|Y))H(Y, Z)] = 0 \tag{3} \end{align} for any $H(Y, Z)$, then, by uniqueness of solutions, it follows that $E(X|Y, Z) = E(X|Y)$ almost surely.

To begin, notice that, if $H(Y, Z) = h(Y)g(Z)$ is separable, then, in using the independence of $(X, Y)$ and $Z$, \begin{align} E[(X-E(X|Y))h(Y)g(Z)] = E[(X-E(X|Y))h(Y)]E[g(Z)] \stackrel{(2)}{=} 0,\end{align} so that $(3)$ again holds. However, how might I proceed in the most general case, where $H(Y, Z)$ is not separable in $Y$ and $Z$?

Answer 1

\begin{align} E(f(X,Y)h(Y,Z)|Z=z) &= \int f(x,y) h(y,z)\mathrm d\mu^{(X,Y)|Z=z}(x,y)\\ &= \int f(x,y) h(y,z)\mathrm d\mu^{(X,Y)}(x,y)\\ &= E(f(X,Y)h(Y,z)) \end{align}

Apply this to $f(X,Y) = X -E(X|Y)$,

$$E((X-E(X|Y))H(Y,Z)|Z=z) = E((X-E(X|Y)H(Y,z)) = 0$$

Take the expectation.

Answer 2

Let $\Omega=\Omega_1\times\Omega_2\times\Omega_3$ be a triple product space. In general, if $P$ is a probability on $\Omega_3$, $Q(\omega_3)$ is a probability on $\Omega_2$ for each $\omega_3$, and $R(\omega_2,\omega_3)$ is a probability on $\Omega_1$ for each $(\omega_2,\omega_3)$, such that for each $(A\times B\times C)\subset \Omega_1\times\Omega_2\times\Omega_3$: $$P_{\Omega}(A\times B\times C)=\int_CdP\int_BdQ(\omega_3)\int_AdR(\omega_2,\omega_3)$$ Then: $\quad P=P_{\Omega_3},\quad Q(\omega_3)=P_{\Omega_2/\Omega_3}(\omega_3),\quad R(\omega_2,\omega_3)=P_{\Omega_1/\Omega_2\circ \Omega_3}(\omega_2,\omega_3)$

In this case: $$P_{X\circ Y\circ Z}(A\times B\times C) =P_{X\circ Y}(A\times B)P_Z(C) =\int_CdP_Z\int_BdP_Y\int_AdP_{X/Y}(y)$$ That is to say: $\quad P_{Y/Z}(z)=P_Y,\quad and\quad P_{X/Y\circ Z}(y,z)=P_{X/Y}(y).\quad $ And so: $$\quad E[X|Y,Z]=\int xdP_{X/Y\circ Z}=\int xdP_{X/Y}=E[X|Y]$$ A simpler but less interesting alternative would be: \begin{gather*} E\big[ E[X|Y]I_{B\times C}(Y,Z)\big]=E\big[E[X|Y]I_B(Y)I_C(Z)\big] =E\big[E[X|Y]I_B(Y)\big]E[I_C(Z)]\\ =E[XI_B(Y)].E[I_C(Z)] =E[XI_B(Y)I_C(Z)]=E[XI_{B\times C}(Y,Z)] \end{gather*}