Conditional and Marginal Probabilities

Question

Consider 3 RVs $X_1,X_2,X_3$ and RV $C$. We are given that $X_1\perp \!\!\! \perp X_2\perp \!\!\! \perp X_3|C$ (i.e. $X$s are conditionally mutually independent given $C$). Consider $p(x_1| x_2, x_3)$. $$p(x_1 | x_2, x_3)=\mathbb{E}_{c}\left[p(x_1 | x_2, x_3, c)\right].$$ By conditional independence, it should hold that $p(x_1|x_2,x_3,c)=p(x_1|c)$, but then we end up with $$p(x_1|x_2,x_3)=\mathbb{E}_c[p(x_1|c)]=p(x_1),$$ i.e. the conditional equal to the marginal, but that only happens when $X_1\perp \!\!\! \perp (X_2, X_3)$ (that I don't think holds). Is anything wrong here?

Answer 1

The first equation is wrong. E.g. $C\in\{0,1\}$ with probability $\frac12$ each; if $C=0$ then $X_i=1$; whereas if $C=1$ then the $X_i$ are independent throws of six-sided dice. Then

\begin{eqnarray} p(X_1=1\mid X_2=X_3=1) &=& \frac{\frac12+\frac12\cdot\frac1{6^3}}{\frac12+\frac12\cdot\frac1{6^2}} \\ &=& \frac{217}{222}\;, \end{eqnarray}

whereas

\begin{eqnarray} &&\mathbb E_c[p(X_1=1\mid X_2=X_3=1,C=c)] \\ &=& \frac12\left(p(X_1=1\mid X_2=X_3=1,C=0)+p(X_1=1\mid X_2=X_3=1,C=1)\right) \\ &=& \frac12\left(1+\frac16\right) \\ &=& \frac7{12}\;. \end{eqnarray}