Conditional independence proof I want to prove that
$\mathbb{P}(X|U,P) = \mathbb{P}(X|U) \implies \mathbb{P}(X|U,P,T) = \mathbb{P}(X|U,T)$
Where all the letters denote random variables. I'm not sure that this is right, but it seems to be an intuitive result.
Thanks.
 A: Without further assumptions, the result does not hold.
Here's a counterexample.
Let $\mathbb{P}$ be the uniform probability on the Boolean cube $\{0,1\}^3$. Let $X,Y,Z$ be the first, the second, and the third coordinate function of the Boolean cube, respectively. Define $U:=\max(X,Y)$, $P:=Z$, and $T :=\min(Y,Z)$. Since $X,Y,Z$ are $\mathbb{P}$-independent of each other, we have
\begin{equation}
\mathbb{P}(X\mid U,P) = \mathbb{P}\big(X\mid \max(X,Y),Z\big) = \mathbb{P}\big(X\mid \max(X,Y)\big)=\mathbb{P}(X\mid U).
\end{equation}
On the other hand,

*

*$\{U=1\}\cap\{P=1\}\cap\{T=0\} = \big\{(1,0,0)\big\}$,

*$\{U=1\}\cap\{T=0\} = \big\{(1,0,0),(0,1,0),(1,1,0),(1,0,1)\big\}$,

*$\{X=1\}\cap\{U=1\}\cap\{T=0\} = \big\{(1,0,0),(1,1,0),(1,0,1)\big\}$,

from which it follows that
\begin{equation}
\mathbb{P}(X=1 \mid U=1,P=1,T=0) = 1\neq \frac{3}{4} = \frac{\frac{3}{8}}{\frac{4}{8}} = \mathbb{P}(X=1 \mid U=1,T=0),
\end{equation}
which implies $\mathbb{P}(X\mid U,P,T)\neq\mathbb{P}(X\mid U,T)$.
