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.



1 Answer 1


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 $\mathbb{P}(X\mid U,P,T)\neq\mathbb{P}(X\mid U,T)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.