conditional probability within another condition

Question

If given $P(Y|X)$, $P(X|Z)$, and $P(Z)$, can we find $P(Y|X,Z)$?

I know that $P(X,Z)=P(X|Z)P(Z)$ and $P(Y|X,Z)=\frac{P(X,Z|Y)P(Y)}{P(X,Z)}$ but since $P(Y)$ is not given, it looks difficult for me to solve.

Thanks in advance!

Answer 1

No, you can't do it. To see why, let's set up a specific example. Let $X, Z$ be independent variables both distributed as $\text{Bernoulli}(0.5)$, and let $Y=Z$. Then we have $$\begin{align} P(Y|X) &= P(Y) = 0.5 \\ P(X|Z) &= P(X) = 0.5 \\ P(Z) &= 0.5 \\ P(Y|X,Z) &= Z \end{align}$$ where we note for example that $P(Y|X)=P(Y)$ because $X,Y$ are independent.

On the other hand, we also could have chosen $X, Y, Z$ to all be independent random variables distributed as $\text{Bernoulli}(0.5)$. In that case all the "givens" would come out the same (all equal to 0.5 again), but this time $P(Y|X,Z) = 0.5$. Therefore it must not be possible to correctly compute $P(Y|X,Z)$ with only the givens you listed.

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If we wanted to add an extra given so the problem would become solvable, we'd basically need to add $P(X,Y,Z)$ or something that lets us compute that quantity. For example, your equation shows that if we also knew $P(X,Z|Y)$ then that would also be sufficient.

Side note: In your question you mentioned "since $P(Y)$ is not given..." but actually you do have a way to compute $P(Y)$ by using $P(Y) = P(Y|X) P(X) = P(Y|X) P(X|Z) P(Z)$. So you do have access to the probability of each individual variable; the issue is that you don't have enough information about how they depend on each other.