# Prove conditional probability relation of random variables

Let there be three random variables $X$, $Y$ and $Z$.

How can I prove the folowing?

$P(X|Y) = \sum\limits_{z} P(X,z|Y)$

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This is nothing more than an application of the definition of conditional probability. – Alex R. Mar 20 '12 at 19:24
Do you mean this? $P(b | a) = \frac{P(a, b)}{P(a)}$ ? – tzsjqnpn Mar 20 '12 at 20:17
Yep, and now remember that P(A)=sum_x P(A, x), or more generally that P(A)= sum_i (A, B_i) where B_i are pairwise disjoint and $\cup_i B_i = \Omega$ ($\Omega$ is the whole space). – Alex R. Mar 20 '12 at 22:52
How is $P(X|Y)$ defined for random variables $X$ and $Y$? (I am not familiar with that notation, and it is of course not the same as $P(A|B)$ for measurable sets $A$ and $B$.) – Jens Mar 21 '12 at 10:21

A correct formulation would be that $\mathrm P(X=x\mid Y=y)=\sum\limits_z\mathrm P(X=x,Z=z\mid Y=y)$ for every $y$ such that $\mathrm P(Y=y)\ne0$.
This formula is an example of the fact that $\mathrm P(A\mid C)=\sum\limits_k\mathrm P(A\cap B_k\mid C)$ for every events $A$, $(B_k)_k$ and $C$ such that $\mathrm P(C)\ne0$ and such that $(B_k)_k$ is a partition of the underlying probability space. In turn, this fact follows from the observation that the events $A\cap B_k\cap C$ are disjoint and that their union over $k$ is $A\cap C$.