Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

How can I prove the folowing?

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

share|improve this question
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
add comment

1 Answer

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 $A\cap C$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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