1
$\begingroup$

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

How can I prove the folowing?

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

  • 0
    This is nothing more than an application of the definition of conditional probability.2012-03-20
  • 0
    Do you mean this? $P(b | a) = \frac{P(a, b)}{P(a)}$ ?2012-03-20
  • 1
    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).2012-03-20
  • 0
    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 $B$.)2012-03-21

1 Answers 1