Proof with Summation of P(X|Y)

Question

I am trying to show that when p(x) > 0 and p(y) > 0, $\sum_{x \in X} p(x|y) = 1$, $\forall y \in Y$. Any hints/ideas? Any ways to visualize this so I can try to explain it to myself aloud? I should add that X and Y are two random variables.

Thank you, The Ginger Kitty Lover

Answer 1

What is $\textrm{P}(X \mid Y)$? It's $$\frac{\textrm{P}(X \cap Y)}{\textrm{P}(Y)}$$ In other words, $$\textrm{P}(\omega \in A \mid \omega \in B) = \frac{\textrm{P}(\omega \in X \cap Y)}{\textrm{P}(\omega \in Y)}$$ Also, $$\sum_{\omega \in \Omega}\textrm{Pr}(\omega) = 1$$

Can you do it from there?