joint distribution marginalization proof, is this right?

Question

prove: $$p(x\mid z) = \sum_y p(x\mid y,z)p(y\mid z)$$

I understand a bit about marginalization. I think my prove should look like this: $$ p(x\mid z) = \sum_y p(x,y\mid z) = \sum_y p(x\mid y,z)p(y) $$

My first portion, where I add the sum of y in, seems like simple application of marginalization. However I don't quiet understand why the second part works or if it works. Anyone able to shed some light on to why this works?

score 2 · Accepted Answer · answered Mar 29 '15 at 01:18

2

Imagine all conditional probabilities with their explicit definitions: $$\sum_{y} \frac {p (x, y, z)}{p (y, z)} \frac {p (y, z)}{p (z)}$$ You will obtain $\sum_{y} p (x, y|z) $ and it is easy to see how this sums over $ y $.

answered Mar 29 '15 at 01:18

Ufuk Can Bicici

3,026

I don't see it. – still learning Mar 29 '15 at 02:19
Think it that way: $ p (x, y|z)=\frac {p(x, y, z)}{p (z)} $ So $\sum_y p (x, y|z)=\sum_y \frac {p(x, y, z)}{p (z)} $ Does this make it clear? – Ufuk Can Bicici Mar 29 '15 at 07:57
Yes I got it! Thanks for sticking with me. – still learning Mar 29 '15 at 19:13

joint distribution marginalization proof, is this right?

1 Answers1