# Bayes' rule and multiple conditioning

This is going to be a stupid question, but I know I am missing something, so here it goes:

My textbook (Probabilistic Robotics, p.17) trivially states that Bayes' rule gives:

$p(x |y,z) = \frac{p(y |x, z)p(x|z)}{p(y|z)}$

Are there some conditionig rule in probability that easily lets you arrive there from the simple Bayes' rule?:

$p(x |y) = \frac{p(y |x)p(x)}{p(y)}$

B/c for me to even get close to something similar I have to do:

$$p(x | z,y) = \frac{ p(y,z|x) p(x)} {p(y,z)}$$ $$= \frac{ \frac{p(y|x)p(z|x)p(x)}{p(z)} } { \frac{p(y,z)}{p(z)} }$$ $$= \frac{p(x|z)p(y|x)}{p(y|z)}$$ Where I had to use that y and z are conditionally independent...

So what am I missing?

Thank you

## 1 Answer

You're missing the definition of conditional probability: $P(x|y) = P(x,y)/P(y)$

• I don't understand how that answers my question...? – luffe Nov 7 '14 at 7:43
• Well, $$P(x|y,z) = \frac{P(x,y,z)}{P(y,z)} = \frac{P(y|x,z)P(x,z)}{P(y,z)} = \frac{P(y|x,z)P(x|z)P(z)}{P(y|z)P(z)} = \frac{P(y|x,z)P(x|z)}{P(y|z)}$$ You asked what you were missing... – Arthur B. Nov 7 '14 at 21:08