0
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I don't understand how that answers my question...? $\endgroup$ – luffe Nov 7 '14 at 7:43
  • $\begingroup$ 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... $\endgroup$ – Arthur B. Nov 7 '14 at 21:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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