Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hi I'm working with particle filters at the moment however my maths isnt so strong i was wondering given the $P(X_n|Y_0,....Y_{n-1})$ and $P(Y_n|X_n)$ how do you obtain $P(Y_n|Y_0,...,Y_{n-1})$? i.e. the equations that links the two?


share|cite|improve this question
What are your thoughts? What have you tried? – Peter Apr 8 '13 at 15:49
Hi @Peter honestly my intuition for probability is terrible, I've just spent a good 10 minutes tying to come up with an answer that might not make me look like a complete idiot, but I honestly have no idea, possibly it might be a multiple of the two :S – Jono Brogan Apr 8 '13 at 16:03
Think of all the set $Y_0 ... Y_{n-1}$ as a single variable $A$. Then, you get the equivalent problem: given $P(C| A)$ and $P(B |C)$ obtain $P(B | A)$ – leonbloy Apr 8 '13 at 16:44
up vote 1 down vote accepted


$P(B) = \sum_C P (B C) = \sum_C P(B | C) P(C)$

Hence (conditioning everything over A),

$P(B | A) = \sum_C P(B | C A ) P(C |A)$

You can't simplify this further, but in some cases (I'd bet in yours), depending on extra properties of the variables, you can assume that $P(B|C A) = P(B | C)$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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