# Probability Question - Conditional Probability

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?

Thanks

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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

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