# Rules from probability theory

Question 1:

From some slides I saw that:

$p(x_{0:t+1}|y_{1:t+1}) = p(x_{0:t}, x_{t+1}|y_{1:t}, y_{t+1})$, and according to the product rule of probability theory $P(a,b) = P(a|b) P(b)$ we will have have:

$$p(x_{0:t+1}|y_{1:t+1}) = p(x_{0:t}, x_{t+1}|y_{1:t}, y_{t+1}) = \frac{p(x_{0:t}, x_{t+1}, y_{t+1}|y_{1:t})}{p(y_{t+1}|y_{1:t})}$$

I don't understand how the product rule of probability theory is applied for this example, what is $a$ and $b$ in this example ?

Note: $x_{0:t}$ is $\{x_0,x_1,x_2, \ldots, x_t\}$

Question 2:

Also, is it always true that: $P(X|Y)= \int P(X|Z) P(Z|Y)\, dz$ ?

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The product rule of probability as you state is $P(A, B) = P(A|B)P(B)$. This rule can also be applied to conditional probabilities as well i.e $P(A, B|C) = P(A|B, C)P(B|C)$. This gives, $P(A|B, C) = \frac{P(A, B|C)}{P(B|C)}$. Comparing the RHS in this equation with the equation that you have, you will see $A = x_{0:t+1}, B = y_{t+1}, C = y_{1:t}$.
As for Question 2, $P(X|Y) = \int P(X, Z|Y) dz$. Now, generally if you factorize the term in the integral, it will factorize as $P(X|Z, Y)P(Z|Y)$. However, in your expression it is factorizing as $P(X|Z)P(Z|Y)$, which implies that $X$ is conditionally independent of $Y$ given $Z$. This need not be always true, hence your expression is not always true.
I didn't know that the product rule can be applied this way to a conditional probability, but it's a little confusing: is $P(A,B|C) = P((A,B)|C) = P(A,(B|C))$ ? – shn Apr 24 '12 at 18:45
It's always $P((A,B)|C)$. – TenaliRaman Apr 24 '12 at 18:51
Ok and can we say that $P((A|B)|C) =$P(A|B,C)$? – shn Apr 24 '12 at 19:40 '$|$' is not an operator so$P((A|B)|C)$does not mean anything. '$|$' is simply a partition to indicate which random variables are unobserved and which ones are observed. Basically, in$P(A_1,A_2,\ldots,A_n|B_1,B_2,\ldots,B_m)$, the '$|$' indicates that$A_1,\ldots,A_n$are unobserved random variables and$B_1,\ldots,B_m\$ are the observed random variables. – TenaliRaman Apr 24 '12 at 21:38