# Bayesian Theorem Update Inference

According to Bayes theorem $P(A|B) = \frac{P(B|A)*P(A)}{P(B)}$

I've found somewhere that: $P(x_t|z_{1:t}) = \frac{P(z_t|x_t)*P(x_t|z_{1:t-1})}{P(z_t|z_{1:t-1})}$ but I don't really understand it, is it expressed according to the bayes theorem ?

This is from page 13 of http://www.igi.tugraz.at/pfeiffer/documents/particlefilters.pdf

According to Bayes theorem $P(x_t|z_{1:t}) = \frac{P(z_{1:t}|x_t)*P(x_t)}{P(z_{1:t})}$ and this is equal to $\frac{P(z_t,z_{1:t-1}|x_t)*P(x_t)}{P(z_t,z_{1:t-1})}$, but since $z_t$ and $z_{t-1}$ are conditionally independent given $x_t$ (according to the above figure), then $P(z_t,z_{1:t-1}|x_t) = P(z_t|x_t)*P(z_{1:t-1}|x_t)$, so we get $\frac{P(z_t|x_t)*P(z_{1:t-1}|x_t)*P(x_t)}{P(z_t,z_{1:t-1})}$ and again according to Bayes theorem we have $P(z_{1:t-1}|x_t) = P(x_t|z_{1:t-1})*P(z_{1:t-1}) / P(x_t)$, so we get $\frac{P(z_t|x_t)*P(x_t|z_{1:t-1})*P(z_{1:t-1})}{P(z_t,z_{1:t-1})} = \frac{P(z_t|x_t)*P(x_t|z_{1:t-1})*P(z_{1:t-1})}{P(z_t|z_{1:t-1})*P(z_{1:t-1})} = \frac{P(z_t|x_t)*P(x_t|z_{1:t-1})}{P(z_t|z_{1:t-1})}$

• The picture misleading: In it the past of X is independent of the present X given the present Z. Your equations (and the slides) are the exact opposite (basically Z and X are switched). Commented Apr 24, 2012 at 14:42
• @ConjugatePrior Oh well I see. So generally, is the $x$ notation which is used for data-points and $z$ for the hidden variables (like in the picture), or is it the opposite (like in my equations) ?
– shn
Commented Apr 24, 2012 at 15:19
• It depends - that's why you can find examples of both. I'd guess $x$ as the state is more common. Commented Apr 24, 2012 at 15:24
• Also, is it always true that: $P(X|Y) = \int P(X|Z)*P(Z|Y) dz$ ?
– shn
Commented Apr 24, 2012 at 15:38
• No it isn't. (That's what additional the conditional independence assumption is giving you.) It might be a good idea to have a look at some Bayesian network work to get a better feel for this sort of thing. I think Russell and Norvig's book has decent introductory coverage. The ML folk around the forum should have some good suggestions too. Commented Apr 24, 2012 at 16:03

To make it look like your first version of Bayes theorem, rewrite $P(z_t | x_t)$ to condition also on $z_{1\ldots t-1}$ as everything else does. That should look more familiar. Now notice that the conditional independence assertion above means that once you condition on $x_t$, $z_t$ no longer actually depends on its past, the part you just added in. So you can take it back out.