1
$\begingroup$

I have a question about how to update posterior probability sequentially when new data comes in sequentially by say $x_1$, then $x_2$, then $x_3$,.... I understand this form when the first data $x_1$ comes in (this very first updating is based on the Baye's formula):

$p(\theta |x_1) = \frac{p(x_1|\theta) * p(\theta)}{\int p(x_1|\theta) *p(\theta) d\theta}$

But how about when $x_2$ is coming in next now? what would be the updating looks like now? could someone gives me step by step procedures?

because I see a form like this when the sixth data ($x_6$) is included:

$p(\theta|x_1,x_2,...,x_6)=\frac{p(x_2,x_3,...,x_6|\theta) *p(\theta|x_1)}{\int p(x_2,x_3,...,x_6|\theta)*p(\theta|x_1)d\theta}$

But I couldn't understand how it arrives.

Could someone shows me how to go from the very first updating, to the form of $p(\theta|x_1,x_2)$ ? because I think once I know how to update the posterior from $x_1$ to $x_2$, then I would be able to figure out how to reach the posterior when the sixth data $x_6$ is included.

Thank you

$\endgroup$
1
  • $\begingroup$ you will get a better answer if you outline how you calculate your posterior or what your model is $\endgroup$ Aug 26, 2019 at 7:04

1 Answer 1

1
$\begingroup$

If the samples are iid given the parameters, e.g. $x_i$ are coin tosses and $\theta=p=P(\text{Head})$, then $p(x_2|x_1,\theta)=p(x_2|\theta)$ and we'll use this fact below: $$\begin{align}p(\theta|x_1,x_2)&=\frac{p(x_2|\theta,x_1)p(\theta|x_1)p(x_1)}{\int p(x_2|\theta,x_1)p(\theta|x_1)p(x_1) d\theta}=\frac{p(x_2|\theta,x_1)p(\theta|x_1)}{\int p(x_2|\theta,x_1)p(\theta|x_1)d\theta}\\&=\frac{p(x_2|\theta)p(\theta|x_1)}{\int p(x_2|\theta)p(\theta|x_1)d\theta}\end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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