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

I have a prior distribution,

$$p(\boldsymbol\theta|\pi)=\prod\limits_{i=1}^K p(\theta_i|\pi).$$

$\theta_i$ can equal $0$ or $1$, so I am using a Bernoulli distribtion so that

$$p(\boldsymbol\theta|\pi)=\prod\limits_{i=1}^K \pi^{\theta_i}(1-\pi)^{1-\theta_i}.$$

I then want to add this distribution onto my marginal likelihood to make up my posterior. Should I solve it as $$p(\boldsymbol\theta|\pi)=\pi^{K\theta_i}(1-\pi)^{K(1-\theta_i)} \, \, ?$$

But then is the product of bernoulli distributions the binomial distribution?

Then should my answer be

$$p(\boldsymbol\theta|\pi)=\left(\begin{array}c K\\ t \end{array}\right)\pi^{t}(1-\pi)^{K-t)} $$

where $K$ is the maximum number of $\theta_i$'s allowed, and $t=\{0, 1\}$ , (i.e. $t=0\, \, \text{or}\, \, 1$)?

What form do I add this prior to my likelihood?

share|cite|improve this question

The equation you have can be represented as follows: $$p(\boldsymbol x|\theta)=\prod\limits_{i=1}^K \theta^{x_i}(1-\theta)^{1-x_i}=\theta^{\sum_i x_i}(1-\theta)^{K-\sum_i x_i}$$

We have the Bayes rule


as $\theta$ is known, we have the joint density $p(x,\theta)=p(\theta,x)$ which specifies all the information we need.

share|cite|improve this answer
But @Seyhmus Güngören, does not $\pi^{\sum\limits_{i=1}^K \theta_i}$ equal $\pi^{K\theta_i}$? (Also, my prior is conditioned on $\pi$, just want to write it down so not using the product symbol.) – Ellie Aug 23 '12 at 14:42
@Ellie in your representation $\theta$ and in my representation $x$ are unknown. They are samples from a set $\{0,1\}$ but finally as $\theta_i\neq\theta\forall i$ you cannot write the equation before the last one in your question. – Seyhmus Güngören Aug 23 '12 at 14:53

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.