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In Bayesian probability, does the prior distribution $\pi(\theta)$ only depend on $\theta$? For example, suppose the prior distribution of the unknown parameter $\theta$ is binomial. Then does $ \pi(\theta) = \binom{n}{\theta} p^{\theta} (1-p)^{n-\theta}$

Whereas if $f(\theta|x_1)$ is binomial then $f(\theta|x_1) = \binom{n}{x_1} p^{x_1}(1-p)^{n-x_1}$

Is this correct?

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    Oh ok it should be $\theta$. But the first is correct?2012-05-04

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Your first statement might be correct, though I doubt it. It is possible that the parameter $\theta$ is an integer in the range $[0,n]$ with that prior distribution, though normally you would give an explicit value for $p$ or describe it as a hyperparameter. But I am guessing you actually expect $\theta$ to be a real number in the range $[0,1]$.

If your second statement is supposed to be an expression for the likelihood or posterior probability (density) of the parameter $\theta$ given the observation $x_1$, then it is almost certainly wrong, as it does not seem to vary with $\theta$.

My guess is that in fact $\theta$ is intended to be a parameter of the binomial distribution where $\Pr(X_1=x_1|\theta,n) = \binom{n}{x_1} \theta^{x_1} (1-\theta)^{n-x_1}$ and with $\theta$ having a (perhaps improper) prior distribution on $[0,1]$.

A common example chosen for the prior in this case, because it is part of a conjugate family, would be a beta distribution proportional to $\theta^{\alpha-1}(1-\theta)^{\beta-1}$ for some given $\alpha $ and $ \beta$, in which case the posterior density, given an observation of $x_1$ out of $n$, is proportional to $\theta^{x_1+ \alpha-1}(1-\theta)^{n-x_1+\beta-1}$ i.e. from the same family.