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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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    Your second equality doesn't make sense. If you'd written either $f(x_1\mid\theta) = \dbinom{n}{x_1} \theta^{x_1}(1-\theta)^{n-x_1}$ or $f(x_1\mid p) = \dbinom{n}{x_1} p^{x_1}(1-p)^{n-x_1}$ it would make some sense, but starting with $f(\theta\mid x_1)$ leaves me wondering what you mean, and then putting an expression with no "$\theta$" in it on the other side of the "$=$" is weird, and it's hard to even guess what you mean. The word "whereas" usually means you're calling attention to a contrast, but I don't know what you have in mind there either.2012-05-04
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    @MichaelHardy: I am talking about the posterior distribution $f(\theta| x_1)$.2012-05-04
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    Oh ok it should be $\theta$. But the first is correct?2012-05-04

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