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Given a random variable $X$ depending on a parameter $\theta$ which itself depends on a parameter $\psi$, how do I compute $p(\theta|X,\psi)$?


A website I have found$^1$ claims that $p(\theta|X,\psi)=\frac{p(\theta|\psi)\cdot p(X|\theta)}{\int_{\Theta}p(\theta| \psi)\cdot p(X| \theta)d\theta}$ but I haven't been able to derive this so far. The only way of doing something remotely useful with the LHS that I have found is to simply plug in the definition of conditional probability. I would also guess that the integral is a result of applying the law of total probability but I'm not sure how exactly to continue from this.

$^1$:https://people.stat.sc.edu/hitchcock/stat535slidesday24.pdf

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2 Answers 2

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This is a one-stage Bayesian model, since the hyperparameter $\psi$ is fixed, rather than being another random variable. Consider the difference between $$\theta \sim \mathcal N(0,1)$$ $$\theta \sim \mathcal N(\psi_1,\psi_2)\qquad \psi_1=0\,,\psi_2=1$$ and $$\theta|\psi_1=0\,,\psi_2=1 \sim \mathcal N(0,1)$$ There is none! The conditional bar "|" is not used in a probabilistic sense there.

This means that, if $$X|\theta\sim f(x|\theta)\qquad \theta|\psi\sim p(\theta|\psi)$$ then $$\theta|X=x,\psi \sim p(\theta|x,\psi) \propto f(x|\theta) p(\theta|\psi)$$

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If $X$ depending on a parameter $θ$ which itself depends on a parameter $ψ$ we should better write this as $p(X;\theta, \psi)$, but it should not be wrong to use $|$ either just if we mention the right thing.

Because there is no indication $\theta$ is random variable.

But Bayes rule should still work $p(\theta|X,\psi) = \frac{p(X|\theta, \psi)p(\theta, \psi)}{p(X)}$

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