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So I'm trying to find the conditional posterior distributions of n given $\theta$ and x as well as $\theta$ given n and x.

These are my priors (poisson and beta)

\begin{equation*} \begin{aligned} & \pi (n) = \frac{ \mu^{n} e^{-\mu}}{n!} \\ & \pi (\theta) = \frac{1}{B(\alpha, \beta)} \theta^{\alpha-1} (1 - \theta)^{\beta-1} \end{aligned} \end{equation*}

and the likelihood function of the parameters given the data is binomial

\begin{equation*} \begin{aligned} L(n, \theta | x) = {n \choose x} \theta^{x} (1- \theta)^{n-x} \end{aligned} \end{equation*}

This is what I have for the first conditional distribution

\begin{equation*} \begin{aligned} & \pi(n | x, \theta) = \frac{\pi(n) L(n | x, \theta)}{ \int \pi(n) L(n | x,\theta) dn} \\ & \propto \pi(n) L(n | x, \theta) \\ & = \frac{ \mu^{n} e^{-\mu}}{n!} \frac{n!}{(n-x)!x!} \theta^{x} (1- \theta)^{n-x} \\ & \propto \frac{ \mu^{n} e^{-\mu}}{(n-x)!} (1- \theta)^{n-x} \\ & \propto \frac{ \mu^{n} \mu^{-x} e^{-\mu} e^{-(1- \theta)}}{(n-x)!} (1- \theta)^{n-x} \\ & = \frac{ \mu^{n-x} e^{-(\mu(1- \theta)}}{(n-x)!} (1- \theta)^{n-x} \\ & = \frac{ (\mu(1-\theta))^{n-x} e^{-(\mu(1- \theta))}}{(n-x)!} \\ \end{aligned} \end{equation*}

I know it should be in the form of x + Y where Y is a poisson random variable so I'm not sure where I'm going wrong.

For my second conditional distribution I have

\begin{equation*} \begin{aligned} & \pi(\theta | x, n) = \frac{\pi(\theta) L(\theta | x, n)}{ \int \pi(\theta) L(\theta | x,n) d\theta} \\ & \propto \pi(\theta) L(\theta | x, n) \\ & = \frac{1}{B(\alpha, \beta)} \theta^{\alpha-1} (1 - \theta)^{\beta-1} \frac{n!}{(n-x)!x!} \theta^{x} (1- \theta)^{n-x} \\ & \propto \frac{1}{B(\alpha+x, \beta + n - x)} \theta^{\alpha + x -1} (1 - \theta)^{\beta + n - x -1} \end{aligned} \end{equation*}

which I think is right because I know it should be some kind of beta distribution.

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Firstly, let's correct some notational mistakes: $L(n,\theta|x)$ is $p(x|n,\theta)$, but in the third line of the derivation of $\pi(n|x,\theta)$, you substitute the expansion of $L(n,\theta|x)$ in place of $L(n|x,\theta)$. However, since in terms of $n$, we have $L(n|x,\theta)=p(x,\theta|n)=p(x|\theta,n)p(\theta|n)=L(\theta,n|x)\pi(\theta)\propto L(\theta,n|x)$ what you've substituted is correct with $\propto$ argument present (not with $=$).

Another typo: $e^{-\mu}e^{-(1-\theta)}$ seems to be equal to $e^{-\mu(1-\theta)}$ in the sixth-line of the same derivation; but it is not. Again, this term doesn't depend on $n$, so $\propto$ argument holds, and we can simply add the term $e^{-\mu(1-\theta)}$ to the multiplication without disturbing anything.

For the final result, i.e. $$\pi(n|x,\theta)\propto \frac{ (\mu(1-\theta))^{n-x} e^{-(\mu(1- \theta))}}{(n-x)!}$$ the PDF (or PMF) is in shifted Poisson form if we let $\lambda=\mu(1-\theta)$, and $n\geq x$, i.e. $n=x+Y$, where $Y$ is Poission.

For the second one, we can substitute the value of $L(n,\theta|x)$ as below, and follow your way to the solution: $$L(\theta | x, n)=p(x,n|\theta)=p(x|n,\theta)p(n|\theta)=p(x|n,\theta)p(n)\propto p(x|n,\theta)=L(n,\theta|x)$$

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  • $\begingroup$ Thanks that makes sense $\endgroup$ – user1992460 Mar 15 at 14:31

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