How to calculate the MH ratio in a birth-death Reversible Jump MCMC?

Question

I know the formula and think I understand it, but I must be doing something wrong.

Assume $J\sim po(\lambda)$ and $X_J=(U_1, \dots, U_J)$ and $U_i\sim U(0,w)$. Then I want to sample from the posterior distribution $(J,X_J)$. So I make a RJ-MCMC move where

• In the birth move, insert randomly a new $U_i\sim U(0,w)$ in $X_j$. That can be done in $j+1$ positions.

• For the death move I remove a random entry in $X_{j+1}$. That can be done to $j+1$ entries.

So the MH-ratio for a birth move becomes

$$ \frac{po(j+1;\lambda)}{po(j;\lambda)}\cdot \frac{\frac{1}{j+1}}{\frac{1}{j+1}}\frac{1}{\frac{1}{w}}\cdot 1=\frac{po(j+1;\lambda)}{po(j;\lambda)}\cdot w $$

However, I don't believe in this formula because it means that the sampling distribution of $J$ will depend on $w$, which I do not think it should.

Answer 1

In the formula I inserted $P(J)$ as the posterior instead of $P(J,X_J)$. So the correct acceptance probability for a birth move is

$$ \frac{po(j+1;\lambda)\left(\frac{1}{w}\right)^{j+1}}{po(j;\lambda)\left(\frac{1}{w}\right)^{j}}\cdot \frac{\frac{1}{j+1}}{\frac{1}{j+1}}\frac{1}{\frac{1}{w}}\cdot 1=\frac{po(j+1;\lambda)}{po(j;\lambda)}. $$

and then there are no problems.