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.


1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.