conditional distribution of $A\mid N = k$

Question

Can someone help me with how to get part 3 of the question?

Answer 1

Hint: Let $f_{\Lambda \mid N}$ be the PDF of $\Lambda \mid N$. Then $$f_{\Lambda \mid N}(\lambda \mid k) = \dfrac{f_{N\mid\lambda}(k\mid\lambda)f_{\lambda}(\lambda)}{\int\limits_{0}^{\infty}f_{N\mid\lambda}(k\mid\lambda)f_{\lambda}(\lambda)\text{ d}\lambda}\text{.}$$

To understand why this is the case, look at Bayes' theorem.

Answer 2

There are unfortunately two conventional usages: $\beta$ is a scale parameter, so the prior distribution of $\Lambda$ is $$ \frac{1}{\Gamma(\alpha)}\cdot \lambda^{\alpha-1} e^{-\lambda/\beta} \, \frac{d\lambda}\beta \text{ for }\lambda>0, \tag 1 $$ or else $1/\beta$ is the scale parameter, so that the prior distribution $$ \frac{1}{\Gamma(\alpha)}\cdot \lambda^{\alpha-1} e^{-\beta\lambda} (\beta\, d\lambda) \text{ for }\lambda>0. \tag 2 $$ The likelihood function is $$ \lambda\mapsto \frac{e^{-\lambda} \lambda^k}{k!}. $$ If we assume $(1)$, the multiplying the prior by the likelihood we get $$ \text{constant}\cdot \lambda^{\alpha+k-1} e^{-(1 + \frac 1 \beta)\lambda} \, d\lambda $$ and if we assume $(2)$, we get $$ \text{constant}\cdot \lambda^{\alpha+k-1} e^{-(1 + \beta)\lambda} \, d\lambda. $$ Either way, it's a Gamma distribution with $\alpha+k$ in place of $\alpha$ and the reciprocal of the scale parameter incremented by $1$.