conditional distribution of $A\mid N = k$ 
Can someone help me with how to get part 3 of the question?
 A: 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.
A: 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 $1$.
