Bayesian statistics: show that corresponding posterior is a proper distribution

Question

We assume that $y_1,...,y_n$ are outcomes of iid. random variables $Y_1,...,Y_n$ with $Y_i \sim poiss(\lambda)$ Now we consider a Bayesian approach. I have found the likelihood to $Pr[\bar{y}=k/n]=Pr[\sum_{i=1}^n Y_i=k]=\frac{(n \lambda)^k e^{-n\lambda}}{k!}$ for $k \in \{0,1,2,...\}$ and shown that $\bar{y}=\frac{1}{n}\sum y_i$ is sucient for $\lambda$ . But now I consider the uniform prior for $\lambda$ i.e. $p(\lambda)\propto 1$. Since has support $(0, \infty)$, this does not integrate, and we say it is an improper prior. We have to show that the corresponding posterior is a proper distribution, namely the Gamma distribution with shape $n\bar{y} + 1$ and rate $n$ (scale $1/n$):

Answer 1

By Bayes's rule, $p(\lambda \mid y)$ is proportional to $p(\lambda) p(y \mid \lambda) \propto \frac{1}{k!} (n\lambda)^k e^{-n \lambda}$ where $k=\sum_{i=1}^n y_i$. Explicitly, the posterior density is $$p(\lambda \mid y) = c (n\lambda)^k e^{-n \lambda}$$ where $c^{-1} = \int_0^\infty (n \lambda)^k e^{-n \lambda} \, d\lambda$. Check that this is the Gamma distribution mentioned in the problem.