I had the expression $$\frac{1}{p(y)}(n\lambda)^k e^{-n \lambda}$$ for $k \in \{0,1,2,...\}$ where we here have that $k=\sum_{i=1}^n y_i$ where I think I can ignore $p(y)$, but if not it is given by $p(y)=\int_0^\infty (n \lambda)^k e^{-n \lambda} \, d\lambda$. And we have that $\bar{y}=\frac{1}{n}\sum y_i$. Now I have to show that this is Gamma distribution with shape $n\bar{y} + 1$ and rate $n$ (scale $\frac{1}{n}$). I have tried many times but can not figure it out how to show it without getting problems. Can anyone help me?
If I calculate backwards I think if I insert the given shape and rate I get: $$f(\lambda,n \bar{y}+1,n)=\frac{n^{n \bar{y}+1}}{\Gamma(n\bar{y} + 1)}\lambda^{n\bar{y}}e^{-n \lambda}$$ But I can figure out how to rewrite the two expressions in order to show they are the same. Can anyone help me?
We have
$$f(\lambda) = \frac{n^k}{p(y)}\lambda^ke^{-n\lambda}$$
is a density function where $\frac{n^k}{p(y)}$ is independent of $\lambda$.
Hence, by comparing with the pdf of gamma distribution, we conclude that it is a gamma distribution with the shape parameter being $k+1$ and the corresponding rate is $n$.
Note that the shape is equal to
$$k+1 = \sum_{i=1}^n y_i + 1 = n\bar{y} + 1.$$