Gamma distribution with respect to the Poisson distribution defined by:
$$P(N=n|\Lambda= \lambda)=\frac{e^{-\lambda}\lambda^n}{n!}$$
Suppose that $\Lambda$ has a scale parameter $\alpha$ and shape parameter $\beta$, the we have the probability distribution for $\Lambda:$
$$G(\lambda)=\alpha^{\beta}\lambda^{\beta-1}e^{-\alpha \lambda}\frac{1}{\Gamma(\beta)}$$
But by the gamma distribution, $$f(x)=x^{k-1}e^{-x}\frac{1}{\Gamma(k)}$$
Question 1. If we plug in $x = \alpha\lambda$, and $k = \beta,$ then why is it:
$$G(\lambda)=\alpha^{\beta}\lambda^{\beta-1}e^{-\alpha \lambda}\frac{1}{\Gamma(\beta)}$$
not;
$$G(\lambda)=\alpha^{\beta-1}\lambda^{\beta-1}e^{-\alpha \lambda}\frac{1}{\Gamma(\beta)}$$
Question 2: Why is it possible to distribute $\Lambda$ as a gamma distribution?
Can anyone please help me clear this confusion? Thank you.
Best Answer
So your change of variable is this \begin{align} x &= \alpha \lambda \\ k &= \beta \end{align} The Jacobian of this change of basis is \begin{equation} J = \begin{bmatrix} \frac{\partial x}{\partial \lambda} & \frac{\partial k}{\partial \lambda} \\ \frac{\partial x}{\partial \beta} & \frac{\partial k}{\partial \beta} \\ \end{bmatrix} = \begin{bmatrix} \alpha & 0 \\ 0 & 1 \end{bmatrix} \end{equation} So $\det J = \alpha$. So, upon going from one basis to another in PDFs, you have to multiply by $\det J$, and that is where your extra $\alpha$ comes from.
Question $2$ is not so clear.