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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.

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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.

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  • $\begingroup$ Thank you, but may I ask why do we need to perform change of variables? Also for the second question, I was wondering why it is possible to consider the mean in the Poisson distribution to make it in to a variable for the gamma distribution? $\endgroup$ Sep 19, 2018 at 5:47
  • $\begingroup$ Welcome. You need to perform change of variables that's what you did when you replaced $x$ with $\alpha \lambda$ and $k$ with $\beta$. Hence, you need to multiply with the appropriate Jacobian factor. As for the second question, this is due to the model. Usually parameters of distributions are also modeled as random variables, like your case. $\endgroup$ Sep 19, 2018 at 11:55

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