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Let $f(x;\theta)$ be the poisson frequency function with mean $\lambda$. and $p(\lambda)$ the Gamma distribution with mean $\mu$, and variance $\mu^2/\alpha$.

I have to show that $g(x)=\frac{\Gamma(x+\alpha)}{x!\Gamma(\alpha)}\left(\frac{\alpha}{\alpha+\mu}\right)^{\alpha}\left(\frac{\mu}{\alpha+\mu}\right)^{x}$.

However, from my calculations:

$\displaystyle g(x)=\int f(x;\theta)p(\lambda) d\lambda=\int\frac{e^{-\lambda}\lambda^x}{x!}\cdot\frac{\mu^{\alpha}\lambda^{\alpha-1}}{\Gamma(\alpha)}e^{-\mu\lambda}d\lambda = \frac{\Gamma(\alpha+x)}{x!\Gamma(\alpha)}\left(1/(\mu+1\right))^{x+\alpha}\mu^{\alpha}$, where the last equality is obtained by doing a coordinate change to $z=\lambda(\mu+1)$

So where did I go wrong?

Any help will be appreciated.

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1 Answer 1

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First make sure your parameters for the Gamma distribution are correct:
$G(a,b)$ has mean $ab$ and variance $ab^2$. Hence you have: $$ ab = \mu \\ ab^2 = \mu^2/\alpha $$ Solve to get: $$ a = \alpha \\ b = \mu/\alpha $$

Hence your integral should be: $$ \int_0^\infty \frac{e^{-\lambda} \lambda^x}{x!} \frac{\lambda^{\alpha-1}e^{-\frac{\alpha\lambda}{\mu}}}{(\mu/\alpha)^\alpha \Gamma(\alpha)} d\lambda $$

Try to solve this from here.

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  • $\begingroup$ Many thanks @ved. ;) $\endgroup$ Commented Jul 1, 2014 at 19:35

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