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$f(Θ)$ is pdf of gamma distribution

$$f(Θ) = \frac{λ^α}{Γ(α)}Θ^{α-1}\exp(-λΘ), $$

$$X\mid Θ \sim \mathrm{poisson}(Θ) \rightarrow \frac{Θ^x\exp(-Θ)}{x!}$$

Suppose that $Θ$ is a random variable that follows a gamma distribution with parameters $λ$ and $α$, where $α$ is an integer, and suppose that, conditional on $Θ$, $X$ follows a Poisson distribution with parameter $Θ$. Find the unconditional distribution of $α + X$ (Hint : Find the mgf by using iterated conditional expectations.

please answer..

Michael Hardy
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Sung Pang
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  • I don't know what the notation $\displaystyle gx\mid\Theta(x) = \cdots$ means. I'm guessing you meant the conditional distribution, given $\Theta$, of $X$, is a Poisson distribution with expected value $\Theta$. One could write $X\mid\Theta \sim \mathrm{Poisson}(\Theta)$. At any rate, one should not use the same symbol, in this case $\Theta$, to refer both to a random variable and to the argument to the density function. One can write $f_\Theta(\theta)=\text{a function of }\theta$. That is fairly standard. Distinguishing, for example, between $Y$ and $y$ makes it possible to${},\dots$ – Michael Hardy Oct 18 '15 at 04:21
  • $\ldots,{}$to understand something like $\Pr(Y\le y)$. ${}\qquad{}$ – Michael Hardy Oct 18 '15 at 04:21

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I will distinguish between $\Theta$ and $\theta$ as described in the comments. We have \begin{align} & \Pr(X=x) = \operatorname{E}(\Pr(X=x\mid\Theta)) = \operatorname{E}\left( \frac{\theta^x e^{-\theta}}{x!} \right) = \int_0^\infty \frac{\theta^x e^{-\theta}}{x!} f_\Theta(\theta)\,d\theta \\[10pt] = {} & \frac 1 {x!} \int_0^\infty \theta^x e^{-\theta} \frac{\lambda^\alpha}{\Gamma(\alpha)}\theta^{\alpha-1}e^{-\lambda\theta} \, d\theta = \frac {\lambda^\alpha} {x!\Gamma(\alpha)} \int_0^\infty \theta^{x+\alpha-1} e^{-\theta(1+\lambda)} \, d\theta \end{align}

The substitution $\eta = \theta(1+\lambda),\quad$ $\dfrac{d\eta}{1+\lambda} = d\theta$ then gets you some constant times a value of the Gamma function.

You will need to know why $\dfrac{\Gamma(x+\alpha)}{\Gamma(\alpha)} = \alpha(\alpha+1)(\alpha+2)\cdots(\alpha+x-1)$.

Michael Hardy
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