I am reading a paper for Expectation Maximization algorithm which can be found here.Is from a greek professor of statistics with the title of the paper "EM ALGORITHM FOR MIXED POISSON AND OTHER DISCRETE DISTRIBUTIONS"
I have a problem in page 7 that says : " Suppose that $$g(\theta | \lambda) = \lambda \exp (– \theta \lambda), \theta, \lambda > 0,$$ i.e. $\theta_{i}$ follows an exponential distribution with mean $ \lambda^{-1}$, then the resulting geometric distribution has probability function given by:
$$P(x| \lambda) = (\frac{ \lambda}{1+ \lambda})(\frac{1}{ 1+\lambda})^{x}$$
$x = 0,1,\dots,\lambda>0$ My questions are:
1) How he ended in the geometric distribution?
2) How someone can use the $$P(x|\phi) = \int_{0}^{\infty} \frac{\exp(-\theta)\theta^{x}}{x!} d G(\theta|\phi)$$ as the probability function of mixed Poisson distribution in the above ? Or it is irrelevant?
The paper initially assumes that $X|\theta \sim \text{Poisson}(\theta)$. Then, under the case (in 4.1) that $\theta|\lambda \sim \text{expo}(\lambda)$, we can compute:
\begin{align*} p(x|\lambda) &= \int_0^\infty p(x,\theta|\lambda) d\theta\\ &= \int_0^\infty p(x|\theta)p(\theta|\lambda) d\theta\\ &= \int_0^\infty \frac{e^{-\theta}\theta^x}{x!} \lambda e^{-\theta \lambda} d\theta\\ &=\frac{\lambda}{x!} \int_0^\infty \theta^x e^{-\theta (\lambda+1)} d\theta\\ \end{align*} The easiest way to solve this integral is to note that the integrand can be rewritten so that it is the pdf of a Gamma random variable with shape parameter $x+1$ and rate parameter $\lambda + 1$. So
\begin{align*} \int_0^\infty \theta^x e^{-\theta (\lambda+1)} d\theta &=\frac{\Gamma(x+1)}{(\lambda + 1)^{(x+1)}} \int_0^\infty \frac{(\lambda + 1)^{(x+1)}}{\Gamma(x+1)}\theta^x e^{-\theta (\lambda+1)} d\theta\\ &=\frac{\Gamma(x+1)}{(\lambda + 1)^{(x+1)}} \int_0^\infty \text{Gamma}(\theta; x+1, \lambda+1) d\theta\\ &=\frac{\Gamma(x+1)}{(\lambda + 1)^{(x+1)}} \end{align*}
Putting this back in to the previous working we get: \begin{align*} p(x|\lambda) =\frac{\lambda}{x!} \frac{\Gamma(x+1)}{(\lambda + 1)^{(x+1)}} =\frac{\lambda}{\lambda+1} \frac{1}{(\lambda + 1)^{x}} \end{align*}