Mixed Distributions Homework: Showing Marginal Distribution of X

[cse63146]

Homework Statement

Let X ~ Poission ($$\lambda \theta$$). Suppose $$\lambda$$ ~ gamma (h,h^-1) and $$\theta$$ ~ generalized Pareto ($$\alpha$$, h^-1,k). Show that the marginal distribution of X is

$$\frac{\Gamma(\alpha + k) \Gamma(\alpha + h) \Gamma(\alpha + x) \Gamma(k + x)}{\Gamma(\alpha) \Gamma(h) \Gamma(k) \Gamma(\alpha + h + k + x)x!}$$

Homework Equations

The Attempt at a Solution

My prof didnt really explain it.

From what I've gathered in the book it's:

$$\int^{\infty}_{- \infty} d \lambda d \theta$$ of the distribution of X, lambda, and theta.

 

[mighty2000]

Firstly, I would like to clarify that X is not a distribution, but rather a random variable. The distribution in question is the Poisson distribution, denoted by X ~ Poisson(\lambda \theta).

To find the marginal distribution of X, we need to integrate out the variables \lambda and \theta from the joint distribution of X, \lambda, and \theta.

Using the definition of the Poisson distribution, we have:

P(X = x) = \frac{(\lambda \theta)^x e^{-\lambda \theta}}{x!}

To integrate out \lambda, we use the fact that \lambda ~ gamma(h,h-1). We know that the gamma distribution has the following probability density function:

f(\lambda) = \frac{h^{h-1}}{\Gamma(h)} \lambda^{h-1} e^{-h \lambda}

Therefore, the integral becomes:

\int^{\infty}_{0} \frac{h^{h-1}}{\Gamma(h)} \lambda^{h-1} e^{-h \lambda} \frac{(\lambda \theta)^x e^{-\lambda \theta}}{x!} d \lambda

We can simplify this using the fact that \Gamma(h) = (h-1)! and \Gamma(h-1) = (h-2)!. This gives us:

\frac{h^{h-1}}{(h-1)!} \int^{\infty}_{0} \lambda^{h-1} e^{-\lambda (h + \theta)} \frac{(\lambda \theta)^x}{x!} d \lambda

Next, we integrate out \theta using the fact that \theta ~ generalized Pareto(\alpha, h-1, k). The probability density function for the generalized Pareto distribution is given by:

g(\theta) = \frac{(\alpha + k - 1)(\alpha + h - 1)(\alpha + x - 1)}{(\alpha - 1)(h - 1)(k - 1)} \theta^{-\alpha - 1} (1 + \frac{\theta}{k-1})^{-(\alpha + h + x)}

Therefore, the integral becomes:

\frac{h^{h-1}}{(h-1)!} \frac{(\alpha + k - 1)(\alpha + h - 1)(