# Thread: Distribution of a + X

1. ## Distribution of a + 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, Χ follows a poisson distribution with parameter Θ. Find the unconditional distribution of α + Χ. (Hint: Find the mgf by using iterated conditional expectations)

2. Use the hint.

$E(e^{(a+X)t}) =E(E(e^{(a+X)t})|\theta)$

$=e^{at}E(E(e^{tX})|\theta) =e^{at}E(e^{\theta(e^t-1)})$

$=e^{at}M_{\theta}(e^t-1)=e^{at}M_{\theta}(s)$

SO, plug in $s=e^t-1$ into the MGF of your gamma distribution.

3. Did you get an interesting result?
I didn't finish this, because I think that's your job.
But I would like to see what your final result is.

4. I'll post the result when I'm finished.

One interesting thing I've found:

$X \sim Negative Binomial(\alpha, \frac{1}{1+\lambda})$

5. I just did this in the past fifteen minutes so please let me know if things go awry.

$M_{a+X} = E[e^{(a+X)t}] = e^{at}M_{\theta} (s) = e^{at} (\frac{\lambda}{\lambda -(e^t - 1)})^a$

Now,

$M'_{a+X} (0) = a(\frac{\lambda + a}{\lambda}) = E[a+X]$

Also,

$E[a+X] = \int_{-\infty}^{\infty} (a+x)f_{a+X} (x) dx$

Therefore,

$\int_{-\infty}^{\infty} (a+x)f_{a+X} (x) dx = a(\frac{\lambda + 1}{\lambda})$

And I'm stuck here. Can I take the derivative of both sides or something?

$f_{a+x} (x) = \frac{1}{(a+x)}$

6. $\alpha + X \sim Negative Binomial$