Answer to Question #317002 in Statistics and Probability for Dinor123

Question

For each fixed λ > 0, let X have a Poisson distribution with parameter λ. Suppose λ

itself is a random variable with the gamma distribution

f(λ) =







1

Γ(n)

λ

n−1

e

−λ

, λ ≥ 0

0, λ < 0

where n is a fixed positive constant. Show that

P(X = k) = Γ(k + n)

Γ(n)Γ(k + 1) 

1

2

k+n

, k = 0, 1, 2

Answer 1

$P\left( X=k \right) =\int_0^{+\infty}{P\left( X=k|\lambda =t \right) f_{\lambda}\left( t \right) dt}=\\=\int_0^{+\infty}{\frac{t^ke^{-t}}{k!}\frac{t^{n-1}e^{-t}}{\varGamma \left( n \right)}dt}=\\=\frac{1}{k!\varGamma \left( n \right)}\int_0^{+\infty}{t^{k+n-1}e^{-2t}dt}=\left[ 2t=x \right] =\\=\frac{1}{k!\varGamma \left( n \right)}\int_0^{+\infty}{\frac{1}{2^{k+n}}x^{k+n-1}e^{-x}dx}=\frac{\varGamma \left( k+n \right)}{\varGamma \left( k+1 \right) \varGamma \left( n \right) 2^{k+n}}$