Question

Suppose that a random variable X has a Poisson distribution with parameter λ. The

parameter λ itself is a random variable with the exponential distribution with mean 1

c ,

where c is a constant. Show that

P(X = k) =

c

(c + 1)k+1

Answer 1

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