help with poisson distribution...?

Question

let y be a random variable having a poisson distribution with parameter λ and that the conditional distribution of X given Y =y is binomilly distributed with parameters y and p. Show that that the distribution of X is Poisson with parameter λp.


thanks for the help... :D

Answer 1

This is called the "thinned" Poisson process
P(XY)=ΣP(X|Y)P(Y) summed over the Y terms to get
P(X)
=ΣxCk*p^k* (1-p)^(x-k) exp^(-λ) λ^x /x!
sum from x=k to

simplifying we have

Σ1/[(x-k)!k!]*p^p*(1-p)^x*/ (1-p)^k λ^x* exp(-λ)

Let x=y+k
plugging in and simplifying

Σ1/(y!k!)p^k[(1-p)^(y+k) λ^(y+k)]/ (1-p)^k* exp(-λ)

Σ1/(y!k!)p^k(1-p)^yλ^y* λ^k * exp(-λ)

now take all of the none-y terms out from the sum

exp(-λ)p^k*λ^k/k! Σ1/y! (1-p)^y λ^y
exp(-λ)p^k*λ^k/k!Σ 1/y! ([1-p]λ)^y

now what's in the sum is exp(λ-pλ)

so we have exp(-λ)*exp(λ-pλ)*p^k*λ^k/k!
which simplifies to
exp(-pλ)*[pλ]^k/k! QED