# Poisson problem with unconditional probability

• October 16th 2012, 05:11 PM
kingcobra
Poisson problem with unconditional probability
Hi all,

lets say that, the number of defects X per metre in a piece of fabric has a Poisson distribution with parameter $\lambda$.
This condition on $\lambda$ being known, that is, we could say that:
$P(X=k|\lambda)=f_{X|\lambda}(X|\lambda)=\frac{\lam bda^{k}}{k!}e^{-k}$.
And we suppose that $\lambda$ is also a random variable with probability density function given by
$f(\lambda)=e^{-\lambda}, \lambda > 0$.
how would you show that the unconditional marginal probability of X is
$P(X=k)=(\frac{1}{2})^{k+1} .......k=0,1,2..$
• October 16th 2012, 08:13 PM
chiro
Re: Poisson problem with unconditional probability
Hey kingcobra.

Recall Conditional Probability which says P(A|B) = P(A and B)/P(B) so P(X = k, lambda = whatever) = P(X = k|lambda)*P(lambda)

Now you have the joint distribution for X and lambda, so now given this can you sum out (or integrate out in continuous distribution) lambda to get P(X = k)?

Hint:

Marginal distribution - Wikipedia, the free encyclopedia
• October 17th 2012, 12:35 AM
sher84
Re: Poisson problem with unconditional probability
P[X=k|λ].f(λ) = (e-2λ)(λk)/(k!)

P[X=k] = ∫(e-2λ)(λk)/(k!) dλ

= (1/k!) * ∫(λke-2λ)dλ

Use Gamma function to find ∫(λ^k)(e^-2λ)dλ
or integration by parts

= (1/k!) * ∫(λk)(e-2λ)dλ

= (1/k!) * k!/2k+1

= (1/2)^k+1