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 $\displaystyle \lambda$.

This condition on $\displaystyle \lambda$ being known, that is, we could say that:

$\displaystyle P(X=k|\lambda)=f_{X|\lambda}(X|\lambda)=\frac{\lam bda^{k}}{k!}e^{-k}$.

And we suppose that $\displaystyle \lambda$ is also a random variable with probability density function given by

$\displaystyle f(\lambda)=e^{-\lambda}, \lambda > 0$.

how would you show that the unconditional marginal probability of X is

$\displaystyle P(X=k)=(\frac{1}{2})^{k+1} .......k=0,1,2..$

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

Re: Poisson problem with unconditional probability

P[X=k|λ].f(λ) = (e^{-2λ})(λ^{k})/(k!)

P[X=k] = ∫(e^{-2λ})(λ^{k})/(k!) dλ

= (1/k!) * ∫(λ^{k}e^{-2λ})dλ

Use Gamma function to find ∫(λ^k)(e^-2λ)dλ

or integration by parts

= (1/k!) * ∫(λ^{k})(e^{-2λ})dλ

= (1/k!) * k!/2^{k+1}

= (1/2)^^{k+1}