Thread: 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 .
This condition on being known, that is, we could say that:
.
And we suppose that is also a random variable with probability density function given by
.
how would you show that the unconditional marginal probability of X is

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

P[X=k|λ].f(λ) = (e^-2λ)(λ^k)/(k!)

P[X=k] = ∫(e^-2λ)(λ^k)/(k!) dλ

= (1/k!) * ∫(λ^ke^-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