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

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