Geometric Distribution, Poisson

[Mad Scientists]

The problem is the following;

N has a geometric distribution with Pr(N=0)>0. M has a Poisson distribution. You are given:

E(N) = E(M); Var(N) = 2Var(M)

Calculate Pr (M>1).

From general knowledge we know that the expected value of a variable in a geometric distribution E(N) = q/p, and Var(N) = q/(p^2).
Also; the expected value of a variable in a Poisson distribution E(M) = lambda and Var(M) also = lambda.

I believe that the answer is 1 - pr(M=0) - pr(M=1) which is the equivalent of
1-e^(-lambda)-lambda*e^(-lambda).

But this would require solving for lambda, a feat I have not yet accomplished.


Any pointers?..

Thanks in advance,

Teddy

 

[EnumaElish]

You should be able to solve for p and \lambda, from \lambda = (1-p)/p, and 2\lambda = (1-p)/(p^2).

Note that Pr(N=0) = p > 0 so \lambda < +\infty.

 

[Mad Scientists]

Thanks Enuma, I was able to solve for it.

 

[jubs]

NM i saw the reply.