The problem is the following;(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Geometric Distribution, Poisson

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