Geometric Distribution, Poisson

  • #1

Main Question or Discussion Point

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
 

Answers and Replies

  • #2
EnumaElish
Science Advisor
Homework Helper
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You should be able to solve for p and [itex]\lambda[/itex], from [itex]\lambda[/itex] = (1-p)/p, and 2[itex]\lambda[/itex] = (1-p)/(p^2).

Note that Pr(N=0) = p > 0 so [itex]\lambda < +\infty[/itex].
 
Last edited:
  • #3
Thanks Enuma, I was able to solve for it.
 
  • #4
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0
NM i saw the reply.
 

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