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Robout

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posted by Robout on 11/5/2009 12:15:41 AM | status: Live | Earned Karma: 25

Poisson random variables.

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Statistics and Probability	N/A	N/A	N/A	11/5/2009 at 11:00:00 PM

Question Details:

Suppose that Y₁and Y₂are independent Poisson random variables with mean m₁ and m₂ respectively. MGF of a Poisson random variable Y with mean λ is

m_y(t) = exp[λ(e^t- 1)]

Find the probabilty function p(u) of U = Y₁+ Y₂ using MGF method.

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Orca

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posted by Orca on 11/5/2009 2:58:08 AM | status: Live

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MGF of the sum of two iid random variables is the product of their MGF's. Hence

since Y₁ and Y₂ are poisson with mean m₁ and m₂ we have

MGF_U=exp[m₁(e^t-1)]exp[m²(e^t-1)]=exp[(m₁+m₂)(e^t-1)]

Which is the MGF for a poisson distribution with mean m₁+m_2.

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