2. Let X be a random variable with the Poisson distribution, parameter 2. Show that, for w =1, 2, 3, ...,P(X > w) = P(Y < 2),where Y is a random variable having the gamma distribution with parameters w and 1.

X~Poisson(λ) Therefore P(X=x)=e-λλxx! ; x=0,1,2,… PX≥w=∑i=w∞P(X=i)=∑i=w∞e-λλii!=e-λ∑i=0∞e-λλii!-∑i=0w-1e-λλii!=e-λeλ-eλΓw,λΓw Simplify further. PX≥w=1-Γw,λΓwNow consider Y~Gammaw,1. fλ=e-λλw-1Γw for λ>0 P(Y≤λ)=∫0λe-λλw-1Γwdλ=Γ(w)-Γ(w,λ)Γ(w)=1-Γ(w,λ)Γ(w) Therefore PX≥w=PY≤λ.

Math

Probability

2. Let X be a random variable with the Poisson distribution, parameter 2. Show that, for w = 1, 2, 3, ..., P(X > w) = P(Y < 1), where Y is a random variable having the gamma distribution with parameters w and 1.

2. Let X be a random variable with the Poisson distribution, parameter 2. Show that, for w = 1, 2, 3, ..., P(X > w) = P(Y < 1), where Y is a random variable having the gamma distribution with parameters w and 1.

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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2. Let X be a random variable with the Poisson distribution, parameter 2. Show that, for w =
1, 2, 3, ...,
P(X > w) = P(Y < 2),
where Y is a random variable having the gamma distribution with parameters w and 1.

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