ke for k = 0, 1, 2, . . . . (2.15) Again, this gives rise to

Question:

λke−λ for k = 0, 1, 2, . . . . (2.15)

Again, this gives rise to a mass function since

∞X k=0 1

k!

λke−λ = e−λ ∞X k=0 1

k!

λk = e−λeλ = 1.

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Related Book For  book-img-for-question

Probability An Introduction

ISBN: 9780198709978

2nd Edition

Authors: Geoffrey Grimmett, Dominic Welsh

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