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
Probability An Introduction
ISBN: 9780198709978
2nd Edition
Authors: Geoffrey Grimmett, Dominic Welsh
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