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4.2 The Poisson Distribution
223
Proof We begin by rewriting the binomial probability in terms of λ:
lim
n→∞
n k
p (1 − p)
k
n−k
n λ k λ n−k
1−
= lim
n→∞ k
n
n
1
n!
λ −k
λ n
k
λ
= lim
1−
1−
n→∞ k!(n − k)!
nk
n
n
n
k
1
n!
λ
λ
=
1−
lim
k! n→∞ (n − k)! (n − λ)k
n
But since [1 − (λ/n)]n → e−λ as n → ∞, we need show only that
n!
→1
(n − k)!(n − λ)k
to prove the theorem. However, note that
n!
n(n − 1) · · · (n − k + 1)
=
(n − k)!(n − λ)k (n − λ)(n − λ) · · · (n − λ)
a quantity that, indeed, tends to 1 as n → ∞ (since λ remains constant).
Example
4.2.1
Theorem 4.2.1 is an asymptotic result. Left unanswered is the question of the relevance of the Poisson limit for finite n and p. That is, how large does n have to be and
how small does p have to be before e−np (np)k /k! becomes a good approximation to
the binomial probability, p X (k)?
Since “good approximation” is undefined, there is no way to answer that question in any completely specific way. Tables 4.2.1 and 4.2.2, though, offer a partial
solution by comparing the closeness of the approximation for two particular sets of
values for n and p.
In both cases λ = np is equal to 1, but in the former, n is set equal to 5—in the
latter, to 100. We see in Table 4.2.1 (n = 5) that for some k the agreement between
the binomial probability and Poisson’s limit is not very good. If n is as large as 100,
though (Table 4.2.2), the agreement is remarkably good for all k.
Table 4.2.1 Binomial Probabilities and Poisson
k
0
1
2
3
4
5
6+
Limits; n = 5 and p = 15 (λ = 1)
5
e−1 (1)k
(0.2)k (0.8)5−k
k
k!
0.328
0.410
0.205
0.051
0.006
0.000
0
1.000
0.368
0.368
0.184
0.061
0.015
0.003
0.001
1.000