IJMMS 2003:48, 3041–3046
PII. S0161171203212229
http://ijmms.hindawi.com
© Hindawi Publishing Corp.
A NONUNIFORM BOUND FOR THE APPROXIMATION
OF POISSON BINOMIAL BY POISSON DISTRIBUTION
K. NEAMMANEE
Received 18 December 2002
It is well known that Poisson binomial distribution can be approximated by Poisson
distribution. In this paper, we give a nonuniform bound of this approximation by
using Stein-Chen method.
2000 Mathematics Subject Classification: 60F05, 60G50.
1. Introduction and main result. Let X1 , X2 , . . . , Xn be independent, possibly
not identically distributed, Bernoulli random variables with P (Xi = 1) = 1 −
P (Xi = 0) = pi and let Sn = X1 + X2 + · · · + Xn . The sum of this kind is often
called a Poisson binomial random variable. In the case where the “success”
probabilities are all identical, pi = p, S is the binomial random variable Ꮾ(n, p).
n
Let λ = i=1 pi and let ᏼλ be the Poisson random variable with parameter λ,
that is, P (ᏼλ = ω) = e−λ λω /ω! for all nonnegative integers ω. It has long been
known that if pi ’s are small, then the distribution of Sn can be approximated
by a distribution of ᏼλ (see, e.g., Chen [2]).
In this paper, we investigate the bound of this approximation. As an illustration, we look at the case of p1 = p2 = · · · = pn = p. There are at least three
√
known uniform bounds: Kennedy and Quine [6] showed that, for 0 < λ ≤ 2− 2,
P Sn ≤ ω − P ᏼλ ≤ ω ≤ 2λ (1 − p)n−1 − e−np ,
(1.1)
Barbour and Hall [1] showed that
P Sn ≤ ω − P ᏼλ ≤ ω ≤ min(p, λp),
(1.2)
and Deheuvels and Pfeifer [5] proved that
P Sn ≤ ω − P ᏼλ ≤ ω (a−1)
(a − np) (np)(b−1) (b − np)
−λ (np)
≤ λpe
−
+R
a!
b!
(1.3)
with a = [np + 1/2 + np + 1/4], b = [np + 1/2 − np + 1/4], and |R| ≤
(1/2)(2p)3/2 /(1 − 2p), for 0 < p < 1/2, and [x] is understood to be the integer part of x.
3042
K. NEAMMANEE
For the general case, Le Cam [7] investigated and showed that
∞ n
−λ ω P Sn = ω − e λ ≤ 16
p2 .
ω!
λ i=1 i
ω=0
(1.4)
It can be observed that the constant 16/λ will be large when λ is small. Stein
[10] used the method of Chen [3] to improve the bound and showed that
n
P Sn ≤ ω − P ᏼλ ≤ ω ≤ λ−1 ∧ 1
p2
i=1
i
(1.5)
for ω = 0, 1, 2, . . . , n and λ−1 ∧ 1 = min(λ−1 , 1). In case when λ tends to 0, one
can see that (1.5) becomes
n
P Sn ≤ ω − P ᏼλ ≤ ω ≤
p2 .
i=1
(1.6)
i
In this paper, we consider a nonuniform bound when λ is small, that is,
λ ∈ (0, 1] and ω ∈ {1, 2, . . . , n − 1}. Note that, when ω ∉ {1, 2, . . . , n − 1}, we can
compute the exact probabilities, that is,
n
P Sn = 0 =
1 − pi ,
n
pj ,
P Sn = n =
i=1
P Sn = ω = 0,
j=1
(1.7)
ω = n + 1, n + 2, . . . .
In finding the uniform bound, there are several techniques which can be used;
for example,
(i) the operator method initiated in Le Cam [7],
(ii) the semigroup approach due to Deheuvels and Pfeifer [4],
(iii) the Chen-Stein technique, see Chen [3] and Stein [10],
(iv) direct computations as in Kennedy and Quine [6],
(v) the coupling method, see Serfling [8] and Stein [10].
In the present paper, our argument closely follows the Chen-Stein technique
in Chen [3] and Stein [10]. The following theorem is our main result.
Theorem 1.1. Let λ ∈ (0, 1] and ω0 ∈ {1, 2, . . . , n − 1}. Then
n
P Sn = ω0 − P ᏼλ = ω0 ≤ 1
p2 .
ω0 i=1 i
(1.8)
2. Proof of the main result. Stein [9] gave a new technique to find a bound
in the normal approximation to a distribution of a sum of dependent random
variables. His technique was free from Fourier methods and relied instead on
the elementary differential equation
f (ω) − wf (ω) = h(ω) − N(h),
(2.1)
A NONUNIFORM BOUND FOR THE APPROXIMATION OF POISSON . . .
3043
where h is a function that is used to test convergence and N(h) = E[h(Z)]
where Z is the standard normal. Chen [3] applied Stein’s ideas in the Poisson
setting. Corresponding to the differential equation in the normal case above,
one has an analogous difference equation
λf (ω + 1) − ωf (ω) = h(ω) − ᏼλ (h),
(2.2)
where ᏼλ (h) = E[h(ᏼλ )] and f and h are real-valued functions defined on
Z+ ∪ {0}. Let ω0 ∈ {1, 2, . . . , n − 1} and define h, hω0 : Z+ ∪ {0} → R by
h(ω) =

1,
if ω = ω0 ,
0,
if ω ≠ ω0 ,

1,
hω0 (ω) =
0,
if ω ≤ ω0 ,
if ω > ω0 .
(2.3)
Then we see that the solution f of (2.2) can be expressed in the form
 (ω − 1)!


λω0 −ω ᏼλ 1 − hω−1 ,



ω
!
0


fω0 (ω) = − (ω − 1)! λω0 −ω ᏼ h

λ
ω−1 ,


ω0 !




0,
if ω0 < ω,
if ω0 ≥ ω > 0,
(2.4)
if ω = 0,
λE fω0 Sn + 1 − E Sn fω0 Sn = P Sn = ω0 − P ᏼλ = ω0 .
(2.5)
(i)
Let Sn = Sn − Xi for i = 1, 2, . . . , n. By using the facts that each Xj takes on
values 0 and 1 and that Xj ’s are independent, we have
n
E Sn fω0 Sn =
pi E f Sn(i) + 1
i=1
n
= λE fω0 Sn + 1 +
pi E fω0 Sn(i) + 1 − fω0 Sn + 1
i=1
n
pi E Xi fω0 Sn(i) + 1 − fω0 Sn(i) + 2
= λE fω0 Sn + 1 +
i=1
n
= λE fω0 Sn + 1 +
pi2 E fω0 Sn(i) + 1 − fω0 Sn(i) + 2 ,
i=1
(2.6)
which implies, by (2.5), that
n
pi2 E fω0 Sn(i) + 2 − fω0 Sn(i) + 1 .
P Sn = ω0 − P ᏼλ = ω0 =
i=1
(2.7)
3044
K. NEAMMANEE
From (2.4), it follows that
fω0 (ω + 2) − fω0 (ω + 1)

ω! 

(ω + 1)ᏼλ hω+1 − λᏼλ hω ,
−λω0 −ω−2



ω
!

0



ω! (ω + 1)ᏼλ 1 − hω+1 + λᏼλ hω ,
= λω0 −ω−2

ω
!

0




ω! 

λω0 −ω−2
(ω + 1)ᏼλ 1 − hω+1 − λᏼλ 1 − hω ,
ω0 !
if ω ≤ ω0 − 2,
if ω = ω0 − 1,
if ω ≥ ω0 .
(2.8)
Case 1 (ω ≤ ω0 −2). Since
ω+1
λk
(ω + 1 − k),
(ω + 1)ᏼλ hω+1 − λᏼλ hω = e−λ
k!
k=0
(2.9)
we have


ω+1
λk
fω (ω + 2) − fω (ω + 1) = λ(ω0 −2)−ω ω! e−λ
(ω + 1 − k)
0
0
ω0 !
k!
k=0


ω+1
(ω + 1)!  −λ λk 
≤
e
ω0 !
k!
k=0
(2.10)
ω0 − 1 !
≤
ω0 !
=
1
,
ω0
where we have used the facts that λ ∈ (0, 1] and 0 ≤ ω + 1 − k ≤ ω + 1 in the
ω+1
first inequality and the conditions ω ≤ ω0 − 2 and e−λ k=0 (λk /k!) ≤ 1 in the
second inequality.
Case 2 (ω = ω0 − 1). We have


ω0 −1 k
∞
−1
λ
λk
−λ
fω (ω + 2) − fω (ω + 1) = λ ω0 e−λ

+ λe
0
0
ω0
k!
k!
k=ω +1
k=0
0

ω0 −1
∞
k+1
λ−1  −λ λk
λ

≤
+ e−λ
k
(k + 1)
e
ω0
k!
(k + 1)!
k=ω +1
k=0

0
=
λ−1 E ᏼλ
ω0
=
1
.
ω0
(2.11)
A NONUNIFORM BOUND FOR THE APPROXIMATION OF POISSON . . .
3045
Case 3 (ω ≥ ω0 ). Since
2λω+3
3λω+4
1λω+2
+
+
+···
(ω + 2)! (ω + 3)! (ω + 4)!
ω 0 λ ω0
ω0 + 1 λω0 +1
ω−ω0 +2
≤λ
+
+···
ω0 + 1 ! ω0 + 2 · · · (ω + 3)
ω0 ! ω0 + 1 · · · (ω + 2)


∞
kλk 
λω−ω0 +2

≤
ω0 + 1 ω0 + 2 · · · (ω + 2) k=ω k!
e λ
E ᏼλ
≤
ω0 + 1 ω0 + 2 · · · (ω + 2)
λ
0
ω−ω0 +2
eλ λω−ω0 +3
=
,
ω0 + 1 ω0 + 2 · · · (ω + 2)
∞
(ω + 1)ᏼλ 1 − hω+1 − λᏼλ 1 − hω = −e−λ
λk k − (ω + 1) < 0,
k!
k=ω+2
(2.12)
we have
∞
λk fω (ω + 2) − fω (ω + 1) = λω0 −ω−2 ω! e−λ
k − (ω + 1)
0
0
ω0 !
k!
k=ω+2
λω!
1
≤
≤
.
(ω + 2)!
(ω + 1)(ω + 2)
(2.13)
From Cases 1, 2, and 3, we conclude that
fω (ω + 2) − fω (ω + 1) ≤ 1 .
0
0
ω0
(2.14)
By (2.7) and (2.14), we have
P Sn = ω0 − P ᏼλ = ω0 

n
n
1 2
2
pi E fω0 Sn(i) + 2 − fω0 Sn(i) + 1 ≤
p .
≤
ω0 i=1 i
i=1
(2.15)
Acknowledgment. The author would like to thank the insightful comments from the referees and financial support by Thailand Research Fund.
References
[1]
[2]
[3]
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3046
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[10]
K. NEAMMANEE
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K. Neammanee: Department of Mathematics, Faculty of Sciences, Chulalongkorn University, Bangkok 10330, Thailand
E-mail address: kritsana.n@chula.ac.th