A Non-Uniform Bound on Poisson Approximation for ... - CiteSeerX

0 downloads 0 Views 230KB Size Report
for approximating the distribution of W (for example, see Stein [10], Arratia,. Goldstein and Gordon [1-2], Barbour, Holst and Janson [4], Neammanee [8] and.
Thai Journal of Mathematics Volume 4 (2006) Number 1 : 179–196

A Non-Uniform Bound on Poisson Approximation for Sums of Bernoulli Random Variables with Small Mean K. Teerapabolarn Abstract : In many situations, the Poisson approximation is appropriate for sums of Bernoulli random variables where its mean, λ, is small. In this paper, we give non-uniform bounds of Poisson approximation with small values of λ, λ ∈ (0, 3], by using the Stein-Chen method. These bounds are sharper than the bounds of Teerapabolarn and Neammanee [12]. Keywords : Bernoulli summands, non-uniform bound, Poisson approximation, Stein-Chen method. 2000 Mathematics Subject Classification : 60F05, 60G05.

1

Introduction

P Consider a random variable W that can be written as a sum α∈Γ Xα of Bernoulli random variables, where Γ is an arbitrary finite index set. The random variables Xα may be dependent, and we will be interested in the case where each of success probability pα = P (Xα = 1) = 1 − P (Xα = 0) is small. It is then reasonable to approximate the distribution of W by Poisson distribution with P mean λ = E[W ] = α∈Γ pα . In the past few years, many mathematicians have been developed the method for approximating the distribution of W (for example, see Stein [10], Arratia, Goldstein and Gordon [1-2], Barbour, Holst and Janson [4], Neammanee [8] and Teerapabolarn and Neammanee [11]). In 2006, Teerapabolarn and Neammanee [12] gave three formulas of nonuniform bounds as follows. For w0 ∈ {0, 1, . . . , |Γ|}, a non-uniform bound by using the coupling method is of the form ¯ ¯ ¾X ½ w0 ¯ X λk e−λ ¯¯ eλ ¯ pα E|W − Wα∗ |, ¯P (W ≤ w0 ) − ¯ ≤ λ−1 (1 − e−λ ) min 1, ¯ k! ¯ w0 + 1 k=0

α∈Γ

(1.1) where Wα∗ is a random variable which has the same distribution as W − Xα conditional on Xα = 1, i.e. Wα∗ ∼ (W − Xα )|Xα = 1. Suppose that for each α there

180

Thai J. Math. 4(2006)/ K. Teerapabolarn

is a subset Γα Γ such that Xα is independent of the collection {Xβ : β ∈ / Γα }, a non-uniform bound in this case is in the form of ¯ ¯ w0 ¯ X λk e−λ ¯¯ ¯ ¯ ¯P (W ≤ w0 ) − ¯ k! ¯ k=0   ¾ X ½  λ X X e ≤ λ−1 (1 − e−λ ) min 1, p2α + (pα pβ + E[Xα Xβ ]) .  w0 + 1  α∈Γ

α∈Γ β∈Γα \{α}

(1.2) In the case of independent summands, (1.2) becomes ¯ ¯ ¾X ½ w0 ¯ X λk e−λ ¯¯ eλ ¯ −1 −λ p2α . ¯ ≤ λ (1 − e ) min 1, ¯P (W ≤ w0 ) − ¯ k! ¯ w0 + 1

(1.3)

α∈Γ

k=0

We know that in many situations, the Poisson approximation is appropriate for W with small values of λ. In this paper, we improve the bounds in (1.1), (1.2) and (1.3) to be more accurately when λ is small, i.e. λ ∈ (0, 3]. Let  −2 λ (λ + e−λ − 1), if w0 = 0,        ¾ ½   1 − e−λ (1 + λ) (1 − e−λ )(1 + λ)    , , if w0 = 1, max   λ2 6λ       ( )  2 2  (2 + λ)[1 − e−λ (1 + λ + λ2 )] 4(1 − e−λ )(1 + λ + λ2 ) + λ 4(λ, w0 ) = max , ,  λ3 60λ      if w0 = 2,       ½ ¾    [w0 !(w0 − λ) + e−λ λw0 +1 ](w0 + 2) 2λ−1 (1 − e−λ )eλ + 1   max , ,   w0 (w0 + 2 − λ)(w0 + 1)! 3(w0 + 1)(w0 + 2)    if w0 ∈ {3, ..., |Γ|}. (1.4) Then the followings are our main results. Theorem 1.1 Let w0 ∈ {0, 1, . . . , |Γ|} and λ ∈ (0, 3]. Then (i)

¯ ¯ w0 ¯ X X λk e−λ ¯¯ ¯ pα E|W − Wα∗ |, ¯P (W ≤ w0 ) − ¯ ≤ 4(λ, w0 ) ¯ k! ¯ k=0

α∈Γ

where there exists Wα∗ such that Wα∗ ∼ (W − Xα )|Xα = 1 and

(1.5)

A Non-Uniform Bound on Poisson Approximation for Sums of Bernoulli . . . 181 (ii) ¯ ¯ w0 ¯ X λk e−λ ¯¯ ¯ ¯P (W ≤ w0 ) − ¯ ¯ k! ¯ k=0  X X ≤ 4(λ, w0 ) p2α +  α∈Γ

X

α∈Γ β∈Γα \{α}

  (pα pβ + E[Xα Xβ ]) ,  (1.6)

where there exists Γα such that Xα is independent of {Xβ : β ∈ / Γα } for every α ∈ Γ. From (1.6), if W ≥ Wα∗ or W − Xα ≤ Wα∗ for every α ∈ Γ, then we have the convenience forms in Theorem 1.2.

Theorem 1.2 Let w0 ∈ {0, 1, ..., |Γ|} and λ ∈ (0, 3]. Then (i) ¯ ¯ w0 ¯ X λk e−λ ¯¯ ¯ ¯ ≤ 4(λ, w0 ){λ − V ar[W ]}, ¯P (W ≤ w0 ) − ¯ k! ¯

(1.7)

k=0

where W ≥ Wα∗ a.s. for every α ∈ Γ and (ii) ¯ ¯ ( ) w0 ¯ X X λk e−λ ¯¯ ¯ 2 pα , (1.8) ¯ ≤ 4(λ, w0 ) V ar[W ] − λ + 2 ¯P (W ≤ w0 ) − ¯ k! ¯ α∈Γ

k=0

where W − Xα ≤ Wα∗ a.s. for every α ∈ Γ. If {Xα : α ∈ Γ} is independent, we have Wα∗ = W − Xα and E|W − Wα∗ | = pα . Then the following corollary follows immediately from (1.6). Corollary 1.3 Let λ ∈ (0, 3] and {Xα : α ∈ Γ} be independent Bernoulli random variables. Then, for w0 ∈ {0, 1, ..., |Γ|}, ¯ ¯ w0 ¯ X X λk e−λ ¯¯ ¯ p2α . (1.9) ¯P (W ≤ w0 ) − ¯ ≤ 4(λ, w0 ) ¯ k! ¯ k=0

α∈Γ

Theorem 1.4 For w0 ∈ {0, 1, ..., |Γ|} and λ ∈ (0, 3], we have ¾ ½ eλ . 4(λ, w0 ) < λ−1 (1 − e−λ ) min 1, w0 + 1

(1.10)

182

Thai J. Math. 4(2006)/ K. Teerapabolarn

Remark 1.5 (i) From Theorem 1.4, we see that for λ ∈ (0, 3], the bounds in (1.5), (1.6) and (1.9) are sharper than the bounds in (1.1), (1.2) and (1.3), respectively. (ii) There are many applications such that the Poisson approximation to be more accurately for λ ∈ (0, 1], see these applications in Arratia, Goldstein and Gordon [1-2], Barbour, Holst and Janson [4] and Lange [7]. In this case, we have  −2 λ (λ + e−λ − 1), if w0 = 0,           λ−2 [1 − e−λ (1 + λ)], if w0 = 1,    (1.11) 4(λ, w0 ) = 2   if w0 = 2, λ−3 (2 + λ)[1 − e−λ (1 + λ + λ2 )],        −λ w0 +1   ](w0 + 2)  [w0 !(w0 − λ) + e λ  , if w0 ∈ {3, ..., |Γ|}. w0 (w0 + 2 − λ)(w0 + 1)!

2

Proof of Main Results

We will prove our main results by using the Stein-Chen method. The method was originally formulated for normal approximation by Stein [9] in 1972, and the idea was applied to Poisson case by Chen [5] in 1975. This method started by the Stein’s equation for Poisson distribution which is, given h, defined by λf (w + 1) − wf (w) = h(w) − Pλ (h), where Pλ (h) = e−λ

∞ X l=0

h(l)

(2.1)

λl l!

and f and h are bounded real valued functions defined on N∪{0}. For w0 ∈ N∪{0}, let Cw0 = {0, 1, .., w0 } and hCw0 : N ∪ {0} → R be defined by   1, if w ∈ Cw0 , hCw0 (w) = (2.2)   0, if w ∈ / Cw0 . Following Barbour, Holst and Janson [4] p.7, the solution Uλ hCw0 of (2.1) can be expressd in the form  (w − 1)!λ−w eλ [Pλ (hCw0 )Pλ (1 − hCw−1 )], if w0 < w,     Uλ hCw0 (w) = (2.3)  (w − 1)!λ−w eλ [Pλ (hCw−1 )Pλ (1 − hCw0 )], if w0 ≥ w,    0, if w = 0.

A Non-Uniform Bound on Poisson Approximation for Sums of Bernoulli . . . 183 Let Vλ hCw0 (w) = Uλ hCw0 (w + 1) − Uλ hCw0 (w). Then, from (2.3), we have Vλ hCw0 (w)  −(w+1) λ  e Pλ (hCw0 )[wPλ (1 − hCw ) − λPλ (1 − hCw−1 )], if w ≥ w0 + 1, (w − 1)!λ =

  (w − 1)!λ−(w+1) eλ Pλ (1 − hCw0 )[wPλ (hCw ) − λPλ (hCw−1 )], if 1 ≤ w ≤ w0 .  w ∞ 0 X  λj X λk   (w − 1)!λ−(w+1) e−λ if w ≥ w0 + 1, (w − k) ,   j! k! j=0 k=w+1 = w ∞ X  λk λj X  −(w+1) −λ  (w − 1)!λ (w − k) , if 1 ≤ w ≤ w0 . e   j! k! j=w0 +1

k=0

(2.4) Hence, by(2.4),

  < 0, if w ≥ w0 + 1, Vλ hCw0 (w)

  > 0, if 1 ≤ w ≤ w0 .

The following lemmas are the properties of Vλ hCw0 which are used in the main theorem. Lemma 2.1 Vλ hCw0 is increasing in w for w ∈ {1, . . . , w0 }. Proof. We shall show that 0 < Vλ hCw0 (w + 1) − Vλ hCw0 (w) for 1 ≤ w ≤ w0 − 1. Note that, from (2.4), Vλ hCw0 (w + 1) − Vλ hCw0 (w) = (w − 1)!λ−(w+2) Pλ (1 − hCw0 ) ( ×

w

w+1 X

w

k=0

k=0

X λk λk (w + 1 − k) −λ (w − k) k! k!

and λ

w X k=0

w

X λk λk+1 (w − k) = (w + 1 − (k + 1)) (k + 1) k! (k + 1)! k=0

=

w+1 X

(w + 1 − k)k

k=0

λk . k!

So, we have Vλ hCw0 (w + 1) − Vλ hCw0 (w) = (w − 1)!λ−(w+2) Pλ (1 − hCw0 ) (w+1 ) X λk × (w − k)(w + 1 − k) k! k=0

> 0.

)

184

Thai J. Math. 4(2006)/ K. Teerapabolarn ¤

Lemma 2.2 Let w0 ∈ N ∪ {0} and w ≥ 1. Then |Vλ hCw0 (w)| ≤ 4(λ, w0 ).

(2.5)

Proof. Case 1. w0 = 0. Follows from Teerapabolarn, Neammanee and Chongcharoen[13] p.14. Case 2. w0 = 1. Note that, from (2.4),  ∞ X λk−(w+1)  e−λ (1 + λ)(w − 1)! , (w − k) k! Vλ hC1 (w) = k=w+1   −2 λ [1 − e−λ (1 + λ)],

if w ≥ 2, if w ≤ 1.

Hence, for w ≥ 2, we have 0 < −Vλ hC1 (w)

½

1 2λ 3λ2 + + + ··· (w + 1)! (w + 2)! (w + 3)! ½ ¾ e−λ (1 + λ)(w − 1)! 2λ 3λ2 = 1+ + + ··· (w + 1)! w + 2 (w + 2)(w + 3) ½ ¾ −λ e (1 + λ) 2λ 3λ2 ≤ 1+ + + ··· 6 4 20

¾

= e−λ (1 + λ)(w − 1)!



λ−1 (1 − e−λ )(1 + λ) , 6

which implies that ½ ¾ λ−1 (1 − e−λ )(1 + λ) |Vλ hC1 (w)| ≤ max λ−2 [1 − e−λ (1 + λ)], . 6 Case 3. w0 = 2. Since Vλ hCw0 is positive for 1 ≤ w ≤ w0 , by (2.4) and lemma 2.1, we have

Vλ hC2 (w) ≤

  e−λ (1 + λ +

λ2 2 )(w

− 1)!

  −3 λ (2 + λ)[1 − e−λ (1 +

∞ X

(w − k)

k=w+1 2 λ + λ2 )],

λk−(w+1) , k!

if w ≥ 3, if 1 ≤ w ≤ 2.

A Non-Uniform Bound on Poisson Approximation for Sums of Bernoulli . . . 185 Hence, for w ≥ 3, 0 < −Vλ hC2 (w + 1, w) = ≤ ≤ =

2 ½ ¾ e−λ (1 + λ + λ2 )(w − 1)! 2λ 3λ2 1+ + + ··· (w + 1)! w + 2 (w + 2)(w + 3) 2 ½ ¾ e−λ (1 + λ + λ2 ) 2λ 3λ2 1+ + + ··· 12 5 30 · ¸¾ λ2 ½ −λ e (1 + λ + 2 ) 4λ−1 λ2 λ3 1+ + + ··· 12 5 2! 3! ¾ λ2 ½ −λ −1 λ e (1 + λ + 2 ) 5 + 4λ (e − 1 − λ) 12 5 2

4λ−1 (1 − e−λ )(1 + λ + λ2 ) + e−λ (1 + λ + = 60 2 −1 −λ 4λ (1 − e )(1 + λ + λ2 ) + 1 ≤ . 60

λ2 2 )

So, |Vλ hC2 (w)| ( ≤ max λ

−3

· µ ¶¸ 4λ−1 (1 − e−λ )(1 + λ + λ2 −λ (2 + λ) 1 − e 1+λ+ , 2 60

λ2 2 )

+1

) .

Case 4. w0 ≥ 3. Note that for 1 ≤ w ≤ w0 , we have, by (2.4) and lemma 2.1, ( Vλ hCw0 (w) ≤ (w0 − 1)!λ ≤

−(w0 +1)

w0 !(w0 − λ) + e w0

e

−λ

∞ X

k=w0 +1 ∞ −λ w0 +1 X

λ

w0 !(w0 − λ) + e−λ λw0 +1 = w0

)(

λk k! λ

(w0 − λ)

w0 X λk k=0

λw0 +1 + k! w0 !

)

k−(w0 +1)

k=w0 +1

½

k!

1 λ λ2 + + + ··· (w0 + 1)! (w0 + 2)! (w0 + 3)! ½ ¾ λ λ2 1+ + + ··· w0 + 2 (w0 + 2)(w0 + 3) ½ ¾ λ λ2 1+ + + · · · w0 + 2 (w0 + 2)2

=

w0 !(w0 − λ) + e−λ λw0 +1 w0 (w0 + 1)!



w0 !(w0 − λ) + e−λ λw0 +1 w0 (w0 + 1)!

=

[w0 !(w0 − λ) + e−λ λw0 +1 ](w0 + 2) w0 (w0 + 2 − λ)(w0 + 1)!

¾

186

Thai J. Math. 4(2006)/ K. Teerapabolarn

and for w ≥ w0 + 1, 0 < −Vλ hCw0 (w) ≤ = ≤ ≤ ≤ =

½ ¾ (w − 1)! 2λ 3λ2 1+ + + ··· (w + 1)! w + 2 (w + 2)(w + 3) ½ ¾ 2λ 3λ2 1 1+ + + ··· (w0 + 1)(w0 + 2) (w0 + 3) (w0 + 3)(w0 + 4) ½ ¾ 1 λ λ2 1+ + + ··· (w0 + 1)(w0 + 2) 3 14 ½ · ¸¾ 1 2λ−1 λ2 λ3 1+ + + ··· (w0 + 1)(w0 + 2) 3 2! 3! ½ ¾ 1 3 + 2λ−1 (eλ − 1 − λ) (w0 + 1)(w0 + 2) 3 2λ−1 (1 − e−λ )eλ + 1 . 3(w0 + 1)(w0 + 2)

So, we have ½ |Vλ hCw0 (w)| ≤ max

[w0 !(w0 − λ) + e−λ λw0 +1 ](w0 + 2) 2λ−1 (1 − e−λ )eλ + 1 , w0 (w2 + 2 − λ)(w0 + 1)! 3(w0 + 1)(w0 + 2)

¾

Hence, from case 1 to case 4, we have (2.5).

. ¤

Proof of Theorem 1.1. (i) Teerapabolarn and Neammanee showed in [12] that ¯ ¯ w0 ¯ X X λk e−λ ¯¯ ¯ pα E|W − Wα∗ |. ¯P (W ≤ w0 ) − ¯ ≤ sup |Vλ hCw0 (w)| ¯ k! ¯ w≥1 α∈Γ

k=0

Hence, by lemma 2.2, (1.5) holds. (ii) Since ¯ ¯ ¯ ¯ X X X X ¯ ¯ ∗ ¯ pα E|W − Wα | = pα E ¯ Xβ − Xβ |Xα = 1¯¯ ¯β∈Γ ¯ α∈Γ α∈Γ β∈Γ\{α} ¯ ¯ ¯ ¯ X X X ¯ ¯ ¯ = pα E ¯Xα + Xβ − Xβ |Xα = 1¯¯ ¯ ¯ α∈Γ β∈Γα \{α} β∈Γα \{α} X X X ≤ p2α + {pα pβ + P (Xα = 1)E[Xβ |Xα = 1]} α∈Γ

=

X

α∈Γ β∈Γα \{α}

p2α +

α∈Γ

=

X

α∈Γ

X

X

{pα pβ + E[E[Xα Xβ |Xα ]]}

α∈Γ β∈Γα \{α}

p2α +

X

X

α∈Γ β∈Γα \{α}

(pα pβ + E[Xα Xβ ]),

(2.6)

A Non-Uniform Bound on Poisson Approximation for Sums of Bernoulli . . . 187 so, by (1.5) and (2.6), (1.6) holds.

¤

Proof of Theorem 1.2. From (1.5), ¯ ¯ w0 ¯ X X λk e−λ ¯¯ ¯ pα E|W − Wα∗ |, ¯ ≤ 4(λ, w0 ) ¯P (W ≤ w0 ) − ¯ k! ¯ α∈Γ

k=0

it suffices to show that every α ∈ Γ and

X

X

pα E|W − Wα∗ | = λ − V ar[W ] where W ≥ Wα∗ a.s. for

α∈Γ

pα E|W − Wα∗ | = V ar[W ] − λ + 2

α∈Γ

X

p2α where W − Xα ≤ Wα∗

α∈Γ

a.s. for every α ∈ Γ. (i) If W ≥ Wα∗ , then X

pα E|W − Wα∗ | =

α∈Γ

X

pα E[(W + 1) − (Wα∗ + 1)]

α∈Γ

= λ2 + λ −

X

pα E[(W − Xα + 1)|Xα = 1]

α∈Γ

= λ2 + λ −

X

E[E[Xα W |Xα ]]

α∈Γ

= λ2 + λ −

X

E[Xα W ]

α∈Γ

= λ2 + λ − E[W 2 ] = λ − V ar[W ]. (ii) If W − Xα ≤ Wα∗ , then X

pα E|W − Wα∗ | =

α∈Γ

X

pα E|Xα + (W − Xα ) − Wα∗ |

α∈Γ

=

X

pα {E[Xα ] + E|(Wα∗ + 1) − (W − Xα + 1)|}

α∈Γ

=

X

pα {E[Wα∗ + 1] − E[W + 1] + 2E[Xα ]}

α∈Γ

=

X

pα E[(W − Xα + 1)|Xα = 1] − λ2 − λ + 2

α∈Γ 2

2

= E[W ] − λ − λ + 2 = V ar[W ] − λ + 2

X α∈Γ

X α∈Γ

p2α .

X α∈Γ

p2α

p2α

188

Thai J. Math. 4(2006)/ K. Teerapabolarn ¤

Proof of Theorem 1.4. We shall show that ½ 4(λ, w0 ) < λ−1 (1 − e−λ ) min 1,

eλ w0 + 1

¾ .

Case 1. w0 = 0. Since 0 < λ−2 e−λ (eλ − 1 − λ), we have λ−2 (λ + e−λ − 1) = λ−1 (1 − e−λ ) − λ−2 e−λ (eλ − 1 − λ) < λ−1 (1 − e−λ ) = λ−1 (1 − e−λ ) min{1, eλ }. Case 2. w0 = 1. λ−1 (1 − e−λ )(1 + λ) (1 − e−λ )(1 + λ) = 6λ 6 ½ ¾ eλ −1 −λ < λ (1 − e ) min 1, 2 and 1 − e−λ (1 + λ) e−λ (eλ − 1 − λ) = 2 λ λ2 ∞ X λk−1 = λ−1 e−λ k! k=2 ½ ¾ λ−1 e−λ λ2 λ3 = λ+ + + ··· 2 3 12 ≤

λ−1 (1 − e−λ ) 2

Suggest Documents