Lectures 16-17: Poisson Approximation. Using Lemma (2.4.3) with θ = 1 and then Lemma (2.4.4), which is valid when max m p n,m 1/2, we have

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1 Lectures 16-17: Poisson Approximation 1. The Law of Rare Events Theorem 2.6.1: For each n 1, let X n,m, 1 m n be a collection of independent random variables with PX n,m = 1 = p n,m and PX n,m = = 1 p n,m. Suppose that 1. n p n,m λ; 2. max 1mn p n,m. If S n = X n,1 + + X n,n, then S n Z Poissonλ. Proof: If we let φ n,m t = 1 p n,m + p n,m e it denote the characteristic function of X n,m, then Ee itsn = n 1 + pn,m e it 1. Let p [, 1] and notice that e it 1 2 exp pe it 1 = exp pree it 1 1 p + pe it 1. 1 Using Lemma with θ = 1 and then Lemma 2.4.4, which is valid when max m p n,m 1/2, we have n exp p n,m e it pn,m e it 1 Since n p n,m λ, it follows that exp p n,m e it p n,m e it 1 p 2 n,m e it max p n,m p n,m. 1mn which is the characteristic function of Z. Ee itsn expλe it 1 1

2 2. Rates of Convergence Let µ, ν be measures on a countable set S. Then the total variation distance between µ and ν is µ ν µz νz = 2 sup µa νa. z A S To show that the second expression is equal to the first, observe that for any A S, µz νz µa νa + µa c νa c = 2 µa νa z and that equality holds when A = {z : µz νz}. Lemma 2.6.2: If µ 1 µ 2 denotes a product measure on Z Z, then Proof: We have µ 1 µ 2 ν 1 ν 2 µ 1 ν 1 + µ 2 ν 2. µ 1 µ 2 ν 1 ν 2 = x,y µ 1 xµ 2 y ν 1 xν 2 y x,y µ1 xµ 2 y ν 1 xµ 2 y + ν 1 xµ 2 y ν 1 xν 2 y = y µ 2 y µ 1 ν 1 + x ν 1 x µ 2 ν 2 = µ 1 ν 1 + µ 2 ν 2. Lemma 2.6.3: For convolutions we have: µ 1 µ 2 ν 1 ν 2 µ 1 µ 2 ν 1 ν 2. Proof: In this case, µ 1 µ 2 ν 1 ν 2 = x x,y µ 1 x yµ 2 y ν 1 x yν 2 y y y µ 1 x yµ 2 y ν 1 x yν 2 y = µ 1 µ 2 ν 1 ν 2. Lemma 2.6.4: Let µ be the Bernoullip distribution and let ν be a Poissonp distribution. Then µ ν 2p 2. 2

3 Proof: Since 1 p e p 1 whenever p, we have µ ν = µ ν + µ1 ν1 + n 2 νn = 1 p e p + p pe p + 1 e p 1 + p = e p 1 + p + p1 e p + 1 e p pe p = 2p1 e p 2p 2. Alternative proof of 2.6.1: Let µ n,m be the distribution of X n,m and let µ n be the distribution of S n. Let ν n,m, ν n and ν be Poisson distributions with means p n,m, λ n = n p n.m and λ, respectively. Since µ n = µ n,1 µ n,n and ν n = ν n,1 ν n,n, Lemmas 2.6.3, and imply that In particular, µ n ν n µ n,m ν n,m 2 sup µ n A ν n A A p 2 n,m, p 2 n,m. and so the Poisson approximation theorem follows upon noting that ν n ν. Example: If p n,m = n 1, then S n has a binomial distribution with parameters n, n 1 and we have sup µ n A ν n A 1 A n. 3. An Example with Dependence Let π be a random permutation of {1, 2,, n}, let X n,m = 1 if πm = m is a fixed point and X n,m = otherwise, and let S n = X n,1 + + X n,n be the total number of fixed points. We begin by computing PS n =. Let A n,m = {X n,m = 1} and notice that the inclusion-exclusion formula implies that PS n > = P n A n,m and consequently = m PA n,m l<m n n 2! + 2 n! = n 1 n = 1 m 1 1 m! PS n = = PA n,l A n,m + n 3 k<l<m n 3! + n! 1 m 1 m! e 1 m= 3 PA n,k A n,l A n,m +

4 as n. Similarly, PS n = k = n 1 k nn 1 n k + 1 PS n k = = 1 k! PS n k = e 1 1 k!. These calculations show that S n Z Poisson1 as n. 4. Poisson Processes Theorem 2.6.7: Let X n,m, 1 m n be independent, non-negative integer valued random variables with PX n,m = 1 = p n,m and PX n,m 2 = ɛ n,m that satisfy the following conditions 1. n p n,m λ; 2. max 1mn p n,m ; 3. n ɛ n,m. If S n = X n,1 + + X n,n, then S n Z where Z Poissonλ. Proof: Let Y n,m = 1 if X n,m = 1 and otherwise. If T n = Y n,1 + + Y n,n, then T n Z. Since PS n R n, the result follows from Slutsky s lemma. Poisson distributions often arise in problems involving sequences of events. Let Ns, t be the number of events arrivals that occur in the time interval s, t]. Suppose that 1. the number of events in disjoint intervals are independent; 2. the distribution of Ns, t only depends on s t; 3. PN, h = 1 = λh + oh; 4. PN, h 2 = oh. Theorem 2.6.8: If conditions 1-4 hold, then N, t has a Poisson distribution with mean λt. Proof: This follows from Theorem by letting X n,m = Nm 1t/n, mt/n and noting that N, t = N, t/n + Nt/n, 2t/n + + Nn 1t/n, t. A family of random variables N = Nt; t is called a Poisson process with rate λ if 1. N has independent increments: given < t 1 < t 2 < < t n, the variables Nt k Nt k 1, k = 1,, n are independent; 2. Nt Ns is Poisson with parameter λt s. 4

5 The family of random variables N, t; t given above is a Poisson process with rate λ. Here we give an alternative construction. Let X 1, X 2, be independent exponential random variables with PX i > t = e λt. These represent the times between successive events. Let T n = X X n be the time of the n th event and let N t = sup{n : T n t} be the number of events that have occurred by time t. Claim: Nt : t is a Poisson process. Proof: A calculation using convolutions shows that PT n = s = λn s n 1 n 1! e λs, i.e., T n is a gamma-distributed random variable with parameters n and λ. Notice that and for n 1 PN t = = PT 1 > t = e λt PN t = n = PT n t < T n+1 = = which shows that N t is Poisson with mean λt. PT n = spx n+1 > t sds λ n s n 1 n 1! e λs e λt s λt λtn ds = e n! Our next task is to show that the increments are independent. Observe that The numerator is equal to PT n+1 u N t = n = PT n+1 u, T n t. PN t = n PT n+1 u, T n t = = PT n = spx n+1 u sds λ n s n 1 n 1! e λs e λu s ds = e λu λtn. n! Since the denominator is PN t = n = e λt λt n /n!, we have or equivalently setting u = t + s PT n+1 u N t = n = e λu /e λt = e λu t, PT n+1 t s N t = n = e λs. Let X 1 = T Nt+1 t and X k = T Nt+k T Nt+k 1 for k 2. The preceding calculation shows that X 1 is independent of N t, PX 1 s N t = n = PX 1 s = e λs, 5

6 and has an exponential distribution with rate λ. Also, since P T n t, T n+1 u, T n+k T n+k 1 v k, k = 2,, K K = PT n t, T n+1 u Pξ n+k v k, k=2 it follows that P X 1 s, X k v K k, k = 2,, K N t = n = e λs k=2 e λv k, and so the random variables X 1, X 2, are independent exponential random variables with parameter λ which are jointly independent of Nt. This implies that the arrivals after t are independent of Nt and have the same distribution as the original sequence. To show that the process Nt; t has independent increments, observe that if we let t = t 1 above, then the vector Nt 2 Nt 1,, Nt n Nt n 1 is measurable with respect to σt k, k 1 and therefore independent of Nt 1. For example, we have } Nt 2 Nt 1 = sup {k : X X k t 2 t 1 and P Nt 2 Nt 1 = k Nt 1 = n Then, by induction, we can conclude that k = P i=1 λt λtk = e. k! P Nt i Nt i 1 = k i, i = 1,, n = k+1 X i t 2 t 1, n i=1 i=1 X i > t 2 t 1 Nt 1 = n e λt i t i 1 λt i t i 1 k i, k i! which shows that the increments are independent Poisson random variables with the rates indicated above. 5. Generalizations Compound Poisson processes are constructed in the following way. Let N t ; t be a standard Poisson process with rate λ and let V 1, V 2, be a collection of i.i.d. random variables. If for each t, N t X t = V i, then the process X t ; t is called a compound Poisson process. 1 6

7 A Poisson point process on a measurable space S, S with finite mean measure µ is a random counting measure Ξ : S {, 1, } that satisfies the following property. If A 1,, A n are disjoint measurable subsets of S, then ΞA 1,, ΞA n are independent Poisson random variables with means µa 1,, µa n. If µs <, then Ξ can be constructed as follows. Let N be a Poisson random variable with mean µs, let X 1, X 2, be i.i.d. with distribution ν = µ /µs, and for each set A S, let ΞA be defined by ΞA = #{j N : X j A}. 7