LECTURE 20THE CENTRAL LIMIT THEOREM• Readings: Section 5.4• X1, . . . , Xni.i.d., finite variance σ2• “Standardized” Sn= X1+ ···+ Xn:Zn=Sn− E[Sn]σSn=Sn− nE[X]√nσ– E[Zn]=0, var(Zn)=1• Let Z be a standard normal r.v.(zero mean, unit variance)• Theorem: For every c:P(Zn≤ c) → P(Z ≤ c)• P(Z ≤ c) is the standard normal CDF,Φ(c), available from the normal tablesUsefulness• universal; only means, variances matter• accurate computational shortcut• justification of normal modelsWhat exactly does it say?• CDF of Znconverges to normal CDF– not a statement about convergence ofPDFs or PMFsNormal approximation• Treat Znas if normal– also treat Snas if normalCan we use it when n is “moderate”?• Yes, but no nice theorems to this effect• Symmetry helps a lot0 1 2 3 4 5 6 700.10.20.30.40.50.60.70.80.910 5 10 15 20 25 30 35 4000.020.040.060.080.10.12n = 80 10 20 30 40 50 60 7000.010.020.030.040.050.060.070.08n = 1630 40 50 60 70 80 90 10000.010.020.030.040.050.06n = 320 5 10 15 2000.020.040.060.080.10.120.14n =20 5 10 15 20 25 30 3500.020.040.060.080.1n =4100 120 140 160 180 20000.0050.010.0150.020.0250.030.035n =320 2 4 6 800.050.10.150.20.25The pollster’s problem using the CLT• f: fraction of population that “ ...%%• ith (randomly selected) person polled:Xi=1, if yes,0, if no.• Mn=(X1+ ···+ Xn)/n• Suppose we want:P(|Mn− f| ≥ .01) ≤ .05• Event of interest: |Mn− f| ≥ .01$$$$X1+ ···+ Xn− nfn$$$$≥ .01$$$$$X1+ ···+ Xn− nf√nσ$$$$$≥.01√nσP(|Mn− f| ≥ .01) ≈ P(|Z| ≥ .01√n/σ)≤ P(|Z| ≥ .02√n)Apply to binomial• Fix p, where 0 <p<1• Xi: Bernoulli(p)• Sn= X1+ ···+ Xn: Binomial(n, p)– mean np, variance np(1 − p)• CDF ofSn− np%np(1 − p)−→ standard normalExample• n = 36, p =0.5; find P(Sn≤ 21)• Exact answer:21&k=0'36k()12*36=0.8785The 1/2 correction for binomialapproximation• P(Sn≤ 21) = P(Sn< 22),because Snis integer• Compromise: consider P(Sn≤ 21.5)1920182122De Moivre–Laplace CLT (for binomial)• When the 1/2 correction is used, CLTcan also approximate the binomial p.m.f.(not just the binomial CDF)P(Sn= 19) = P(18.5 ≤ Sn≤ 19.5)18.5 ≤ Sn≤ 19.5 ⇐⇒18.5 − 183≤Sn− 183≤19.5 − 183⇐⇒0.17 ≤ Zn≤ 0.5P(Sn= 19) ≈ P(0.17 ≤ Z ≤ 0.5)= P(Z ≤ 0.5) − P(Z ≤ 0.17)=0.6915 − 0.5675=0.124• Exact answer:'3619()12*36=0.1251Poisson vs. normal approximations ofthe binomial• Poisson arrivals during unit interval equals:sum of n (independent) Poisson arrivalsduring n intervals of length 1/n– Let n →∞, apply CLT (??)– Poisson=normal (????)• Binomial(n, p)– p fixed, n →∞: normal– np fixed, n →∞, p → 0: Poisson• p =1/100, n = 100: Poisson• p =1/10, n = 500:
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