Stat 100A.Small review of CLT and some math tocomplement the proof of the CLT p osted in less onJuana [email protected] Department of StatisticsDecember 5, 2014J. Sanchez Stat 100A. Small review of CLT and some math to complement the proof of the CLT posted in lessonInterpretation of the Central Limit Theorem for Sums of iidRandom VariablesLet X1, X2, ......, Xnbe a set of i.i.d. random variables, each withE(xi) = µxand Var(xi) = σ2x. The Central Limit Theorem (CLT) tells usthat if n is large enough, the sum of the n iid r.v.’s, i.e.,x1+ x2+ ..... + xnorPni=1xi, is normally distributed, regardless of thedistribution of the Xi. The distribution of the sum is called the samplingdistribution of the sum.ExampleLet Xii = 1, 2, ...., n be iid random variables each distributed as Poissondistribution with parameter λ = 3, then µ = E(Xi) = λ = 3 andσ2= Var(Xi) = λ = 3. According to the CLT, i.e.,P100i=1xiis Normal.P(P100i=1xi> 400) = P(z >400−300√300) and you look in the normal tablesfor the final answer.J. Sanchez Stat 100A. Small review of CLT and some math to complement the proof of the CLT posted in lessonExampleLet the Xii = 1, 2, ...., n come from an exponential distribution withparameter λ = 3, then µ = E(Xi) =1λ= 1/3 andσ2= Var(Xi) =1λ2= 1/9 . The sum of 100 exponentials (n=100) isnormally distributed because n is large. I.e.,Pni=1xiis normallydistributed.ExerciseSuppose that Xii = 1, 2, ...., 200 is a set of i.i.d random variables andXi∼ Gamma(α = 4, λ = 3). Describe the shape of the samplingdistribution of the sum of the 200 random variables.J. Sanchez Stat 100A. Small review of CLT and some math to complement the proof of the CLT posted in lessonExp ected Value and Variance of the Normal distributionfor the Sum?For any random variable, and any n, if x1, x2, ....., xnare i.i.d., thenµPni=1xi= nµxand σ2Pni=1xi= nσ2for any n. So this result will also betrue for the sum of a large number of random variables.Thus, the normal distribution of the sum of a large n of i.i.d. randomvariables hasµPni=1xi= nµxσ2Pni=1xi= nσ2ExampleIf the Xi’s come from a Poisson distribution with parameter λ = 3, thenµ = E (Xi) = λ = 3 and σ2= Var(Xi) = λ = 3. Then the sum of 100Poissons (n=100) is normally distributed because n is large, withµP100i=1xi= 100 × 3 = 300 σ2P100i=1xi= 300J. Sanchez Stat 100A. Small review of CLT and some math to complement the proof of the CLT posted in lessonExampleIf the Xi, i = 1, 2, ..., n come from an exponential distribution withparameter λ = 3, then µ = E(Xi) =1λ= 1/3 andσ2= Var(Xi) =1λ2= 1/9 . The sum of 100 i.i.d. exponentials (n=100)is normally distributed because n is large withµP100i=1xi= 100 × 1/3 = 33.3 σ2P100i=1xi= 100 × 1/9 = 11.1ExerciseSuppose that Xii = 1, 2, ...., 200 is a set of i.i.d random variables andXi∼ Gamma(α = 4, λ = 3). Describe the sampling distribution of thesum of the 200 random variables.J. Sanchez Stat 100A. Small review of CLT and some math to complement the proof of the CLT posted in lessonA consequence of the Central Limit Theorem for the sumBecause the sample mean ¯x =Pxinis a linear function of the sum, andthe sum when n is large is normally distributed, ¯x is also normallydistributed withµ¯x= µx, σ2¯x=σ2n, σ¯x=σp(n)J. Sanchez Stat 100A. Small review of CLT and some math to complement the proof of the CLT posted in lessonExampleThe average number of days spent in a North Carolina Hospital for acoronary bypass in 1992 was 9 days and the standard deviation was 4days. What is the probability that a random sample of 30 patients willhave an average stay longer than 9.5 days?P(¯x > 9.5) = P z >9.5 − 94√30!= P(z > 0.6846)= 0.2467J. Sanchez Stat 100A. Small review of CLT and some math to complement the proof of the CLT posted in lessonCentral Limit Theorem for the special case of Bernoullirandom variables.ExampleThe Census Bureau reported that 70% of all americans havehospitalization coverage by a private insurance plan.In a large hospital, there were 200 patients admitted one week. What isthe probability that more than 65% have a private insurance plan?Think about the population first.Define the random variable Y as the hospitalization insurance status of aperson in the population. Patients in this population have insurance(Y=1) or not (Y=0). Thus Y is a Bernoulli Random variable. TheE(Y ) = µy= p = 0.7, and theVar(Y ) = σ2y= p(1 − p) = 0.7 × 0.3 = 0.21, and σ = 0.458 .J. Sanchez Stat 100A. Small review of CLT and some math to complement the proof of the CLT posted in lessonThink of the 200 patients’s insurance status as y1, y2, ....., y200, a set ofi.i.d. random variables, each Bernoulli (p=0.7). Then,X =Pni=1yiis the number of patients, out of 200, with insuranceThus µX= E (X ) = np = 200 × 0.7 = 140 andσ2X= np(1 − p) = 200 × 0.7 × 0.3 = 42, σ = 6.48.Similarly,Xn, which is usually called ˆp (sample proportion) in this context,is also normally distributed with µˆp= p = 0.7 andσˆp=√p(1−p)√n=0.458√200= 0.03238P (ˆp > 0.65) = Pz >0.65 − 0.70.03238= P(z > −1.544)= 0.9387J. Sanchez Stat 100A. Small review of CLT and some math to complement the proof of the CLT posted in lessonHow large must n be?Let X be a random variable and X1, X2, ....., Xna set of i.i.d. randomvariables (called a sample) from the distribution of X . How large must nbe for us to be able to use the Central Limit Theorem? Depends on theprobability distribution of X. In the case that the X is uniform, n can bevery small. The more skewed the distribution of the X is, the larger nmust be.If X is normal, it does not matter how large n is. Sums of normals arenormal and average of normals is normal.In the next slides, we will present more formally the Central LimitTheorem.J. Sanchez Stat 100A. Small review of CLT and some math to complement the proof of the CLT posted in lessonFormal TheoremLet X1, ....Xnbe iid r.v’s, with E (Xi) = µ and V (Xi) = σ2< ∞. DefineYnasYn=√n¯X − µσwhere¯X =1nPXiThen Ynconverges in distribution towards a standard normal randomvariable.J. Sanchez Stat 100A. Small review of CLT and some math to complement the proof of the CLT posted in lessonPreparing for the proof: Moment generating functionsrevisitedA Maclaurin series is a Taylor series expansion of a function g(t) aroundthe point t = 0. In other words, a Maclaurin series is just a way to writea function g(t) as an infinite series, in which the jth term is the
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