# Theorem 48 strong law of large numbers let x 1 x 2 be

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Theorem 4.8. Strong Law of Large Numbers.LetX1, X2, . . .be i.i.d withE|Xi|<andEXi=μ. Then¯Xnμasn→ ∞.A corollary of this result is the frequency interpretation of probability. LetXi= 1 if theeventAoccurs on theith trial and 0 otherwise.EXi=P(A). Theorem 4.8 implies that¯Xn= the fraction of timesAoccurs in the firstntrials converges toP(A). While Theorem4.8 is nice, for practical purposes Theorem 4.7 suffices since it says that ifnis large thesample mean is close to the true mean with high probability.
118CHAPTER 4.LAW OF LARGE NUMBERS4.5Central Limit TheoremAt this point we know that ifX1, X2, . . .are independent and have the same distributionwith meanEXi=μand variance var (Xi) =σ2(0,) then, see (4.14), the sumSn=X1+· · ·+Xnhas meanESn=and var (Sn) =2. Using (1.15) and (4.2) we see thatSn-has mean 0 and variance2soSn-σnhas mean 0 and variance 1.The remarkable fact, called the central limit theorem is that asn→ ∞this scaled variableconverges to the standard normal distribution.Theorem 4.9.SupposeX1, X2, . . .are independent and have the same distribution withmeanEXi=μand variance var(Xi) =σ2(0,). Then for alla < bPaSn-σnbZba12πe-x2/2dxTo apply this result we need to learn how to use the normal table.If we let Φ(x) =P(χx) be the normal distribution function thenP(aχb) = Φ(b)-Φ(a)The values of Φ for positive values ofxare given in the table at the back of the book. BysymmetryΦ(-x) =P(χ≤ -x) =P(χx) = 1-Φ(x) soP(-xχx) = Φ(x)-(1-Φ(x)) = 2Φ(x)-1(4.17)To illustrate the use of Theorem??we will use a small part of the tablex0123Φ(x)0.5000.84130.97720.9986P(-xχx)00.68260.95440.9972In words, for the normal distribution the probability of being within one standard deviationof the mean is 68%, within two standard deviations is 95%, and the probability of beingmore that three standard deviations away is ¡ 0.3%.We begin by considering the situationP(Xi= 1) =P(Xi=-1) which hasEXi= 0 andvar (Xi) =E(X2i) = 1. Takinga=-1 andb= 1 and using our little tableP(-nSnn)P(-1χ1) = 0.6826Thus ifn= 2500 = 502our net winnings will be[-50,50] with probability0.6826.Since 1275-1225 = 50 this means that the number of heads will be[1225,1275] with thatprobability. 1275/2500 = 0.51 so this corresponds to the fraction of heads[0.49,0.51].Using the other two values in the table, ifn= 2500 our net winnings will be in [-100,100]with probability0.9544, and in [-150,150] with probability0.9972.The last resultshows that the bound from Chebyshev’s inequality can be very crude. Chebyshev tells usthatP(|χ| ≥3)var (χ)32=19while the true value ofP(|χ| ≥3) = 1-0.9972 = 0.00281/357.Turning to a smaller value ofn.
4.5.CENTRAL LIMIT THEOREM119Example 4.20.Suppose we flip a coin 100 times. What is the probability we get at least56 heads?

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