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=nμand var (Sn) =nσ2. Using (1.15) and (4.2) we see thatSn-nμhas mean 0 and variancenσ2soSn-nμσ√nhas 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 < bPa≤Sn-nμσ√n≤b→Zba1√2πe-x2/2dxTo apply this result we need to learn how to use the normal table.If we let Φ(x) =P(χ≤x) be the normal distribution function thenP(a≤χ≤b) = 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(χ≥x) = 1-Φ(x) soP(-x≤χ≤x) = 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(-x≤χ≤x)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(-√n≤Sn≤√n)≈P(-1≤χ≤1) = 0.6826Thus ifn= 2500 = 502our net winnings will be∈[-50,50] with probability≈0.6826.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 probability≈0.9544, and in [-150,150] with probability≈0.9972.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 = 0.0028≈1/357.Turning to a smaller value ofn.