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Unformatted text preview: 6.10 The Central Limit Theorem 227 An alternative statement of the law is lim n P [ S n X ] ] = 1 2. The Strong Law of Large Numbers: For each > 0, P bracketleftBig lim n  S n X  > bracketrightBig = An alternative statement of the law is P bracketleftBig lim n  S n X  bracketrightBig = 1 Thus, the strong law states that with probability 1 the sequence of sample means S n converges to a constant value X , which is the population mean of the random variables, as n becomes very large. This validates the relativefrequency definition of probability. The weak law states that the probability is high that the sequence of sample means converges to X as n becomes very large. Thus, the weak law is a convergence statement about a sequence of probabilities; it states that the sequence of random variables { S n } converges in probability to the population mean X as n becomes very large. 6.10 The Central Limit Theorem While the strong law of large numbers helps to validate the relativefrequency definition of probability, it says nothing about the limiting distribution of the sum S n . The central limit theorem achieves this purpose. Let X 1 , X 2 ,..., X n be a sequence of mutually independent and identically distributed random variables each of which has a finite mean X and variance 2 X . Let S n = X 1 + X 2 + + X n The central limit theorem states that for large n the distribution of S n is approxi mately normal, regardless of the form of the distribution of the X k . Now, S n = E [ S n ] = n X 2 S n = n 2 X Converting S n to a standard normal random variable (i.e., zero mean and vari ance = 1) we obtain Z n = S n S n S n = S n n X radicalBig n 2 X = S n n X X n 228 Chapter 6 Functions of Random Variables Then the central limit theorem states that if F Z n ( z ) is the CDF of Z n , then lim n F Z n ( z ) = lim n P [ Z n z ] = 1 2 integraldisplay z e u 2 / 2 du = Phi1( z ) This means that lim n Z n = N ( ; 1 ) . Example 6.22 Assume that the random variable S n is the sum of 48 indepen dent experimental values of the random variable X whose PDF is given by f X ( x ) = braceleftBigg 1 3 1 x 4 otherwise Find the probability that S n lies in the range 108 S n 126. Solution The expected value and variance of X are given by E [ X ] = ( 4 + 1 ) 2 = 2 . 5 2 X = ( 4 1 ) 2 12 = 3 4 Thus, the mean and variance of S n are given by E [ S n ] = 48 E [ X ] = ( 48 )( 2 . 5 ) = 120 2 S n = 48 2 X = ( 48 ) parenleftbigg 3 4 parenrightbigg = 36 Assuming that the sum approximates the normal random variable, which is usu ally true for n 30, the CDF of the normalized random value of S n becomes P [ S n s ] = F S n ( s ) = Phi1 parenleftbigg s E [ S n ] S n parenrightbigg = Phi1 parenleftbigg s 120 6 parenrightbigg Therefore, the probability that S n lies in the range 108 S n 126 is given by...
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 Fall '07
 Carlton
 Central Limit Theorem, Law Of Large Numbers

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