Chapt 6d - 6.10 The Central Limit Theorem 227 An...

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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 relative-frequency 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 relative-frequency 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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Chapt 6d - 6.10 The Central Limit Theorem 227 An...

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