Theorem 48 Strong Law of Large Numbers Let X 1 X 2 be iid with E X i and EXi \u03bc

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. Let X 1 , X 2 , . . . be i.i.d with E | X i | < and EX i = μ . Then ¯ X n μ as n → ∞ . A corollary of this result is the frequency interpretation of probability. Let X i = 1 if the event A occurs on the i th trial and 0 otherwise. EX i = P ( A ). Theorem 4.8 implies that ¯ X n = the fraction of times A occurs in the first n trials converges to P ( A ). While Theorem 4.8 is nice, for practical purposes Theorem 4.7 suffices since it says that if n is large the sample mean is close to the true mean with high probability.
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118 CHAPTER 4. LAW OF LARGE NUMBERS 4.5 Central Limit Theorem At this point we know that if X 1 , X 2 , . . . are independent and have the same distribution with mean EX i = μ and variance var ( X i ) = σ 2 (0 , ) then, see (4.14), the sum S n = X 1 + · · · + X n has mean ES n = and var ( S n ) = 2 . Using (1.15) and (4.2) we see that S n - has mean 0 and variance 2 so S n - σ n has mean 0 and variance 1. The remarkable fact, called the central limit theorem is that as n → ∞ this scaled variable converges to the standard normal distribution. Theorem 4.9. Suppose X 1 , X 2 , . . . are independent and have the same distribution with mean EX i = μ and variance var ( X i ) = σ 2 (0 , ) . Then for all a < b P a S n - σ n b Z b a 1 2 π e - x 2 / 2 dx To 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 then P ( a χ b ) = Φ( b ) - Φ( a ) The values of Φ for positive values of x are given in the table at the back of the book. By symmetryΦ( - x ) = P ( χ ≤ - x ) = P ( χ x ) = 1 - Φ( x ) so P ( - 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 table x 0 1 2 3 Φ( x ) 0.500 0.8413 0.9772 0.9986 P ( - x χ x ) 0 0.6826 0.9544 0.9972 In words, for the normal distribution the probability of being within one standard deviation of the mean is 68%, within two standard deviations is 95%, and the probability of being more that three standard deviations away is ¡ 0.3%. We begin by considering the situation P ( X i = 1) = P ( X i = - 1) which has EX i = 0 and var ( X i ) = E ( X 2 i ) = 1. Taking a = - 1 and b = 1 and using our little table P ( - n S n n ) P ( - 1 χ 1) = 0 . 6826 Thus if n = 2500 = 50 2 our 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 that probability. 1275 / 2500 = 0 . 51 so this corresponds to the fraction of heads [0 . 49 , 0 . 51]. Using the other two values in the table, if n = 2500 our net winnings will be in [ - 100 , 100] with probability 0 . 9544, and in [ - 150 , 150] with probability 0 . 9972. The last result shows that the bound from Chebyshev’s inequality can be very crude. Chebyshev tells us that P ( | χ | ≥ 3) var ( χ ) 3 2 = 1 9 while the true value of P ( | χ | ≥ 3) = 1 - 0 . 9972 = 0 . 0028 1 / 357. Turning to a smaller value of n .
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4.5. CENTRAL LIMIT THEOREM 119 Example 4.20. Suppose we flip a coin 100 times. What is the probability we get at least 56 heads? Let X i = 1 if the i th flip is heads, 0 otherwise, and S 100 = X 1 + · · · + X 100 . EX i = 1 / 2 and var ( X i ) = 1 / 4 so ES 100 = 50 and σ ( S 100 ) = p 100 / 4 = 5. To apply the central limit theorem we write P ( S 100 56) = P S 100 - 50 5 6 5 P ( χ 1 . 2) = 1 - P ( χ 1 . 2) = 1 - 0 . 8849 = 0 . 1151 If the question has been “What is the probability of at most 55 heads?” we would have computed P ( S 100 55) = P S 100 - 50 5 5 5 P ( χ 1 . 0) = 0 . 8413
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