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

This preview shows page 16 - 19 out of 39 pages.

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.
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 .
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

#### You've reached the end of your free preview.

Want to read all 39 pages?