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12 slutskys theorem and the continuous map ping

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12 Slutsky’s theorem and the continuous map- ping theorem The law of large numbers and central limit theorem can often be used to tell us that a sequence of random variables converges in probability to some constant, or converges in distribution to a normal distribution with some mean and variance. But what can we say about the behavior of a sequence of random numbers formed from two or more different sequences, each of which is converging in probability or distribution? Slutsky’s theorem helps us to answer this question. Suppose that X 1 , X 2 , . . . and Y 1 , Y 2 , . . . are sequences of random variables, with X n p c and Y n d Y for some constant c and some random variable Y . Slutsky’s theorem says that 1. X n + Y n d c + Y . 2. X n Y n d cY . 3. Y n /X n d Y/c , provided that c 6 = 0. For instance, suppose that X n p 2 and Y n d N (0 , 1). Then Slutsky’s theorem implies that (1) X n + Y n d 2+ N (0 , 1) = N (2 , 1); (2) X n Y n d 2 N (0 , 1) = N (0 , 4); and (3) Y n /X n d N (0 , 1) / 2 = N (0 , 1 / 4). The continuous mapping theorem tells us something similar about the behavior of a function of a sequence of random variables that is converging in probability. If X n p c as above, and if g is a function that is continuous at the point c , then the continuous mapping theorem says that g ( X n ) p g ( c ). For instance, if X n p 2, then the continuous mapping theorem implies that X 2 n p 4, because the function g ( x ) = x 2 is continuous. 15
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The requirement that the function g is continuous at the limit point c is im- portant. To see why, consider the following example. For each n , let X n U (1 - 1 n , 1 + 1 n ), so that X n is uniformly distributed on the interval (1 - 1 n , 1 + 1 n ). It is simple to show that X n p 1. Let the function g be given by g ( x ) = 0 for x < 1 1 for x 1 . For every n , the random variable g ( X n ) is equal to zero with probability 1 / 2, and equal to one with probability 1 / 2. Therefore, the distribution of g ( X n ) does not change with n , and so it cannot be true that g ( X n ) p g (1) = 1. We cannot apply the continuous mapping theorem here because g is not continuous at 1. Let us consider an example of how Slutsky’s theorem and the continuous map- ping theorem may be used together. Suppose that X n p σ 2 , and that Y n d N (0 , σ 2 ), with σ > 0. How does the random variable Z n = Y n / X n behave as n → ∞ ? The function g ( x ) = x is continuous (for positive x , which is enough for our purposes), so the continuous mapping theorem implies that X n p σ . Next, applying part 3 of Slutsky’s theorem, we obtain Y n /X n d N (0 , σ 2 ) = N (0 , 1). Therefore, Z n d N (0 , 1). 13 Asymptotic inference about the mean Let X 1 , X 2 , . . . be an infinite sequence of iid random variables with common ex- pected value μ = E ( X ) and common variance σ 2 = Var( X ). Suppose we observe just the first n random variables, X 1 , . . . , X n . What can we say about the expected value μ ? The law of large numbers tells us that the sample mean ¯ X n should be close to μ when n is large. The central limit theorem can be used to get a more precise idea of just how close ¯
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