1 for x 1 to verify convergence in distribution we

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1 for x > 1 . To verify convergence in distribution, we need to check that lim n →∞ F n ( x ) = F ( x ) for each x . We will consider three cases separately: x < 0, 0 x 1, and x > 1. In the first case, F n ( x ) = F ( x ), so trivially lim n →∞ F n ( x ) = F ( x ). In the second case, F n ( x ) = nx/ ( n + 1), which converges to F ( x ) = x as n → ∞ . In the third case, we have F n ( x ) = F ( x ) = 1 for all n > 1 / ( x - 1), implying once again that lim n →∞ F n ( x ) = F ( x ). This formally establishes that Z n d U (0 , 1). 11 The central limit theorem Suppose we have an iid sequence of random variables X 1 , X 2 , X 3 , . . . with common expectation E ( X i ) = 0 and common variance Var( X i ) = 1. As before, let ¯ X n denote the average of the first n of the random variables, and imagine a sequence of averages ¯ X 1 , ¯ X 2 , ¯ X 3 , . . . formed using progressively more X i ’s. The law of large numbers tells us that ¯ X n p 0, since the expected value of each X i is zero. But consider the following question: what happens to n ¯ X n as n → ∞ ? Here we are considering the limiting behavior of the product of two terms, n and ¯ X n , where the first term increases to infinity as n increases, while the second term converges in probability to zero. Which term will dominate in the limit? The central limit theorem tells us that neither term dominates, and in fact they balance one another perfectly so that their product converges in distribution to the standard normal distribution. That is, n ¯ X n d N (0 , 1) . Using the definition of convergence in distribution, we may restate this result as lim n →∞ P ( n ¯ X n x ) = 1 2 π Z x -∞ e - y 2 / 2 d y for all x. This is a remarkable result. The mathematical constants e and π seem to have appeared out of nowhere. Our assumption that each of the X i ’s has expected value zero and variance one may appear restrictive. In fact, we can dispense with these conditions if we are willing to state the central limit theorem in a slightly different form. Suppose that each X i has mean μ and variance σ 2 . In this case, the central limit theorem tells us that 1 n n X i =1 X i - μ σ d N (0 , 1) . 14
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This second form of the central limit theorem follows immediately from the first. To see this, let Y i = ( X i - μ ) , and let ¯ Y n = 1 n n i =1 Y i . Then 1 n n X i =1 X i - μ σ = n ¯ Y n . Since each Y i has expected value zero and variance one, the first version of the cen- tral limit theorem implies that n ¯ Y n d N (0 , 1), establishing the second version. The simplest proof of the central limit theorem requires knowledge of complex numbers and some basic Fourier analysis, and therefore falls outside the scope of this course. But applications of the central limit theorem are ubiquitous in applied statistical work. Nearly every hypothesis test or confidence interval reported in applications relies on the central limit theorem for its large sample validity.
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