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The Mathematical and Statistical Foundations of Econometrics in .NET Printer barcode 128 in .NET The Mathematical and Statistical Foundations of Econometrics




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The Mathematical and Statistical Foundations of Econometrics using .net toadd code 128 for asp.net web,windows application USPS POSTNET Barcode If each of the co Code 128A for .NET mponents of a sequence of random vectors converges in distribution, then the random vectors themselves may not converge in distribution. As a counterexample, let Xn = X 1n X 2n N2 0 1 ( 1)n /2 , n ( 1) /2 1 0 .

(6.16). Then X 1n d N (0 , 1) and X 2n d N (0, 1), but X n does not converge in distribution. Moreover, in general X n d X does not imply that X n p . For example, if we replace X by an independent random drawing Z from the distribution of X , then X n d X and X n d Z are equivalent statements because they only say that the distribution function of X n converges to the distribution function of X (or Z ) pointwise in the continuity points of the latter distribution function.

If X n d X implied X n p X , then X n p Z would imply that X = Z , which is not possible because X and Z are independent. The only exception is the case in which the distribution of X is degenerated: P(X = c) = 1 for some constant c: Theorem 6.16: If X n converges in distribution to X , and P(X = c) = 1, where c is a constant, then X n converges in probability to c.

Proof: Exercise. Note that this result is demonstrated in the left-hand panels of Figures 6.1 6.

3. On the other hand, Theorem 6.17: X n p X implies X n d X .

Proof: Theorem 6.17 follows straightforwardly from Theorem 6.3, Theorem 6.

4, and Theorem 6.18 below. Q.

E.D. There is a one-to-one correspondence between convergence in distribution and convergence of expectations of bounded continuous functions of random variables: Theorem 6.

18: Let X n and X be random vectors in Rk . Then X n d X if and only if for all bounded continuous functions on Rk limn E[ (X n )] = E[ (X )]. Proof: I will only prove this theorem for the case in which X n and X are random variables.

Throughout the proof the distribution function of X n is denoted by Fn (x) and the distribution function of X by F(x). Proof of the only if case: Let X n d X . Without loss of generality we may assume that (x) [0, 1] for all x.

For any > 0 we can choose continuity points a and b of F(x) such that F(b) F(a) > 1 . Moreover, we can. Modes of Convergence choose continuity points a = c1 < c2 < < cm = b of F(x) such that, for j = 1, . . .

, m 1,. x (c j ,c j+1 ]. (x) . x (c j ,c j+1 ]. (x) .. (6.17). Now de ne (x) =. x (c j ,c j+1 ]. (x). x (c j , c j+1 ], (6.18). j = 1, . . . , m 1, (x) = 0 elsewhere. Then 0 (x) (x) for x (a, b], 0 (x) (x) 1 for x (a, b]; / hence, limsup E[ (X n )] E[ (X n )]. n limsup n x (a,b] n (x) (x). dFn (x) . x (a,b] /. (x) (x). dFn (x) + + 1 lim ( Fn (b) Fn (a)) = + 1 (F(b) F(a)) 2 . Moreover, we have E[ (X )] E[ (X )]. 2 , and (6.19). (6.20). lim E[ ( X n )] = E[ (X )].. (6.21). If we combine (6. .NET Code128 19) (6.

21), the only if part easily follows. Proof of the if case: Let a < b be arbitrary continuity points of F(x), and let if x b, = 0 = 1 if x < a, (x) = (6.22) b x = b a if a x < b.

Then clearly (6.22) is a bounded continuous function. Next, observe that E[ (X n )] = (x)dFn (x).

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