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is continuous in y on a little neighborhood of c. in .NET Encode Code 128 Code Set B in .NET is continuous in y on a little neighborhood of c.

is continuous in y on a little neighborhood of c. using .net topaint barcode code 128 on asp.net web,windows application interleaved 25 The Mathematical and Statistical Foundations of Econometrics 2 of large numbers (Theorem 6 .2) we have plimn Yn = E(U1 ) = 1. Let (x, y) = x/ y.

Note that (x, y) is continuous on R (1 , 1 + ) for 0 < < 1. Thus, by Theorem 6.21, Z n = (X n , Yn ) (X, 1) = U0 N (0, 1) in distribution.

Q.E.D.

. Corollary 6.2: Let U1 . .

. U .net framework Code 128 Code Set B n be a random sample from Nk ( , ), where is nonsingular.

Denote U = (1/n) n U j , = (1/(n 1)) n (U j j=1 j=1 )(U j U )T , and let Z n = n(U )T 1 (U ). Then Z n d 2 . U k Proof: For a k k matrix A = (a1 , .

. . , ak ), let vec(A) be the k 2 1 vecT T tor of stacked columns a j , j = 1, .

. . , k of A : vec(A) = (a1 , .

. . , ak )T = b, 1 ), X n = for instance, with inverse vec (b) = A.

Let c = vec( ), Yn = vec( n(U ), X Nk (0, ), and (x, y) = x T (vec 1 (y)) 1 x. Because is nonsingular, there exists a neighborhood C( ) = {y Rk k : y c < } of c such that for all y in C( ), vec 1 (y) is nonsingular (Exercise: Why ), and consequently, (x, y) is continuous on Rk C( ) (Exercise: Why ). The corollary follows now from Theorem 6.

21 (Exercise: Why ). Q.E.

D. 6.6.

Convergence of Characteristic Functions Recall that the characteristic function of a random vector X in Rk is de ned as (t) = E[exp(itT X )] = E[cos(t T X )] + i E[sin(t T X )] for t Rk , where i = 1. The last equality obtains because exp(i x) = cos(x) + i sin(x). Also recall that distributions are the same if and only if their characteristic functions are the same.

This property can be extended to sequences of random variables and vectors: Theorem 6.22: Let X n and X be random vectors in Rk with characteristic functions n (t) and (t), respectively. Then X n d X if and only if (t) = limn n (t) for all t Rk .

Proof: See Appendix 6.C for the case k = 1. Note that the only if part of Theorem 6.

22 follows from Theorem 6.18: X n d X implies that, for any t Rk ,. lim E[cos(t T X n )] = E[cos( t T X )]; lim E[sin(t T X n )] = E[sin(t T X )];. hence,. lim n (t) = lim E[cos(t T X n )] + i lim E[sin(t T X n )]. = E[cos(t X )] + i E[sin(t .net vs 2010 Code 128C X )] = (t)..

n T Modes of Convergence Theorem 6.22 plays a key role in the derivation of the central limit theorem in the next section. 6.

7. The Central Limit Theorem The prime example of the concept of convergence in distribution is the central limit theorem, which we have seen in action in Figures 6.4 6.

6: Theorem 6.23: Let X 1 , . .

. , X n be i.i.

d. random variables satisfying E(X j ) = , var (X j ) = 2 < and let X = (1/n) n X j . Then n( X ) d j=1 2 N (0, ).

Proof: Without loss of generality we may assume that = 0 and = 1. Let (t) be the characteristic function of X j . The assumptions = 0 and = 1 imply that the rst and second derivatives of (t) at t = 0 are equal to (0) = 0, (0) = 1, respectively; hence by Taylor s theorem applied to Re[ (t)] and Im[ (t)] separately there exists numbers 1,t , 2,t [0, 1] such that 1 (t) = (0) + t (0) + t 2 Re[ ( 1,t t)] + i Im[ ( 2,t t)] 2 1 2 = 1 t + z(t)t 2 , 2 for instance, where z(t) = (1 + Re[ ( 1,t t)] + i Im[ ( 2,t t)])/2.

Note that z(t) is bounded and satis es limt 0 z(t) = 0. Next, let n (t) be the characteristic function of n X . Then n n (t) = (t/ n) 1 = 1 t 2 /n + z(t/ n) t 2 /n 2 n 1 = 1 t 2 /n 2.

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