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Derivation of the Asymptotic Distribution of bn 797

and B be nonrandom a * m and b * m matrices, and let d be a nonrandom a * 1 vector.
Then

d + AV is distributed N(d + AMV, A VA′); (18.74)

cov (AV, BV ) = A VB′; (18.75)

if A VB′ = 0a * b, then AV and BV are independently distributed; and (18.76)

(V - MV)′ V-1(V - MV) is distributed xm2 . (18.77)

Let U be an m-dimensional multivariate standard normal random variable with distribu-
tion N(0, Im). If C is symmetric and idempotent, then

U′CU has a xr2 distribution, where r = rank(C). (18.78)

Equation (18.78) is proven as Exercise 18.11.

Appendix

18.3 Derivation of the Asymptotic Distribution of βn

This appendix provides the derivation of the asymptotic normal distribution of 2n(Bn - B)
given in Equation (18.12). An implication of this result is that Bn ¡p B.
First consider the “denominator” matrix X′X > n = n1 g in=1XiX′i in Equation (18.15). The
n1 n
(j, l) element of this matrix is g i=1 XjiXli. By the second assumption in Key Concept 18.1,

Xi is i.i.d., so XjiXli is i.i.d. By the third assumption in Key Concept 18.1, each element of

Xi has four moments, so, by the Cauchy–Schwarz inequality (Appendix 17.2), XjiXli has two
n
moments. Because XjiXli is i.i.d. with two moments, n1 g i=1 XjiXli obeys the law of large

numbers, so n1 g n 1 Xji Xli ¡p E(Xji Xli). This is true for all the elements of X′X > n, so
i=
X′X > n ¡p E(XiX′i) = QX.
n
Next consider the “numerator” matrix in Equation (18.15), X′U > 2n = 2n1 g i= 1Vi,

where Vi = Xiui. By the first assumption in Key Concept 18.1 and the law of iterated

expectations, E(Vi) = E[XiE(ui|Xi)] = 0k+1. By the second least squares assumption,

Vi is i.i.d. Let c be a finite k + 1 dimensional vector. By the Cauchy–Schwarz inequality,

E[(c′Vi)2] = E[(c′Xiui)2] = E[(c′Xi)2(ui)2] … 2E[(c′Xi)4]E(u4i ), which is finite by the

third least squares assumption. This is true for every such vector c, so E(ViV′i) = V is

finite and, we assume, positive definite. Thus the multivariate central limit theorem of Key

Concept 18.2 applies to 2n1 g n 1Vi = 1 X′U; that is,
i= 2n

1 X′U ¡d N(0k + 1, V). (18.79)
2n