Biomedical Engineering Reference
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By (9.27) and assumption (A2), using the Cauchy{Schwarz inequality, the
last four terms on the right side in the above equation converge to zero in
probability. So we have
P` 2 (; b n ; b n ) = P` 2 (; 0 ; 0 ) + o p (1) ! p I( 0 ):
This combined with assumptions (A1) and (A2) imply that
P n ` 2 (; b n ; b n ) = (P n P)` 2 (; b n ; b n ) + P` 2 (; b n ; b n ) ! p P` 2 (; 0 ; 0 ) = I( 0 ):
This completes the proof for the case of d = 1.
When d > 1, let ` j (x; ;h) be the j-th element of `(x; ;h). We need to
show that
P n ` j (; b n ; b n )` k (; b n ; b n ) ! p P` j (; 0 ; 0 )` k (; 0 ; 0 ); 1 j;k d: (9.28)
When j = k, this follows from the proof for d = 1. When j 6= k, then by
assumption (A1), we have
(P n P)` j (; b n ; b n )` k (; b n ; b n ) ! p 0:
(9.29)
Write
P[` j (; b n ; b n )` k (; b n ; b n ) ` j (; 0 ; 0 )` k (; 0 ; 0 )]
Pf` j (; b n ; b n )[` k (; b n ; b n ) ` k (; 0 ; 0 )]g
+Pf` k (; 0 ; 0 )[` j (; b n ; b n ) ` j (; 0 ; 0 )]g
=
T 1n + T 2n :
By Equation (9.24), P` j (; b n ; b n ) = O p (1). Thus, by the Cauchy-Schwarz
inequality,
T 1n P` j (; b n ; b n )P[` k (; b n ; b n ) ` k (; 0 ; 0 )] 2 ! p 0
(9.30)
and
T 2n P` k (; 0 ; 0 )P[` j (; b n ; b n ) ` j (; 0 ; 0 )] 2 ! p 0:
(9.31)
Now (9.28) follows from (9.29) to (9.31). This completes the proof.
 
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