Biomedical Engineering Reference
In-Depth Information
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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