(i)c The statistic Zn,m,r can be rewritten as

√— L (n-1) rJ
— £ (S2n(Хг/п) - σ2(Хг/п))
i =1
s--------------------------------S/--------------------------------'
An,m,r
—

L ( m— 1) rJ
Σ
j=1
'm - Xj/m) - σ σ2(Xs)dS
JO
S/
Bm,r
√— l ( n-1)rJ r
+— ∑ σ2(Xi/n) - V— I σ2(Xs)ds.
n i=1 0o
'-----------------------------------v-----------------------------------'
C n,m,r
(23)
Note that An,m,r = op(1) by Lemma 2.
We first need to show that Cn,m,r = oa.s.(1). Given Assumption 1(a), Lemma 1, and recalling
the modulus of continuity of a diffusion (see McKean, 1969, pp.95-96),
—— l ( n- i)rJ
— 52 σ2(Xi/n) -V— σ2(Xs)ds
n TX -ʃo
l— l ( n-1)rJ
— Σ σ 2( Xi/n )
i=1
l ( n-1)rJ r( i+1) /n
— ∑ σ2 ( Xs )d s
i=1 i/n
(σ2(Xi/n) - σ2(Xs)) ds
l ( n-1)rJ r( i+1) /n
— ∑
i=1 i/n
L ( n-1)rJ r( i +1) /n

- σ2(Xs)∣ ds
V— sup ∣σ2(Xs) - σ2(Xτ)∣ ≤ V— sup ∣Vσ2(Xτ)∣ sup ∣Xs - Xτ∣
|s—τ ∣≤ 1 /n τ ∈ [0 ,r ] |s—τ ∣≤ 1 /n
s∈ [θ ,r ] s∈ [θ ,r ]
V—iC>(1.S( (nε/2)Oa.s. (n 1 /2 log n) = 0a.s. (1),
as n1 /2 ε/2n 1 /2 log n → 0. Thus,
Zn,m,r
Bm,r + 0a.s.(1).
The statement then follows from the proof of Step 2 in part i(a).
(i)d The statement follows by the same argument as the one used in part (i)b and by the continuous
mapping theorem.
23
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