A Proofs
Before proving Theorem 1, we need the following Lemmas.
Lemma 1. Let Assumption 1 hold. Then
sup ∣μ(Xs) | = Oa.s. (nε/4),
s∈[0,1]
sup ∣σ2(Xs)∣ = Oa.s.(nε/2),
s∈[0,1]
sup ∣g( fs) | = Oa.s. (nε/2),
s∈[0,1]
for any ε > 0, arbitrarily small.
A.1 Proof of Lemma 1
We start from the case when Xt follows (1). Define Rl = {inf t : |Xt| >l}. Thus, Rl is an
Ft-measurable stopping time. Let
min(t,Rl)
min(t,Rl) σ2 (Xs)dW1,s.
Xmin(t,Rl )
= μ ( Xs )d s + /
00
Obviously, for all t ≤ Rl, Xmin(t r) = Xt. Now let Ωl = {ω : Rl > 1} and l = ln = nε/4. Thus, given
the growth conditions in Assumption 1(a), Xt is a non-explosive diffusion, and so Pr(Ωln → 1) = 1.
By a similar argument, given Assumptions 1(a), 1(b), the same holds when the volatility process
follows (2). Therefore, the statement follows. ■
Lemma 2. Let Assumption 1 hold. Under H0, if, as n →∞, nξn →∞, nξn2 → 0 and, for any
ε > 0 arbitrarily small, m/n1 -ε → 0, then, pointwise in r,
√m
n
Sn2(Xi/n)
— σ2(Xi/n)) -→ 0.
i=1
A.2 Proof of Lemma 2
By Ito’s formula
√ l ( n-1)rj
m £ ( S2n ( Xi/n ) — σ 2 ( Xi/n ))
i=1
'------------------------------------------------------------------'
An,m,r
√m
n
l (n- 1)rj
i=1
∑j=11 {∣Xj∕n-Xi∕n∣<ξn}n (x(j+1)/n Xj/n) 2
vn-1 1 σ (Xi/n)
j=j =1 1 {∣Xj∕n-Xi∕n ∣<ξn}
18