r- L(n-1)rJ / yn-11 rι
+ √m v j=1 {∣X∙∕∙
n ⅛ I
i =1 \
4
-X√n∣<ξn} 2 n ʃ/n') / (Xs Xj/n ) μ ( Xs )d S
yn-1 1
X3=1 1 {∣Xj∕n-Xi∕n∣<ξn}
S/
S n,m,r
ч.
L ( n-1) rJ ∕y n-1 1
= 1 {lχj∕n-χi∕n∣<ξn
∑n-1
j =1
}n (ʃj/r+1)/n ( g( fs )- g( fi∕n)) d s) ʌ
} v J/. (25)
1 {IX∙∕n-Xi∕n∣<ξn}
S
Qn,m,r
Now, Pn,m,r is Op(1), again for the same argument used in the proof of part (i)a Step 1, and
similarly Sn,m,r = oP (1), since it has the same behavior under both Ho and H a . Then, we
need to show that Qn,m,r, in absolute value, diverges. Now,
L ( n-1) rJ y n— 11
1 Z) _ 1 v 3=1 υX∙∕n
√.∙ = n I
i=1
ч
L ( n- 1) rJ yn-11
+1 v j=1 { χ
n I
i =1 \
ч_______________________________________________
L (n-1) rJ ( y
1 v
n
i=1
4
n-1
= = 1 ^χj∕n-
-χ,∕,∣<tn}n (Cυ/n (g (χs) - g (Xi/n))ds)
y n-1
=1 1 {∣Xj∕n-Xi∕n∣<ξn}
V
T n,m,r
-Xi,,∖<ξn}n (Cj/n (g ( fs) - g (Xs))ds)
y n-1
=1 1 {∣X∙∕n-Xi∕n∣<ξn}
V
U n,m,r
∙Xi∕n∣<ξn} (g (fi/n) - g (Xi/n))
∑n-1
j = 1 1 {∣Xj∕n-Xi∕n∣<ξn}
--
V n,m,r
(26)
Note that Тптг is oP(1), by the same argument as the one used in part (i)a, Step 1.
As for Vn,m,r, it can be rewritten as
1 L(n-1) rJ / r r ʌ
^ ∑ (g (fi∕n) - g (χi∕n)) = Çyθ g(χs)ds- yɔ g(fs)dSJ + OP(n-1 /2),
(27)
where the first term of the right hand side of (27) is almost surely different from 0, given that
Xs and fs have different occupation density. Also, in the case in which fs is one-dimensional,
we have that
1 l ( n 1)rJ ∞ ∞
- ∑ (g (fi/n) - g (Xi/n)) = g(a)Lx(r,a)da - g(a)Lf (r,a)da + Op(n-1 /2),
n /_i J-OQ -∞
where Lf (r,a) (resp. Lx(r,a)) denotes the standardized local time of the process ft (resp.
Xt) evaluated at time r and at point a, that is it denotes the amount of time spent by the
25
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