SLIDE 1
Martingale Difference Central Limit Theorem Yichen Zhou May 9, 2016 - - PowerPoint PPT Presentation
Martingale Difference Central Limit Theorem Yichen Zhou May 9, 2016 - - PowerPoint PPT Presentation
Martingale Difference Central Limit Theorem Yichen Zhou May 9, 2016 Intuition Why martingale difference CLT. Statisticians rely on CLTs to make inference. CLTs we have seen before all require independence. Martingale difference CLT
SLIDE 2
SLIDE 3
Setup
◮ Martingale array:
{Sni, Fni, 1 ≤ i ≤ kn} zero-mean, square-integrable martingales for each n ≥ 1.
◮ It can be derived from an ordinary martingale
{Sn, Fn, 1 ≤ n} by setting kn = n, Fni = Fn, Sni = s−1
n Si
sn = (var(Sn))
1 2 .
SLIDE 4
◮ Martingale difference:
Xni = Sni − Sn,i−1.
◮ Notice
E(Xni|Fn,i−1) = 0.
◮ Conditional variance and squared variation of Sni:
V 2
ni = i
- j=1
E(X2
nj|Fn,j−1),
U2
ni = i
- j=1
X2
nj.
SLIDE 5
Martingale CLT
Theorem (Martingale CLT I)
Follow the notations above. Suppose η2 is an a.s. finite r.v., and max
i
|Xni|
p
− → 0,
- i
X2
ni p
− → η2, E
- max
i
X2
ni
- < M < ∞,
Fni ⊆ Fn+1,i. Therefore Snkn =
- i
Xni
d
− → Z, where Z has the characteristic function E(eitZ) = E
- exp(−1
2η2t2)
- .
SLIDE 6
Martingale CLT
Theorem (Martingale CLT II)
Follow the notations above. Suppse E
- max
i
|Xni|
- → 0,
- i
X2
ni p
− → σ2, then Snkn
d
− → N(0, σ2).
SLIDE 7
Proof of Martingale CLT
Main idea of the proof:
◮ Truncate the martingale array by a stopping time. ◮ Show that the truncation results in a negligible difference. ◮ Show that the truncated array converges in distribution to
normal.
SLIDE 8
Truncation
◮ WLOG, σ2 = 1. ◮ Define the stopping time
Tn = inf{t :
t
- i=1
X2
ni > 2} ∧ kn. ◮ Consider the new martingale array
{ ˜ Sni = Sn,i∧Tn, Fni, 1 ≤ i ≤ kn} with its differences Zni = Sn,i∧Tn − Sn,(i−1)∧Tn = Xni✶{i−1
j=1 X2 nj≤2}
= Xni✶{Tn>i−1}.
SLIDE 9
Truncation is Negligible
◮ Recall kn
- i=1
X2
ni p
− → σ2 = 1.
◮ After truncation,
P( ˜ Snkn = Snkn) = P(Xn,i = Zn,i for some i ≤ kn) = P(Tn ≤ kn − 1) ≤ P kn
- i=1
X2
ni > 2
- → 0.
◮ Therefore
˜ Snkn − Snkn
p
− → 0,
kn
- i=1
Z2
ni p
− → 1.
SLIDE 10
Convergence
◮ Suffices to show
E exp(it ˜ Snkn) → exp
- −t2
2
- .
◮ Taylor expansion
exp(ix) = (1 + ix) exp
- −x2
2 + r(x)
- ,
where |r(x)| < |x|3. Thus exp(it ˜ Snkn) =
kn
- j=1
exp(itZnj) =
kn
- j=1
(1 + itZnj) exp −t2 2
kn
- j=1
Z2
nj + kn
- j=1
r(tZnj) .
SLIDE 11
Martingale CLT
Lemma
For n ≥ 1, let {Un}, {Vn} be random variables satisfying the following conditions:
- 1. Un
p
− → a,
- 2. {Vn} and {VnUn} are uniformly integrable sequences,
- 3. EVn → 1.
Then EVnUn → a.
◮ Based the lemma, let
Vn =
kn
- j=1
(1+itZnj), Un = exp −t2 2
kn
- j=1
Z2
nj + kn
- j=1
r(tZnj) .
SLIDE 12
Martingale CLT
◮ Proof of the lemma:
Vn(Un − a) is u.i.. Suffices to show Vn(Un − a)
p
− → 0. P(|Vn(Un − a)| > ǫ) ≤ P(|Un − a| > ǫ/K) + P(|Vn| > K) → 0.
◮ |VnUn| ≤ 1, so {VnUn} is u.i.
SLIDE 13
Martingale CLT
◮ To show
Un = exp −t2 2
kn
- j=1
Z2
nj + kn
- j=1
r(tZnj) p − → exp
- −t2
2
- ,
notice kn
i=1 Z2 ni p
− → 1 and
- kn
- j=1
r(tZnj)
- ≤ |t|3
kn
- j=1
|Znj|3 ≤ |t|3
kn
- j=1
|Xnj|3 ≤ |t|3 max
j
|Xnj|
kn
- j=1
X2
nj p
− → 0.
SLIDE 14
Martingale CLT
◮ To show
Vn =
kn
- j=1
(1 + itZnj) =
Tn
- j=1
(1 + itXnj) is u.i., by |1 + iz|2 ≤ exp(z2), |Vn| =
j<Tn
|1 + itXnj| |1 + itXnTn| ≤ exp t2 2
- j<Tn
X2
nj
(1 + |t||XnTn|) ≤ exp(t2)(1 + |t| max
j
|Xnj|). Recall E(maxj |Xnj|) → 0, so maxj |Xnj| is u.i. So is Vn.
SLIDE 15
Martingale CLT
◮ To show EVn → 1, notice
EVn = E
kn
- j=1
(1 + itZnj) = E E
kn
- j=1
(1 + itZnj)
- Fn,kn−1
= E E (1 + itZnkn|Fn,kn−1)
kn−1
- j=1
(1 + itZnj) = E
kn−1
- j=1
(1 + itZnj) = 1.
SLIDE 16