Questions on Martingales Problem 1

Questions on Martingales

Problem 1. Show that P > Q if and only if for every « > 0 there is a 6 > 0 such that P(F) < 6
implies Q(F) < «. Not to turn in: What does this say about the absolute continuity lemma?
Problem 2. Prove Kolmogorov’s maximal inequality using Martingales.
Problem 3. Let (2,7, P) = ((0,1],8,dx) be the “standard” probability space where here, dz,
denotes the Lebesgue measure. Let
Inn =[ AK oR+ 1)
kn Qn’ Qn ;
Let An = Un}e_o let F, denote the filtration formed by looking at all of these intervals for
fixed n, ie., Fy = o(An).
1. What is ¥, ? [Just explain in words, no proof needed]
2. (Lebesgue’s differentiation theorem) Prove that for any x € (0,1), if we let J, € A, be such
that x € J,,, then for any f € L'(dr) we have
f(z) = jim Tf study
n—00 | In| In ,
almost surely. Here |Jn| = o denotes the length of Jn.
Problem 4. (The law of the iterated logarithm) Recall that by the CLT we know that if X; are
iid N(0,1) random variables then the CLT says that
X;
» Vn”
in law, where Z ~ N(0,1). Now we're going to do one better:
im ——2-*i__ =}
V 2n log log n
to this end, let h(k) = /2k log log k. In this problem, we'll just do the upper bound, the matching
lower bound is not to be turned in and will be problem 5.
1. Consider the Exponential martingale: for 9 > 0, let M, = e9Sn-"> Show that My is a
martingale

2. Use Doob’s submartingale inequality to conclude that
P(sup
k?n
Sk t) ? e ?t
en ?2
2
3. Optimizing the choice of ?, show that in fact
P(sup
k?n
Sk t) ? e
t2
2n
4. Let K > 1 real, let cn = Kh(Kn 1). Show that P(supk?Kn Sk > cn) ? 1
((n 1) log K)K .
5. Conclude that lim Sk
p
2k log log k
? 1 .
Problem 5. (Not to be turned in) The matching lower bound in the LIL. This part’s not so
bad bit a little messy, so I’ll leave it for you to think about. Do the following:
1. For the other side, fix N > 1 and integer let S(r) = Sr and define
Fn = S(Nn+1
) S(Nn
) > (1 ?)h(Nn+1
Nn
)
Show that if we let y = (1 ?)
p
2 log log(Nn+1 Nn) we have that
P(Fn)
1
p
2?
1
y + 1
y
exp( y2
/2).
[Hint: recall Mill’s inequality which says that if Z ? N(0, 1) we have that for every y > 0,
P(Z y)
1
y + 1
y
e y2/2
p
2?
]
2. convince yourself that the lower bound scales like 1/(n log N)(1 ?)2
which is not summable
3. using this fact conclude that
lim
Sk
h(k)
(1 ?)
q
1 1
N
2
p
N
send N ! 1 to get the result.