Solution for HW 7
(a) and (b). Note that cfw_min(S, T ) n = cfw_S n cfw_T n Fn and cfw_max(S, T )
n = cfw_S ncfw_T n Fn . (c). We have cfw_S+T n = nk=0 cfw_S = kcfw_T = nk Fn .
2. (a).The identity is not true in general. Consider Xi i.i.d s.t. P(Xi = 0)
Solution for HW 8
Assume that , , are probability measures on (S, S) satisfying
. On one
d
hand, implies that (A) = A d d for all A S . On the other hand,
gives
d
that S f d = S f d d for all bounded measurable f 0. In particular, take f = 1A d for
d
d
Solution for HW 11
Dene N := infcfw_n; Xn > M for M > 0. According to Theorem 5.2.6, (XnN )nN is
+
+
also submartingale. Observe that XnN M + supn (Xn Xn1 )+ and thus supn EXnN
M + E supn (Xn Xn1 )+ < . By Theorem 5.2.8 (martingale convergence theorem),
Solution for HW 6
1. Let Aj := cfw_|Sj | > 2a and |Sk | 2a for k < j and Bj := cfw_|Sn Sj | a. Note that
Aj Bj cfw_|Sn | > a for j n. Moreover, (Aj )jn are pairwise disjoint () ; Aj and Bj are
()
independent for j n (), we have then P(|Sn | > a) P(n (Aj B
Solution for HW 5
Fix > 0. Using the argument in HW3, Q1(iv) we get that P(Xn n )
n=1
EX1
. The rst Borel-Cantelli lemma gives that P(Xn n i.o.) = 0, which implies that
lim supn Mn a.s. Since is arbitary small, we have lim supn Mn = 0. Similarly, lim inf
Solution for HW 7
1. (a) and (b). Note that cfw_min(S, T ) n = cfw_S n cfw_T n Fn and cfw_max(S, T )
n = cfw_S ncfw_T n Fn . (c). We have cfw_S+T n = n cfw_S = kcfw_T = nk Fn .
k=0
2. (a).The identity is not true in general. Consider Xi i.i.d s.t. P(Xi =
Solution for HW 3
1. (1). Suppose EXn0 < . This implies that a.s. Xn0 is nite. Since Xn X a.s., we have
for n n0 , Xn0 Xn Xn0 X a.s. and is positive. By monotone convergence theorem,
E(Xn0 Xn ) E(Xn0 X), which implies EXn EX. (2). Note that |X|1|X|>n 0 a.
205A Homework #7, due Tuesday 21 October.
1. Suppose S and T are stopping times. Are the following necessarily
stopping times? Give proof or counter-example.
(a) min(S, T )
(b) max(S, T )
(c) S + T .
2. Let (Xi ) be i.i.d. with EXi2 < . Let Sn =
bounded s
205A Homework #3, due Tuesday 23 September.
1. Use the monotone convergence theorem to prove the following.
(i) If Xn 0, Xn X a.s. and EXn < for some n then EXn EX.
(ii) If E|X| < then E|X|1(|X|>n) 0 as n .
(iii) If E|X1 | < and Xn X a.s. then either EXn
205A Homework #11, due Tuesday 25 November.
In each question, there is some xed ltration (Fn ) with respect to which
martingales are dened.
1. Let (Xn ) be a submartingale such that supn Xn < a.s. and E supn (Xn
Xn1 )+ < . Show that Xn converges a.s.
2.
Solution for HW 2
1. (a). Note that (Q (0, 1) = (Qc (0, 1) = 0 but (0, 1) = 1. Thus is not nitely
additive on B. (b). Consider B = (a1 , b1 ] (an , bn ] and B = (a1 , b1 ] (an , bn ] disjoint.
We can check that at most one of them is equal to 1 (otherwise
Solution for HW 1
1. (a). It is obvious that F. Take A F, then n N s.t. A Fn . This implies
Ac Fn F. Now take B F, then m N s.t. B Fm . We have A B Fmn since
(Fn )nN is increasing. Therefore, F is a eld. (b). Consider = N and Fn = (cfw_k; k n)
(i.e. all t
Solution for HW 9
We rst consider h(x, y) = 1xA 1yB where A, B are measurable sets in R. On one hand,
()
E(h(X, Y )|G)(w) = 1Y (w)B P(X A|G)(w) = 1Y (w)B (w, A), where () is due to the fact
that Y is G -measurable. On the other hand, h(x, Y (w)(w, dx) = 1
Solution for HW 10
1. Consider A := cfw_maxmn |Sm | > x and N := infcfw_m; |Sm | > x or m = n. Note that N is
2
a stopping time which is a.s. bounded by n. Observe that (Sn s2 )nN is martingale. Apply
n
2
Theorem 5.4.1 in Durrett, 0 = E(SN ss ) (x + K)2 P
Solution for HW 4
We
show directly (ii). By pairwise independence, we get EDn = 0 and EDn2 = n1 [ 01 f 2 (x)dx
R1
2
2
n
( 0 f (x)dx)2 ] := n . Using Chebyshev's inequality, we obtain P(|Dn | > ) V arD
= n
2
2.
R
x
X
S(xy)T ( y )f (x)
2. Take S and T two m
Solution for HW 2
(a). Note that (Q (0, 1) = (Qc (0, 1) = 0 but (0, 1) = 1. Thus is not nitely
0
0
0
0
additive on B. (b). Consider B = (a1 , b1 ] (an , bn ] and B 0 = (a1 , b1 ] (an , bn ] disjoint.
We can check that at most one of them is equal to 1 (ot
Solution for HW 3
(1). Suppose EXn0 < . This implies that a.s. Xn0 is nite. Since Xn X a.s., we have
for n n0 , Xn0 Xn Xn0 X a.s. and is positive. By monotone convergence theorem,
E(Xn0 Xn ) E(Xn0 X), which implies EXn EX . (2). Note that |X|1|X|>n 0 a.s.
Solution for HW 6
Let Aj := cfw_|Sj | > 2a and |Sk | 2a for k < j and Bj := cfw_|Sn Sj | a. Note that
Aj Bj cfw_|Sn | > a for j n. Moreover, (Aj )jn are pairwise disjoint () ; Aj and Bj are
() P
independent for j n (), we have then P(|Sn | > a) P(nj=1 (Aj
205A Homework #4, due Tuesday 30 September.
1. Monte Carlo integration [cf. Durr. 2.2.3] Let f : [0, 1] R be such
1
that 0 f 2 (x) dx < . Let (Ui ) be i.i.d. Uniform(0, 1). Let
n
Dn := n
1
1
f (Ui )
f (x) dx.
0
i=1
(i) Use Chebyshevs inequality to bound
205A Homework #8, due Tuesday 4 November.
[Theorem 7 and Corollary 8 refer to the notes linked from the week 8
row of the schedule.]
1. Suppose probability measures satisfy
. Show that
d
d d
=
.
d
d
d
2. In the setting of Theorem 7 [hard part], where S2
205A Homework #10, due Tuesday 18 November.
1. Let Sn = n i , where the (i ) are independent, Ei = 0 and var i <
i=1
n
2
2
. Let s2 =
n
i=1 var i . So we know that (Sn sn ) is a martingale.
Suppose also that |i | K for some constant K. Show that
max |Sm |
205A Homework #1, due Tuesday 9 September.
1. [Bill. 2.4] Let Fn be classes of subsets of S. Suppose each Fn is a eld,
and Fn Fn+1 for n = 1, 2, . . . Dene F = Fn . Show that F is a
n=1
eld. Give an example to show that, if each Fn is a -eld, then F need
205A Homework #5, due Tuesday 7 October.
1. Let (Xn ) be i.i.d. with E|X1 | < . Let Mn = max(X1 , . . . , Xn ). Prove
that n1 Mn 0 a.s.
2. [Durr. 2.3.2] Let 0 X1 X2 . . . be r.v.s such that EXn an and
var(Xn ) Bn , where 0 < a, B < and 0 < < 2 < . Prove t
205A Homework #2, due Tuesday 16 September.
1. [similar Bill. 2.15] Let B be the Borel subsets of R. For B B dene
(B) = 1
=0
if (0, ) B for some > 0
if not
(a) Show that is not nitely additive on B.
(b) Show that is nitely additive but not countably addit
205A Homework #6, due Tuesday 14 October.
1. [Durr. 2.5.9] Let (Xi ) be independent, Sn =
Prove that
P (Sn > 2a)
n
i=1 Xi ,
Sn = maxin |Si |.
P (|Sn | > a)
, a > 0.
minjn P (|Sn Sj | a)
[Hint. If |Sj | > 2a and |Sn Sj | a then |Sn | > a.]
2. [Durr. 2.5.1
Solution for HW 4
1
1
2
1. We show directly (ii). By pairwise independence, we get EDn = 0 and EDn = n [ 0 f 2 (x)dx
2
1
2
( 0 f (x)dx)2 ] := . Using Chebyshevs inequality, we obtain P(|Dn | > ) V arDn = n 2 .
2
n
2. Take S and T two measurable bounded fu