MATH 507b ASSIGNMENT 3 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 5: (1.2) 1 2 in two steps means 1 3 2, so p2 (1, 2) = p(1, 3)p(3, 2) = (.9)(.4) = .36. 2 3 in 3 steps means 2 1 3 3 or 2 2 1 3 or 2 1 1 3, so p3 (2, 3) = (.7)(.9)(.6) +
MATH 507a ASSIGNMENT 10 SOLUTIONS
FALL 2012
Prof. Alexander
(5.1.9) In Exercise 1.8 we can take G = cfw_, so that E(X | G) = EX. Then (continuing the notation from that exercise) we have E(U + V )2 = E(X EX)2 = var(X).
By definition, var(X | F) = E(U 2 |
MATH 507a ASSIGNMENT 1 SOLUTIONS
FALL 2012
Prof. Alexander
(1.1.1)(i) Suppose A1 , A2 , . i Fi . Then for each i we have A1 , A2 , . Fi so n An Fi .
Therefore n An i Fi .
Suppose now that A i Fi . Then A Fi for all i, so Ac Fi for all i, so Ac i Fi .
Thus
MATH 507a ASSIGNMENT 6 SOLUTIONS
FALL 2012
Prof. Alexander
(3.1.2) Let k = k(n) be a sequence for which the quantity xn = (k n)/ n converges to
some x. Then nk = xnn + 1 1. We use an bn to mean the ratio converges to 1. By
Stirlings formula,
nk
2nP (Sn =
MATH 507a ASSIGNMENT 7 SOLUTIONS
FALL 2012
Prof. Alexander
(3.3.1) Let be the p.m. with characteristic function , and let X be the corresponding
random variable. Let X 0 be another r.v. independent of X, also with distribution . Then
0
0
Eeit(XX ) = EeitX
MATH 507a ASSIGNMENT 4 SOLUTIONS
FALL 2012
Prof. Alexander
(2.3.8)(a) Let g(x) = x/(1 + x) for x 0. Then g 0 (x) = 1/(1 + x)2 is decreasing, so for
a, b 0,
Z a+b
Z b
0
g(a + b) g(a) =
g (x) dx
g 0 (x) dx = g(b) g(0) = g(b),
a
0
so g(a + b) g(a) + g(b). S
MATH 507a ASSIGNMENT 5 SOLUTIONS
FALL 2012
Prof. Alexander
(2.5.5) First show (ii) (iii). Let
(
x for x 1,
g(x) =
,
1 for x > 1,
h(x) =
x
.
1+x
Then P
(checking separately for x 1 and
Pfor x > 1) we readily see that h g 2h. Now (ii)
n Eg(Xn ) < and (iii)
MATH 507a ASSIGNMENT 2 SOLUTIONS
FALL 2012
Prof. Alexander
(1.3.4)(ii) Suppose F is a -field in Rd for which all continuous functions are measurable. To
show Rd F its enough to show that V F for every open V Rd . For this, in turn, it
is enough to show th
MATH 507a ASSIGNMENT 8 SOLUTIONS
FALL 2012
Prof. Alexander
= c > . Then the real part has the same limit, so by
(3.3.19) Suppose limt&0 (t)1
t2
Fatous Lemma,
Z
Z
Z 2
1
1 cos tx
1 cos tx
x
> c = lim
(dx) = EX 2 . (1)
(dx)
lim
(dx) =
2
2
t&0 R
t
t
2
R t&0
MATH 507a
SAMPLE FINAL EXAM
FALL 2012
Prof. Alexander
(1) Let X1 , X2 , . . . be iid with characteristic function , and suppose 0 (0) = ia for some
real a. Show that (X1 + + Xn )/n a in probability. HINT: In place of convergence in
probability, prove anot
MATH 507a
MIDTERM EXAM SOLUTIONS
Fall 2012
Prof. Alexander
(1) For every Borel A,
P (f (X) A) = P (f (X) A, X f 1 (A)
= P (f (X) A)P (X f 1 (A) by independence
= P (f (X) A)2 ,
so P (f (X) A) = 0 or 1. In particular, the d.f. of f (X) takes only the value
MATH 507a ASSIGNMENT 3 SOLUTIONS
FALL 2007
Prof. Alexander
(2.1.8) (i) By Fubinis Theorem,
Z Z
P (X + Y = 0) =
1cfw_x+y=0 (dx) (dy)
Z Z
1cfw_x=y (dx) (dy)
=
R
R
Z
=
(cfw_y) (dy).
R
R
R
Let A = cfw_y : (cfw_y) > 0. Then A is countable so this integral is
Z
MATH 507a ASSIGNMENT 9 SOLUTIONS
FALL 2012
Prof. Alexander
P
P
P
(3.4.13)(i) If > 1 then j P (Xj 6= 0) = j j < so P (Xj 6= 0 i.o.) = 0 so
j=1 Xj
converges a.s., that is, Sn S .
(ii) For < 1 we use Lindeberg-Feller, with Xnj = Xj /n(3)/2 . We need a stand
MATH 507b ASSIGNMENT 4 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 5: (2.1) Let A (X0 , ., Xn ) and B (Xn , Xn+1 , .). Using Theorem 1.2 Chapter 4 (p. 224) and the Markov property, since 1A Fn we get P (A B | Xn ) = E (E (1A 1B | Fn ) | Xn ) =
MATH 507b ASSIGNMENT 2 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 4: (7.1) Let N = min{n : Xn > }, with N = if there is no such n. By Theorem 7.6, EX0 EXN
{XN >}
XN dP P (XN > ) = P (sup Xn > ),
n
so P (supn Xn > ) EX0 /. (7.2) By Theorem 4
MATH 507b ASSIGNMENT 1 SOLUTIONS SPRING 2008 Prof. Alexander (4.4) Fix n and define the bounded stopping time N = n min{k : |S)k| x}. Since 2 |m | K we have |SN | x + K, so ESN (x + K)2 . Also Es2 E(s2 1{N =n} ) = s2 P (N = n) s2 P (max |Sm |
MATH 507b ASSIGNMENT 8 SOLUTIONS SPRING 2008 Prof. Alexander (1) The mean of I(f, t) = 0 f (s) dWs is 0 since it's a martingale and I(f, 0) = 0, so the t variance is EI(f, t)2 = 0 Ef (s)2 ds. In particular: (a) For f (s) = |Ws |1/2 , we have Ef (s)2
MATH 507b ASSIGNMENT 6 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 7: (1.3) Let be N (0, t), so m,n has the distribution of 2-n/2 . We have E 2 = t, E( 2 - t)2 = 2t2 , so the r.v.'s 2 - 2tn are i.i.d. with mean 0. Therefore m,n
2n 2 2n 2
E
m=1
2
MATH 507b ASSIGNMENT 7 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 7: (4.1)(i) Let u < v a and Ys () = 1{(t-s)(u,v)} for s < t, so we have YTa (Ta )1{Ta <t} = 1{(t)(u,v)} 1{Ta <t} . Note that (2a - v, 2a - u) is the "mirror image" of (u, v) relati
MATH 507b ASSIGNMENT 5 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 5: (4.3) The stationary measure is unique up to constant multiples, which means the ratio of the stationary measure at any two points y and z doesn't depend on the base point, say y
MATH 507b TAKE-HOME MIDTERM SOLUTIONS SPRING 2008 Prof. Alexander (1)(a) Since E(Yk | Fk-1 ) Yk-1 we have for k > n that E(Yk | Fn ) = E(E(Yk | Fk-1 ) | Fn ) E(Yk-1 | Fn ), so {E(Yk | Fn ), k n} is monotone nondecreasing a.s., so limk E(Yk | Fn )
MATH 507b TAKE-HOME FINAL EXAM SOLUTIONS SPRING 2008 Prof. Alexander (1)(a) Suppose an a. Then {Tan } is a bounded increasing sequence so Tan S for some finite S. Then an = BTan BS since Bt is continuous, so BS = a, and therefore S = Ta . Thus Tan T
MATH 507a
FINAL EXAM SOLUTIONS Fall 2007 Prof. Alexander
(1) Sn /n (t) = Ee For fixed t, as n , t n
n itSn /n
= Sn
t n 1 n
n
=
t n
n
.
=
t 1 + (0) + o n
e (0)t = eiat ,
so Sn /n a in distribution, hence also in probability. (2)(a) P
MATH 507a
SAMPLE MIDTERM SOLUTIONS
Fall 2012
Prof. Alexander
(1) Since (x) , we can find 0 a1 < a2 < . . . with (ak ) 2k . Then let X take the
value ak with probability 2k , so that
E(X) =
X
(ak )P (X = ak )
k=1
X
1 = .
k=1
(2)(a) Let K = supx |g 0 (x)|.