MA414 SOLUTIONS 5. 13.2.2012
Q1. Proof (Doobs Submartingale Inequality). Let
F := cfw_max Xk c,
Fk := cfw_X0 < ccfw_X1 < c. . . cfw_Xk1 < ccfw_Xk c.
Then F is the disjoint union F = F0 . . . Fn . Also Fk Fk , and Xk c
on Fk . So
E [Xn I
Lecture 3. 26.1.2012
9. The Borel-Cantelli lemmas and the zero-one law.
First, recall from Real Analysis the denition of the upper and lower
limit, lim sup and lim inf, of a real sequence xn :
lim sup xn := inf sup xk , = lim sup xk
Lecture 4. 2.2.2012
II. STOCHASTIC PROCESSES
1. Conditional expectations.
Suppose that X is a random variable, whose expectation exists (i.e.
E |X | < , or X L1 ). Then EX , the expectation of X , is a scalar (a
number) non-random. The expecta
Lecture 5. 9.2.2012
5. Martingales: discrete time. We refer for a fuller account to [W]. The
classic exposition is Ch. VII in Doobs book [D] of 1953.
Denition. A process X = (Xn ) in discrete time is called a martingale
(mg) relative to (cfw_F
Lecture 6. 16.2.2012
Corollary (Doob). A non-negative supermg Xn is a.s. convergent.
Proof. As Xn is a supermg, EXn decreases. As X 0, E [Xn ] 0. So
E [|Xn |] = E [Xn ] is decreasing and bounded below, so (convergent and)
bounded: Xn is L1 -bo
Lecture 7. 23.2.2012
A non-negative right-continuous submartingale is of class (D). So it has
a Doob-Meyer decomposition. We specialize this to X 2 , with X cM2 :
X 2 = X0 + M + A,
with M a continuous martingale and A a
Lecture 8. 1.3.2012
n W t =
cfw_(W (ti+i )W (ti )2 (ti+1 ti ) =
cfw_(i W )2 (i t) =
where since i W N (0, i t), E [(i W )2 ] = ti , so the Yi have zero mean,
and are independent by independent increments of W . So
E [(n W
Lecture 9. 8.3.2012
Example. We calculate W (u)dW (u). We start by approximating the integrand by a sequence of simple functions.
W (0) = 0
Xn (u) = .
W (n 1)t/n)
0 u t/n,
t/n < u 2t/n,
(n 1)t/n < u t.
Lecture 10. 15.3.2012
Theorem (Brownian Martingale Representation Theorem). Let M =
(M (t)t0 be a RCLL local martingale with respect to the Brownian ltration
(Ft ). Then
M (t) = M (0) +
H (s)dW (s), t 0
with H = (H (t)t0 a progressively m
MA414 SOLUTIONS 2. 26.1.2012
k := X k E [X k ] = k
a.s. (n ).
The k th central moment is 0 := (X X )k . Then
X i ()ki (X )ki
(X i )()ki (X )ki .
By SLLN, as n this tends a.s. to
MA414 SOLUTIONS 3. 7.2.2011
Q1. We are using almost all for numbers on the line; this means under
Lebesgue measure (unless otherwise stated). We are talking about decimal
expansions; this relates to the fractional part, which is in [0, 1];
Lecture 2. 19.1.2012.
5. Modes of convergence.
We need (at least) four modes of convergence two strong, one intermediate, one weak. We begin with the strong modes.
We say that Xn X almost surely, or a.s., if Xn X with probability
1: P (Xn X )