and Stochastic Calculus 3rd Edition
by René Schilling, Böttcher, All 23 Chapters Covered
TEST BANK
,Contents
1 Robert Brown’s new thing 5
2 Brownian motion as a Gaussian process 15
3 Constructions of Brownian motion 29
4 The canonical model 39
5 Brownian motion as a martingale 49
6 Brownian motion as a Markov process 63
7 Brownian motion and transition semigroups 77
8 The PDE connection 99
9 The variation of Brownian paths 111
10 Regularity of Brownian paths 119
11 Brownian motion as a random fractal 125
12 The growth of Brownian paths 131
13 Strassen’s functional law of the iterated logarithm 137
14 Skorokhod representation 145
15 Stochastic integrals: L2–theory 147
16 Stochastic integrals: Localization 161
17 Stochastic integrals: Martingale drivers 165
18 Itô’s formula 169
19 Applications of Itô’s formula 183
20 Wiener Chaos and iterated Wiener–Itô integrals 195
21 Stochastic differential equations 207
22 Stratonovich’s stochastic calculus 225
23 On diffusions 227
,1 Robert Brown’s new thing
Problem 1.1. Solution:
a) We show the result for Rd-valued random variables. Let ξ, η ∈ Rd. By
ξ X ξ X
assumption, lim E exp [i c( ) , ( n))] = E exp [i c( ) , ( ))]
n→∞ η Yn η Y
⇐⇒ lim E exp [i⟨ξ, X n⟩+ i⟨η, Yn ⟩] = E exp [i⟨ξ, X ⟩ + i⟨η, Y ⟩]
n→∞
If we take ξ = 0 and η = 0, respectively, we see that
lim E exp [i⟨η, Y n ⟩] = E exp [i⟨η, Y ⟩] or Yn —
→
d
Y
n→∞
d
lim E exp [i⟨ξ, ⟩] = E exp [i⟨ξ, X ⟩] or → X.
—
n→∞
Xn Xn
Since Xn ı Yn we find
E exp [i⟨ξ, X ⟩+ i⟨η, Y ⟩] = lim E exp [i⟨ξ, Xn⟩+ i⟨η, Y n⟩]
n→∞
= lim E exp [i⟨ξ, X n⟩]E exp [i⟨η, Y n ⟩]
n→∞
= lim E exp [i⟨ξ, Xn ⟩] lim E exp [i⟨η , Yn ⟩]
n→∞ n→∞
= E exp [i⟨ξ, X ⟩] E exp [i⟨η, Y ⟩]
and this shows that X
ı Y
.
b) We have
1 almost sur d
Xn = X + ——————— X =⇒ → X
—
n n→∞
→
ely X n
1 almost surel y d
Y = 1 − X = 1 − −X ———————→ 1Y −X =⇒ —
→ 1 −X
n n n
n n→∞
almost surely d
Xn + = 1— 1 =⇒ + → 1.
—
n→∞
Yn → Xn Yn
, R.L. Schilling: Brownian Motion (3rd edn)
A simple direct calculation shows that 1 −X ∼21 (δ0 +δ1) ∼ Y . Thus,
d d d
X —
→ X, Y —
→ Y ∼ 1 −X, X + Y —
→ 1.
n n n n
Assume that (Xn, Y n) —
→d(X, Y ). Since X ı Y , we find for the distribution of X +Y :
X + Y ∼2 1 (δ0 + δ1)∗ 12 (δ 0 + δ1) = 14(δ 0 ∗ δ 0 + 2δ1 ∗ δ 0 + δ 1 ∗ δ1) = 1 (δ40 + 2δ 1 + δ2).
Thus, X + Y ∼/ δ0 ∼ 1 = limn(X n + Y n ) and this shows that we cannot have that
d
(X ) —→ (X, Y ).
+ Yn d—
→ X + Y : this follows since we have
c) If Xn ı Yn and X ı Y , then we have
Xn
for all ξ ∈ R:
lim E eiξ(Xn+ Yn ) = lim E eiξXn E eiξYn
→∞
n→∞ n
= lim E eiξXn lim E eiξYn
n→∞ n→∞
= E eiξX E eiξY
= E [eiξX eiξY ]
a )
= E eiξ(X+Y ).
A similar (even easier) argument works if (X n, Y n )d—
→ (X, Y ). Then we have
f (x, y) ∶ =eiξ(x+y)
is bounded and continuous, i.e. we get directly
lim E e iξ(Xn+Yn) lim E f (X n, Y n) = E f (X, Y ) = E eiξ(X +Y ).
n→∞ n→∞
For a counterexample (if Xn and Yn are not independent), see part b).
Notice that the independence and d-convergence of the sequences Xn, Yn already
implies X Yı and the d-convergence of the bivariate sequence( Xn , Y n) . This is a
consequence of the following
Lemma. Let (X n )n and ( Y)En n 1 be sequences of random variables (or
1
random vectors) on
E the same probability space (Ω , A , P). If
Xn ı Yn for all n E 1 and Xn ——→d
X and Y — —→ d
Y,
n
n→∞ n→∞
then (Xn , Yn) — —→
d
(X, Y ) and X ı Y .
n→∞
Proof. Write φ X , φ Y , φ X,Y for the characteristic functions of X , Y and the
pair
(X, Y ). By assumption
lim (ξ ) = lim E eiξXn = E eiξX = φX (ξ ).
φ→∞Xn
n→∞
n
6