2 Wiener Process or Brownian motion

Question 1

RECAP
probability space
RVs

F\B(R) measurable

Answer 1

prob space
(Ω,F,P)
rvs
functs X : Ω → R which are F\B(R) meausrable

this means
inverse image X₋₁(A)∈F

Question 2

recap
lemma linking sigma algebra is generated by a collection of sets
which are open intervals

{X<a}

Answer 2

equivalent so saying
inverse image of all elements in this generating class
X₋₁(A)∈F ∀ A∈[ (a,infinity) : a∈(R ]

{X<a}={w∈Ω : X(w)<a}
subset of Ω so is an event

this is a RV if ∈F for any a

Question 3

RECAP CONDITION EXPECTATION

Answer 3

G ⊂ F be a sub-σ-algebra of F
provided
||x|| =E[|X|] < infinity

then we can define new RV,
the conditional expectation
E[X|G] which has 2 properties:

1) E[X|G] G measurable: meaning {X>a}⊆ G⊆ F for all a in R. In particular, if I take an X which is G-measurable and thus it is F-measurable

2) E[1_A X] = E[1_A E[X|G]] ∀ A∈G

cond exp is itself a RV measurable wrt sigma algebra G

Question 4

Consider a fixed prob space
(Ω,F,P)
stochastic processses will

Answer 4

describe evolutions in time

time will be an added parameter to w in Ω

Question 5

Consider:
probability space tossing a coin n times

im interested in first outcome:
RV X

Answer 5

Ω={ (a_1,…,a_n): a_i∈{H,T}}

each outcome of the form
(H,T,H,H,…T)
tells you what happens at each coordinate index

if we are interested only in say second outcome we will need to state a function which outputs the second coord

Define RV X₁ which tells you what happened in the first toss: for w in Ω
Xₙ:Ω to {H,T} or to R
X₁(w)=X_1((a₁,…aₙ)) = a₁
Xₙ(w)=X_1((a₁,…aₙ)) = aₙ

Question 6

considering DICRETE time as well as
Ω={ (a_1,…,a_n): a_i∈{0,1}}

Define RV X giving nth

Answer 6

X:Ω to {0,1}
instead we consider with to vars
X_t(w)
X_t : Ω × {1,…n} → R
Time variable {1,…n}

if i fix time a variable it becomes:
X : Ω → R
a random var as a function of Ω

If I fix an w in Ω
X : {1,…n} → R
n choices in the time var
X_t(w) = (X₁(w),…,Xₙ(w))
if w is fixed I get real numbers, giving a vector corresp to coordinates (not RV)

X₁(w)=X_1((a₁,…aₙ)) = a₁
Xₙ(w)=X_1((a₁,…aₙ)) = aₙ

Question 7

remark : “to make my conscious clear”

Answer 7

X_t : [0,1] ×Ω → R
You can actually put a sigma algebra on this set,
taking the collection A of sets of the form

A={ (a,b) X A : a,b in [0,T], A ∈F}

sigma algebra generated by this collection
σ(A)=:B([0,T]) ⊗ F

It follows that we can fix t or w

Question 8

Definition 2.1.1. A stochastic process

Answer 8

A stochastic process is a map X : Ω × [0, T] → R which is F ⊗ B([0, T])-measurable.

Notice that a stochastic process is a function of two variables (ω, t) → X(ω, t).

(a stochastic process is a collection of random vars, uncountably many?)

Question 9

A stochastic process is a map X : Ω × [0, T] → R which is F ⊗ B([0, T])-measurable.
If we fix t ∈ [0, T]

Answer 9

function X(·, t) : Ω → R is F-measurable, hence it is a random
variable.
That is, we can see a stochastic process as a collection of random variables X(t) indexed
by t ∈ [0, T].

Question 10

notation stochastic process

Answer 10

(X(t))_{t∈[0,T]}
.

Question 11

A stochastic process is a map X : Ω × [0, T] → R which is F ⊗ B([0, T])-measurable.
, if we fix ω ∈ Ω,

Answer 11

, if we fix ω ∈ Ω, then we have a function of time t 7→ X(ω, t), which will be called
the trajectory or path corresponding to the fixed ω.

if we fix time then we have a rv corresponding to this

Question 12

continuous stochastic process

Answer 12

a stochastic process X is continuous if for all ω ∈ Ω if the trajectory t → X(ω, t) is a continuous function of time.
From now on, unless otherwise indicated, we drop the ω dependence as usual

(if with probability 1 the map is continuous
P(t → Xₜ, is continuous)=1
{ω ∈ Ω | t → Xₜ(ω,) is continuous}

Question 13

note on stochastic processes

Answer 13

trajectories correspond to individual ω, and these are continuous

each outcome is a function not a number

Question 14

Remark 2.1.2. Functions Y : Ω × [0, T] → R which satisfy the following…
Caratheodory functions

Answer 14

Functions Y : Ω × [0, T] → R which satisfy the following
* for each ω ∈ Ω, the map t → Y (ω, t) is either left or right continuous
* for each t ∈ [0, T], the map ω → Y (ω, t) is F-measurable,
are called Caratheodory functions. It can be checked that Caratheodory functions are F ⊗B([0, T])-
measurable.

Question 15

Definition 2.1.3. Wiener process

Brownian motion

Answer 15

Definition 2.1.3. A continuous stochastic process
W_t or B_t
(W(t))t∈[0,T] is called Wiener process if
1. W(0) = 0
2. For all s < t, W(t) − W(s) ∼ N (0, |t − s|)
3. For all 0 ≤ t₁ < t₂ < … < tₙ ≤ T, the random variables
W(t₂) − W(t₁), …, W(tₙ ) − W(tₙ₋₁)
are independent.

Question 16

summary brownian motion

Answer 16

(all trajectories start from origin)
(differences/ which are RVS themselves have Normal distribution with mean and variance)
(increments along intervals are independent , time intervals are disjoint wouldnt be true otherwise)

In summary:
Brownian motion is a stochastic process with continuous trajectory, continuous funct of time
starts from 0 probability 1
distribution for intervals is gaussian

The distributuion of RV W_t~N(0,t)

Question 17

Exercise 2.1.4. Let (W(t))_{t∈[0,T]} be a Wiener process,
let c > 0 and set

W~(t) = cW(t/c²) for
t ∈ [0, c²T].

Show that W~ is a Wiener processes on [0, c²T].

Answer 17

W~(t) is a stochastic process
We need to check:
1. W~(0)=cW(0/c²) =c W(0)= 0
2. For all s < t, W(t) − W(s) ∼ N (0, |t − s|)
3. For all 0 ≤ t₁ < t₂ < … < tₙ ≤ T, the random variables
W(t₂) − W(t₁), …, W(tₙ ) − W(tₙ₋₁)
are independent.

Notes…

LECTURE:
1)Due to property of wiener process
2)we verify it has a normal distribution, w~(T)-W~(s) =. cw(t/c^2)-cw(s/c^2)=c[w(t/c^2)-w(s/c^2)] by definition we know this is normally distributed w(t/c^2)-w(s/c^2)] ~N(0, (t/c^2)-(s/c^2))
which means that this is multiplied by c mean is multiplied by the constant variance multiplied by the constant squared
~N(0, t-s)
3) looking at the increments and finding the sequence w~(t_3)-w~(t_2)=c[….] in terms of a sequence of c times w(t_n/c^2)This is an increasing sequence, multiplying by the constant doesn’t change the property of the Weiner process and we can conclude they are independent

by conclusion we have verified all properties of Weiner process-

Question 18

Can you find a function of time which has this property? That if you scale it you get same object
f:(0,infinity) to R
f(x)=c f(x/c)

Answer 18

if you choose

f(x)=x scaled xc divide by c is is x

???

what about if you divide by c^2=
f(x)=c sqrt(X/c^2)

Question 19

Exercise 2.1.5. Let (W (t))_t∈[0,T ] be a Wiener process and let c ∈ (0, T ). Set W~(t) = W (c + t) − W(c) for t∈[0,T −c].

Show that W~ is a Wiener process on [0,T −c].

Answer 19

solution: Very similar to the exercise above.

verifying properties again
1)w~(0)=W(0+c)-W(c)=W(c)-W(c)=0
2)For s<t increments W~(t)-W~(s)= W(c+t)-W(c)-W(c+s)+W(c)=W(c+t)-W(c+s)
~N(0, c+t-c-s)
This has normal distribution N(0, t-s)

3)for increasing times compute the increments for these times
t_1<t_2<…<t_
w~(t_2)-w~(t_1),…, w~(t_n)-w~(t_(n-1))
w(c)-W(c+t_2)-w(c)+w(c+t_1,….
these are still increments over disjoint intervals and so are indepenentend

Question 20

def 2.1.7 FILTRATION

Answer 20

A family F := (Ƒₜ)ₜ∈[0,T ] of σ-algebras Ƒₜ ⊂ Ƒ is called filtration if Ƒₛ ⊂ Ƒₜ for s < t.

The filtration F is called complete if Ƒ_0 contains all the null sets of Ƒ.

A filtration is called right continuous if
Ƒₜ:= ∩_ₛ>ₜƑₛ

(knowledge always increasing never decreasing)

Question 21

fix filtration

Answer 21

fix filtration
F := (Ƒₜ)ₜ∈[0,T ]

Question 22

Def 2.1.7
adapted to the filtration

Answer 22

Stochastic process is called adapted to the filtration F if X(t) is Ƒₜ measurable for each t ∈ [0, T ],

Notice that every stochastic process X is adapted to its own filtration, that is, the family of σ-algebras given by Ƒˣ(t) = σ(X(s),s ≤ t).

Question 23

Def 2.1.8 F-Wiener process

Answer 23

We will say that (W (t))t∈[0,T ] is a F-Wiener process if
1. (W (t))_t∈[0,T ] is a Wiener process
2. (W (t))_t∈[0,T ] is adapted to the filtration F
3. for all 0 ≤ s ≤ t ≤ T, the random variable W(t) − W(s) is independent from Ƒ_s.

If we have F-wiener process then it will be a martingale

Question 24

DEF 2.1.9 F-martingale

Answer 24

A stochastic process (X(t))ₜ∈[0,T] is an F-martingale if
1. (X(t))ₜ∈[0,T] is F-adapted,
2. E[|X(t)|] < ∞ for all t ∈ [0,T],
3. E[X(t)|Ƒ_s] = X(s) for all s ≤ t.

Question 25

EX 2.2.10
Let (W (t))_t∈[0,T ] be a F-Wiener process. Show that it is a martingale with respect to F.

Answer 25

W_t is an F-wiener process, then this process is also a martingale

verifying the properties for martingale
1)By the definition of F Weiner process adapted to F

2) For t ∈ [0, T ].Then using Hölder’s inequality, and that since W is a Wiener process, W(t) ∼ N(0,t), we have

(recall X has nORMAL DIST then distribution function f stated …property of distribution discussed)

Expected values E|W(t)|<= (E[|W(t)^2|])^0.5 = t^0.5< ∞

3)For0≤s<t:

E[W (t)|Fs] = E[W (t) − W (s) + W (s)|Ƒ_s]

= E[W (t) − W (s)|Ƒ_s] + E[W (s)|Ƒ_s]
=: (⋆).

W (t) − W (s) is independent of F_s , as W is an F-Wiener process
* W(s) is Ƒ_s-measurable, as W is adapted (again, because W is an F-Wiener process). so expected values aren’t affected

(⋆) = E(W(t) − W(s)) + W(s) = W(s).
as W (t) − W (s) ∼ N (0, t − s):

Question 26

Exercise 2.1.11. Let (W (t))t∈[0,T ] be an F-Wiener process. Show that ((W (t))^2 − t)t∈[0,T ] is a martingale with respect to F.

Answer 26

If W_t is f_t measurable then this function may be? X_T will be, Adapted to F?

Let (W (t))t∈[0,T ] be a an F-Wiener process. Then (we show X is a martingale s.t X= W^2-t)

(i) (CHECK X IS F ADAPTED)
Let t ∈ [0, T ]. By assumption W (t) is Ƒ_t-measurable. So since the map x → x² − t is
continuous and thus Borel, the composition (W (t))² − t is also Ƒ_t -measurable.
Thus X_t is F adapted

(ii) (CHECK E[|X_t|] finite
Let t ∈ [0, T ]. Then (using the triangle inequality and that W (t) ∼ N (0, t)), we have
E[|(W (t))² − t|] ≤ E[(W (t))²] + t =t+t
< ∞.
by considering the variance and expectation

(iii) (We want to show the martingale property …)
For0≤s<t:

E[(W(t))² − t|Ƒ_s] = E[(W(t))² − 2W(t)W(s) + (W(s))² + 2W(t)W(s) − (W(s))² − t|Ƒ_s] = E[(W (t) − W (s))² |Ƒ_s] + E[2W (t)W (s)|Ƒ_s] − E[(W (s))² |Ƒ_s] − t
=: (⋆)

Note the following:
* W(t)−W(s) is independent of Ƒ_s, therefore
E[(W (t) − W (s))²|F_s] = E[(W (t) − W (s))²]
as W (t) − W (s) ∼ N (0, t − s):
=t−s

• W (s) is Ƒ_s-measurable, hence
  E[2W (t)W (s)|Ƒ_s] = 2W (s)E[W (t)|Ƒ_s]
  since F-Wiener processes are F-martingales
  = 2W (s)W (s)
  = 2(W (s))²
• W (s) is Ƒ_s-measurable, hence E[(W (s))2|Ƒ_s] = (W (s))²

Substituting these into (⋆), we get
(⋆)=t−s+2(W(s))² −(W(s))² −t
That is:
= (W (s))² − s.
E[(W (t))² − t|Fs] = (W (s))² − s.

constant
is indeed a martingale

Question 27

Theorem 2.1.12 (Doob’s inequality)

Answer 27

Let (M (t))t∈[0,T ] be a continuous martingale. Then, for any
p ∈ (1,∞) we have
E[supₜ∈[₀,ₜ ] |M(t)|ᵖ≤ (p/(p−1))ᵖ E [|M(T)|ᵖ]

missed out?

Question 28

Recall property
for conditional expectation

If Y is g measurable and Z is a RV
then
E[YZ|g]

Answer 28

If we have RV Y measurable wrt sigma algebra g

then conditional expectation of Y given g
E[Y|g]= Y

when y indep of g also true
E[Y|g]=E[Y]

g is the event it is raining
g doesnt influence map at all

If Y is g measurable and Z is a RV
then
E[YZ|g] = yE[Z|g]

Question 29

Exercise
Let (W (t))t∈[0,T ] be an F-Wiener process. Define another process Y_t= (W_t)^3 - 3(W_t)

is a martingale with respect to F.

Answer 29

“One we need to practice”
similar to previous

Question 30

Definition 2.3.1

MODIFICATION
for 2 stochastic processes

Answer 30

Let X, X˜ : Ω × [0, T] → R be stochastic processes. We call X˜ a modification of X if for all t ∈ [0, T] we have
P(X˜(t) = X(t))= 1

Question 31

E.G
Consider RV R with continuous distribution .

X_t=0 for all t
X~_t
is defined by
=
{0 if t not equal to R
{1 if t=R

Answer 31

These are modifications of each other.

Because for each t, P(X_t=X_t)= P(t not equal to R?)= 1

R is a rv with continuous dist so P(R=t)=0

Question 32

Def 2.3.2 α-Hölder continuous

α in [0.1]

Answer 32

Let α ∈ (0, 1]. A function f : [0, T] → R is called α-Hölder continuous if there exists a constant C𝒻 such that |f(t) − f(s)| ≤ C𝒻 |t − s|ᵃ for all t, s ∈ [0, T].

The collection of all α-Hölder continuous functions on [0, T] will be denoted by Cᵃ

The closer α to 1 the more regular function is
if α=1: differentiable at almost every point

Question 33

An example of Holder continuous funct

Answer 33

f(t)=t
satisfies

|f(t)-f(s)|=|s-t| for every s and t in domain

f is 1-Holder continuous
(satisfies Lipshitz)

Question 34

alpha-holder funct?

f(t)=|t|

Answer 34

higher exponent is more regular
e.g.
if f(t)=|t|

if t,s >0

|f(t)-f(s)|= |t-s|

if α=1: differentiable at almost every point C^1 function

Now, let’s consider two cases:

If t ≥ 0 and s ≥ 0:
|t| = t and |s| = s, so | |t| - |s| | = |t - s| = |t - s|.
If t < 0 and s < 0:
|t| = -t and |s| = -s, so | |t| - |s| | = |(-t) - (-s)| = |-(t - s)| = |t - s|.
In both cases, |f(t) - f(s)| simplifies to |t - s|

Question 35

alpha-holder funct?

f(t)=sqrt(t)

Answer 35

need to show that: there exists a constant C𝒻>0 such that
|f(t) − f(s)| ≤ C𝒻|t − s|^α for all t, s ∈ [0, T].

e.g.
if f(t)=sqrt(t)

if t,s >0

|f(t)-f(s)|= |sqrt(t)-sqrt(s)|
≤C|√(t-s)|

(squaring ????
|√t - √s|²= t +s - 2 (√st) ≤ t+s
(C|√(t-s)|)²= C² |t-s|

α=1/2

Question 36

example of alpha-holder continuous

F(t)=tᵃ

Answer 36

F(t)=tᵃ
|f(t)-f(s)|=||tᵃ-sᵃ|≤ C_𝒻 |t − s|ᵃ for all t, s ∈ [0, T].

easy if alpha=0.5

o/w requires calculus

f belongs to Cᵃ
(element of)

Question 37

example of holder continuous

Answer 37

sqrt funt is C^0.5

|f(t)-f(s)|
<=C x sqrt(t-s)

Question 38

sample path

Answer 38

Of wiener process

trajectory of process

Question 39

Theorem 2.3.3 (Kolmogorov).

continuity …

there exists a **continuous modification **

Answer 39

Let X : Ω×[0, T] → R be a process such that there exists positive
constants α, K and β, such that

E[|X(t) − X(s)|^α]
≤ K|t − s|^{1+β}

for all s, t, ∈ [0, T].

Then, there exists a **continuous modification **
X˜ : Ω × [0, T] → R of X
such that for all γ < β/α, we have P(X˜ ∈ C^γ) = 1.

If X is continuous then it holds that
P(X(t) = X˜(t), ∀t ∈ [0, T]) = 1
and for all γ < β/α
P(X ∈ C^γ) = 1

Question 40

REMARK
Theorem 2.3.3 (Kolmogorov).

continuity …

there exists a **continuous modification **

Answer 40

If X Was deterministic funct Expectation doesnt do anything so
E[|X(t) − X(s)|^α] = |X(t) − X(s)|^α
≤ K|t − s|^{1+β}

raising both sides to 1/a

|X(t) − X(s)|
≤ K|t − s|^{(1+β)/α}

{(1+β)/α} could be >1
thus if something is a-holder continuous with exponent >1
has to be constant

Question 41

{(1+β)/α} could be >1
thus if something is a-holder continuous with exponent >1
has to be constant

Answer 41

suppose we have funct f

|f(t) − f(s)|<= C_f |t − s|

Question 42

Kolmogorov thm meaning

Answer 42

For every t
the P(X(t)=X~(t))=1?

Proof is technical so skipped

thm will help see properties of trajectories of wiener process

thm quantifies trajectory continuity

{1+β}
means exponent >1

Question 43

Applying Kolmogorov thm to wiener process

Answer 43

cant apply kolmogorov thm as beta =0 not >0

We can apply this to see that the Wiener process is γ-Hölder continuous for all γ < 1/2.

Indeed, since W(t) − W(s) ∼ N (0, |t − s|) for s < t, we have that

moments: normal distribution with mean 0 and variance t-s

E[|W(t) − W(s)|²ᵐ]
=(1/sqrt(2π|t − s|
)) * integral_[0,∞]
|x|²ᵐ exp(− x²/(2|t−s|)) dx

raising powers we can see apply Kolmogorov to see the wiener process is a continuous modification

Question 44

trajectories of BM are continuous but we quantify how continuous they are

Answer 44

trajectories of BM are continuous but we quantify how continuous they are

differentiable from the right and left but maybe not at origin e.g |x|

square root funct deriv goes to infinity as x go to 0

sqrt funct similar to BM:
has the same scaling
if BM w_t
cW(t/c^2) this is also BM
BM behaves like sqrt does at 0

Question 45

f is bounded and a holder cont

The closer α to 1 the more regular function is

Answer 45

then it is B holder cont for any B<a

bounded using another constant

do as exercise by splitting up powers

Question 46

Considering continuity of functions

Answer 46

If we consider functions such as
sin
cos
absolute value
a trajectory of Brownian motion
they are continuous
square root function
which ones are more regular?

cos/sin trig
as they do not have a point at which you cant differentiate

if we consider
absolute value
and square root
their differentials/derivatives
abs: -1 and 1 except for at 0
sqrt: 0.5x^-0.5 approaches infinity as we get closer to 0

sqrt funct has things in common with brownian motion

Question 47

sqrt funct has things in common with brownian motion

Answer 47

it can be scaled
Take B, W_t W(t)
and scale
cW(t/c^2) is also brownian motion

we will see more rigourously is that every point behaves like the sqrt behaves at 0

Question 48

RECALL A function is α-Hölder continuous if

Answer 48

Let α ∈ (0, 1]. A function f : [0, T] → R is called α-Hölder continuous if there exists a constant C𝒻>0 such that
|f(t) − f(s)| ≤ C𝒻|t − s|^α for all t, s ∈ [0, T].

The collection of all
α-Hölder continuous functions on [0, T] will be denoted by C^α.

the higher/closer to 1 α the more regular a function is

α=1 implies differentiable at every point

Question 49

Suppose that f is bounded and a-holder continuous
then its also B-holder continuous for all B <=a

TRUE OR FALSE

Answer 49

TRUE

show by using that |f(s)-f(t)|= |f(s)-f(t)|ʸ||f(s)-f(t)|¹⁻ʸ

Question 50

Kolmogorov’s continuity Criterion

(Theorem 2.3.3 )

Answer 50

If you have stochastic process X_t s.t
there exists a K:

E[|X(t) − X(s)|^α]
≤ K|t − s|^{1+β}

for all s, t, ∈ [0, T].
β,α>0

we will assume: β<α
Now if we have this, we will assume that with Probability 1:
map t→X(t) ∈Cˠ for all Ɣ< β/α
(Ɣ-holder continuous)

Note that: if it was a deterministic function we would have exponent (1+β)/α. Which is greater than β/α so there is a trade off.

we considered a non-deterministic function, allowed to be stochastic, we check the condition with the expectation. This tells you that in fact your function is Ɣ-holder continuous for such a Ɣ above. Trade off? 1/α. In order to consider stochasticity with probability

Question 51

Kolmogorov’s continuity Criterion
there exists a K:

E[|X(t) − X(s)|^α]
≤ K|t − s|^{1+β}

WHAT HAPPENS IF X WAS DETERMINISTIC

Answer 51

the expectation does nothing,
E[|X(t) − X(s)|^α]
≤ K|t − s|^{1+β}

satisfied becomes

|X(t) − X(s)|^α
≤ K|t − s|^{1+β}

becomes
|X(t) − X(s)|
≤ K|t − s|^{{1+β}/α}

for a {{1+β}/α} holder continuous funct

can be something greater than 1? if something is a-continuous with exponent greater than 1

If something is a-holder continuous with a greater than 1 then it has to be continuous

Question 52

If something is a-holder continuous with a greater than 1 then it has to be continuous

suppose we have

a function f with a= 3/2

Answer 52

suppose we have f with a= 3/2

|f(t) − f(s)| ≤ C_f|t − s|³/²

I claim function f is constant:

take any two such points
I can take a partition of n pieces
|t-s|/n
and re write
tₖ= s+ [k|t-s|/n]
t ₀=s
tₙ=t

note that:
|tₖ − tₖ₋₁|= | [k|t-s|/n] − [(k-1)|t-s|/n]|
= ||t-s|/n|

|f(t) − f(s)| =
|Σₖ₌₁ⁿ [(f(tₖ)-f(tₖ₋₁)]|
(by triangle inequality)
≤ Σₖ₌₁ⁿ |[(f(tₖ)-f(tₖ₋₁)]|
(by holders condition)
≤ C_f[Σₖ₌₁ ⁿ |tₖ − tₖ₋₁|³/²]

= C_f[Σₖ₌₁ⁿ [|t-s|/n|³/²]
=C_f|t-s|³/²(1/n)³/² n

if i choose a smaller and smaller pieces this converges to 0, thus
0≤|f(t) − f(s)|≤0
and function is constant

Thats why we only talk about holder continuous functions with exponent between 0 and 1

Question 53

We will apply kolmogorov thm to see that the Wiener process is γ-Hölder continuous for all γ < 1/2.

Answer 53

For BM W(t):
E[|W(t)-W(s)|²]
= |t − s|
since W(t) − W(s) ∼ N (0, |t − s|) for s < t,
its the variance

Looking at the form we want we need 1 + beta: so instead look at, for m>1

E[|W(t) − W(s)|²ᵐ]
= integral_[0,∞][
(1/sqrt(2π|t − s|
)) *
|x|²ᵐ exp(− |x|²/{2|t−s|}) ]
dx

from the gaussian distribution
(density function of W(t)-W(S)*|x|²ᵐ as its the expectation of random var to the power ²ᵐ )

Then we have:
=(|t-s|ᵐ/sqrt(2π|t − s|
)) * integral_[0,∞]
(|x|²ᵐ/|t-s|ᵐ)* exp(− |x|²/{2|t−s|}) dx

by change of vars u= x/√ |t-s| we have

= Cₘ |t-s|ᵐ

thus
E[|W(t) − W(s)|²ᵐ]≤ Cₘ |t-s|ᵐ
m arbitrary so i can choose any m>1

The exponent alpha = 2m
beta =m-1
γ< (m-1)/(2m)
true for all m, for m close to 1 this is close to 0
for large m this is close to 1/2
so
γ<1/2
and almost behaves like a square root

Question 54

SHOW THAT
=(|t-s|ᵐ/sqrt(2π|t − s|
)) * integral_[0,∞]
(|x|²ᵐ/|t-s|ᵐ)* exp(− |x|²/{2|t−s|}) dx

by change of vars u= x/√ |t-s| we have

= Cₘ |t-s|ᵐ

Answer 54

=(|t-s|ᵐ/sqrt(2π) * integral_[0,∞]
(u²ᵐ)* exp(− u²/2) .du

But this integral which only depends on m, thus is a constant
which dep only on m
= Cₘ |t-s|ᵐ

Question 55

(density function of W(t)-W(S))

Answer 55

f(x) = (1 / sqrt(2 * pi * |t - s|)) * exp(-x^2 / (2 * |t - s|))

Question 56

True or false?
Brownian motion W
P(W∈ Cˠ)=1 for all Ɣ< 1/2

Answer 56

TRUE
prev example

We can show that P( W∈ Cᴮ)=0 for all B> 1/2 also
A bit technical for now

Question 57

TRUE OR FALSE
BM W cannot be continuously differentiable

Answer 57

TRUE cannot be continuously differentiable

shown using the concept of quadratic variation

Question 58

Theorem 2.3.4 (Quadratic Variation).

Answer 58

Let s < t and let (tᵢⁿ)ᵢ₌₁ ᵐ⁽ⁿ⁾, for n ∈ N be a sequence of partitions of [s, t]

such that max ᵢ≤ₘ₍ₙ₎
|tᵢ₊₁ⁿ - tᵢⁿ| → 0 as n → ∞. Then,

Σᵢ₌₁ⁿ |W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|² → (t − s),
in L₂(Ω), as n → ∞

(take increment for neighbouring points of BM and sum)

Question 59

let (tᵢⁿ)ᵢ₌₁ ᵐ⁽ⁿ⁾, for n ∈ N be a sequence of partitions of [s, t]

such that max ᵢ≤ₘ₍ₙ₎
|tᵢ₊₁ⁿ - tᵢⁿ| → 0 as n → ∞.

Answer 59

ie
first partition from s to t
t₀¹,t₁¹,t₂¹,t₃¹,t₄¹…

t₀²,t₁²,t₂²,t₃²,t₄²..
i take partition such that the distance between partitions converges to 0

Question 60

PROOF
Theorem 2.3.4 (Quadratic Variation).

Let s < t and let (tᵢⁿ)ᵢ₌₁ ᵐ⁽ⁿ⁾, for n ∈ N be a sequence of partitions of [s, t]

such that max ᵢ≤ₘ₍ₙ₎
|tᵢ₊₁ⁿ - tᵢⁿ| → 0 as n → ∞. Then,

Σᵢ₌₁ⁿ |W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|² → (t − s),
in L₂(Ω), as n → ∞

Answer 60

Σᵢ₌₁ⁿ |W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|² → (t − s),
in L₂(Ω), as n → ∞
MEANS that the L₂ norm converges
this is
E[
(Σᵢ₌₁ᵐ⁽ⁿ⁾⁻¹ |W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|² -(t-s))²
]
=
E[
(Σᵢ₌₁ᵐ⁽ⁿ⁾⁻¹ |W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|² -(tᵢ₊₁ⁿ - tᵢⁿ) )²
]
=
Σᵢ₌₁ᵐ⁽ⁿ⁾⁻¹ E[||W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|² -(tᵢ₊₁ⁿ - tᵢⁿ))²]
≤
4*
Σᵢ₌₁ᵐ⁽ⁿ⁾⁻¹ E[||W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|⁴ -(tᵢ₊₁ⁿ - tᵢⁿ))²]
≤
N Σᵢ₌₁ᵐ⁽ⁿ⁾⁻¹ (tᵢ₊₁ⁿ - tᵢⁿ)²
≤
N MAXᵢ|(tᵢ₊₁ⁿ - tᵢ)| *[Σᵢ₌₁ᵐ⁽ⁿ⁾⁻¹ (tᵢ₊₁ⁿ - tᵢⁿ)
=
NMAXᵢ|(tᵢ₊₁ⁿ - tᵢ)| *[Σᵢ₌₁ᵐ⁽ⁿ⁾⁻¹ (tᵢ₊₁ⁿ - tᵢⁿ)|t-s|
which conv to 0
by assumption that our seq of terms gets closer

as n → ∞.
. For the second equality we have used that if i≠ j, then
E[|W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|² -(tᵢ₊₁ⁿ - tᵢⁿ)(|W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|² -(tᵢ₊₁ⁿ - tᵢⁿ))]
=0

Question 61

For the second equality we have used that if i≠ j, then
E[|W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|² -(tᵢ₊₁ⁿ - tᵢⁿ)(|W(t ⱼ₊₁ⁿ) − W(t ⱼⁿ)|² -(t ⱼ₊₁ⁿ - t ⱼⁿ))]
=0

Answer 61

BY defn. of brownian motion increment |W(t ⱼ₊₁ⁿ) − W(t ⱼⁿ)|is independent of|F_{t ⱼⁿ} so conditional expectation is just the expectation

we use properties of cond exp:
WLOG assume i<j
E[|W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|² -(tᵢ₊₁ⁿ - tᵢⁿ)(|W(t ⱼ₊₁ⁿ) − W(t ⱼⁿ)|² -(t ⱼ₊₁ⁿ - t ⱼⁿ))]
The same as taking cond expectation:
E[E[|W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|² -(tᵢ₊₁ⁿ - tᵢⁿ)(|W(t ⱼ₊₁ⁿ) − W(t ⱼⁿ)|² -(t ⱼ₊₁ⁿ - t ⱼⁿ))]
|F_{t ⱼⁿ}]

j is greater than i thus this implies i+1≤j
so we have that
W(tᵢ₊₁ⁿ) and W(tᵢⁿ) are F_{t ⱼⁿ} measurable (F_{t ᵢ₊₁ⁿ} and F_{t ᵢⁿ} )
t ᵢ₊₁ⁿ and t ᵢⁿ :these are constants
so the first term is F_{t ⱼⁿ} measurable and we can take it out of the conditional expectation
thus
=
E[|W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|² -(tᵢ₊₁ⁿ - tᵢⁿ)]E[(|W(t ⱼ₊₁ⁿ) − W(t ⱼⁿ)|² -(t ⱼ₊₁ⁿ - t ⱼⁿ))]
|F_{t ⱼⁿ}]

E[|W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|² -(tᵢ₊₁ⁿ - tᵢⁿ)]E[(|W(t ⱼ₊₁ⁿ) − W(t ⱼⁿ)|² -(t ⱼ₊₁ⁿ - t ⱼⁿ))]]
=0

as E[**(|W(t ⱼ₊₁ⁿ) − W(t ⱼⁿ)|² ]= (t ⱼ₊₁ⁿ - t ⱼⁿ))] thus second exp is 0

so we have shown i≠ j, are 0 thus used prev

Question 62

also used
(a-b)^2

Answer 62

≤ 4(|a|^2+|b|^2|

we use that
Σᵢ₌₁ᵐ⁽ⁿ⁾⁻¹ E[|W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|² -(tᵢ₊₁ⁿ - tᵢⁿ))²]
with
[|W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|² =a
and
(tᵢ₊₁ⁿ - tᵢⁿ))=b
note squared term

Question 63

Theorem 2.3.4 (Quadratic Variation). DISCUSSION

Answer 63

has some negative implications:
From this follows that BM cannot be continuously differentiable
we discuss the concept of variation

Question 64

Given a function f
Variation of funct on interval;
Var_[a,b] f

Answer 64

Take partition of interval [a,b]
P={t₁,t₂,…,tₖ}
for each partition I take
Σᵢ₌₁ᵏ |f(tᵢ₊₁)-f(tᵢ)|
now take supremum of all partitions on the interval

CLAIM:
If we take func f which is cont differentiable: deriv is a cont funct
then for any interval
its variation is finite

(if cont differtiable then f’ is continuous, then sup f’ over any interval is finite by continuity we use this to show)

Question 65

CLAIM:
If we take func f which is cont differentiable: deriv is a cont funct
then for any interval
its variation is finite

Answer 65

(if cont differtiable then f’ is continuous, then sup f’ over any interval is finite by continuity we use this to show)

Take any partition and consider the sum:

Take partition of interval [a,b]
P={t₁,t₂,…,tₖ}
for each partition I take
Σᵢ|f(tᵢ₊₁)-f(tᵢ)|
(from MVT)
=Σᵢ |f’(tᵢ₊₁)||tᵢ₊₁-tᵢ|
≤ sup_{x in [a,b]}(f’(x)) Σᵢ |tᵢ₊₁-tᵢ|
=sup_{x in [a,b]}(f’(x)) (b-a)

then taking supremum of partitions on both sides
the sup is finite if f continuous differentiable

Question 66

Facts shown for BM

Answer 66

Fact 1:

Σᵢ₌₁ⁿ |W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|² → (t − s),
in L₂(Ω), as n → ∞

for partitions defined previously
Fact 2:
If f is continuously differentiable then variation on any interval is finite
Var f _[a,b]]< ∞

we will use these to claim :
fact 1 implies
P(Var W _[a,b]]< ∞)=1

Question 67

CLAIM:
Corollary 2.3.5. Let (W(t))ₜ∈[₀,ₜ] be a Wiener process.

Then P(Var_[₀,ₜ] W = ∞) = 1.

PROOF

Answer 67

ASSUME the contrary:
P(Var_[₀,ₜ] W <∞) > 0
by thm 2.3.4 we have that there is a sequence of partitions
[0,1] (tᵢⁿ)ᵢ₌₁ ᵐ⁽ⁿ⁾, for n ∈ N be a sequence of partitions of [s, t]

such that max ᵢ |tᵢ₊₁ⁿ - tᵢⁿ| → 0 as n → ∞. Then,

Σᵢ₌₁ⁿ |W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|² → (t − s)=
1 (2.2)

almost surely. On the other hand, on the event {Var_{[0,1]} W < ∞} we have

Σᵢ₌₁ᵐ⁽ⁿ⁾⁻¹ |W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|²
take out the max
≤ max_i |W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|Σᵢ₌₁ᵐ⁽ⁿ⁾⁻¹ |W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|
≤
max_i |W(tᵢ₊₁ⁿ) − W(tᵢⁿ)|Var_{[0,1]} W →0
(as we have a partition, the variation is the supremum over all of these)

which contradicts 2.2
as W is continuous, BM, and increments become closer together functions of these values become closer for this continuous funct

Question 68

remark
BM and square root?

Answer 68

We have show P(W is cont diff)=0
stronger statement would be
P( there exists s>0 s.t W is differentiable at s)=0 (at any point)

at whatever point we look at BM not differentible (thm 2.3.6)

There is no single point where BM behaves better than square root

Question 69

Theorem 2.3.6 (Nowhere differentiability)

Answer 69

Let (W(t))t∈[0,1] be a Wiener process. Then with
probability one, W is nowhere differentiable on [0, 1]

proof wont be asked in exam, technical

Question 70

SUMMARY OF BM chapter

Answer 70

BM paths are ALMOST (1/2)-holder continuous

nowhere differentiable
(cannot speak of derivative of BM in sense of being a funct)

Question 71

wiener process

Answer 71