theoretical tutorial session 1 - drexel universitysong/gene golub summer school/song... · >c ˙...

Theoretical Tutorial Session 1

Xiaoming Song

Department of MathematicsDrexel University

July 25, 2016

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Basic properties for Brownian motion

1. Selfsimilarity:For any a > 0, the process a−

12 Bat , t ≥ 0 is also a Brownian

motion.

Proof: Let Yt = a−12 Bat . It is obvious that the process

Yt = a−12 Bat , t ≥ 0 is an Gaussian process with zero mean

and continuous trajectories. It is sufficient to show that for any0 ≤ s ≤ t < +∞ the covariance function E(YtYs) = s which isproved as follows:

E(YtYs) = E(a−12 Bata−

12 Bas) = a−1E(BatBas) = a−1(as) = s.

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2. For any h > 0, the process Bt+h − Bh, t ≥ 0 is a Brownianmotion.

Proof: Let Yt = Bt+h − Bh. We know that the processYt = Bt+h − Bh, t ≥ 0 is an Gaussian process with zero meanand continuous trajectories. Next, we calculate its covariancefunction: for any 0 ≤ s ≤ t < +∞,

E(YtYs) = E((Bt+h − Bh)(Bs+h − Bh))

= E(Bt+hBs+h)− E(BhBs+h)− E(Bt+hBh) + E(BhBh)

= (s + h)− h − h + h= s.

Therefore, the process Bt+h −Bh, t ≥ 0 is a Brownian motion.

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3. The process −Bt , t ≥ 0 is a Brownian motion.

Proof: This process is a Gaussian process with zero mean andcontinuous trajectories, and for any 0 ≤ s ≤ t < +∞,

E((−Bt )(−Bs)) = E(BtBs) = s.

Thus, The process −Bt , t ≥ 0 is a Brownian motion.

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Chebyshev’s inequality

If ξ is a random variable with E(|ξ|p) <∞, then for any λ > 0

P(|ξ| > λ) ≤ E(|ξ|p)

λp .

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Borel-Cantelli lemma

LemmaLet A1,A2,A3, . . . be a sequence of events in a probabilityspace. If

∑∞n=1 P(An) <∞, then the probability that infinitely

many of them occur is 0, that is,

P(lim supn→∞

An) = 0.

Note that

lim supn→∞

An =∞⋂

n=1

∞⋃k=n

Ak

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Strong law of large numbers

TheoremIf ξ1, ξ2, ξ3, . . . is a sequence of iid random variables andµ = E(ξ1) <∞, then∑n

i=1 ξi

n=ξ1 + ξ2 + · · ·+ ξn

n→ µ

almost surely as n→∞.

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4. Almost surely limt→∞

Btt = 0 and the process

Xt =

tB1/t , t > 0;

0, t = 0

is a Brownian motion.

Proof: We first show that limn→∞

Bnn = 0 which is true by using the

strong law of large numbers since

Bn

n=

∑ni=1(Bi − Bi−1)

n

and B1 −B0,B2 −B1, . . . ,Bn −Bn−1, . . . are iid with law N(0,1).

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Then for any t > 0 and t ∈ [n,n + 1) for some n, we have∣∣∣∣Bt

t− Bn

n

∣∣∣∣ =

∣∣∣∣Bt

t− Bn

t+

Bn

t− Bn

n

∣∣∣∣≤

sups∈[n,n+1] |Bs − Bn|t

+ |Bn|(

1n− 1

t

)≤

sups∈[n,n+1] |Bs − Bn|n

+|Bn|

n. (1)

Note that limn→∞

Bnn = 0 implies

limn→∞

|Bn|n

= 0. (2)

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For the process X , using limt→∞

Btt = 0 almost surely we can

prove that the trajectories are continuous almost surely.

The process X is a Gaussian process with zero mean. For any0 ≤ s ≤ t <∞, we have

E(XtXs) = E((tB1/t )(sB1/s)) = (t)(s)E(B1/tB1/s) = (t)(s)(1/t) = s.

Therefore, X is also a Brownian motion.

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Fatou’s lemma

LemmaLet ξ1, ξ2, ξ3, . . . be a sequence of non-negative randomvariables on a probability space (Ω,F ,P), then

E(lim infn→∞

ξn) ≤ lim infn→∞

E(ξn).

Let ξ1, ξ2, ξ3, . . . be a sequence of non-negative randomvariables on a probability space (Ω,F ,P). If there exists anon-negative integrable random variable η such that ξn ≤ ηfor all n, then

lim supn→∞

E(ξn) ≤ E(lim supn→∞

ξn).

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Hewitt-Savage 0-1 law

TheoremLet ξ1, ξ2, ξ3, . . . be a sequence of independent andidentically-distributed (iid) random variables taking values in aset E . Then, any event whose occurrence or non-occurrence isdetermined by the values of these random variables and whoseoccurrence or non-occurrence is unchanged by finitepermutations of the indices, has probability either 0 or 1.

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5. lim supt→∞Bt√

t=∞ and lim inft→∞ Bt√

t= −∞.

Proof: It suffices to show that lim supn→∞Bn√

n =∞ and

lim infn→∞ Bn√n = −∞. For any positive constant C, we have

P(

lim supn→∞

Bn√n> C

)= P

(lim sup

n→∞

Bn√

n> C

)≥ lim sup

n→∞P(

Bn√n> C

)(Fatou’s lemma)

= lim supn→∞

P (B1 > C) (Scaling property)

>0.

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Let ξn = Bn − Bn−1, n ≥ 1. Then ξ1, ξ2, ξ3, . . . is a sequence ofiid random variables. Note that the event

lim supn→∞

Bn√

n> C

= lim sup

n→∞

∑nk=1 ξk√

n> C

is unchanged by finite permutations of the indices. Hence, byHewitt-Savage 0− 1 law together with the last inequality onprevious page, we have

P(

lim supn→∞

Bn√

n> C

)= 1.

Since C is arbitrary, we can show that lim supn→∞Bn√

n =∞.

Since −Bt , t ≥ 0 is also a Brownian motion, we can show that

lim infn→∞

Bn√n

= lim infn→∞

−Bn√n

= − lim supn→∞

Bn√n

= −∞.

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Remark: From the Exercise 5, we obtain the following result

P(lim supt→∞

Bt = +∞) = 1, (4)

andP(lim inf

t→∞Bt = −∞) = 1. (5)

Furthermore,

P(lim supt→∞

Bt = +∞, lim inft→∞

Bt = −∞) = 1. (6)

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6. Show that for each fixed t ≥ 0

P(ω ∈ Ω : lim sup

h→0+

Bt+h − Bt

h=∞

)= 1.

Proof: Given this Brownian motion B, we construct anotherBrownian motion X defined in Exercise 4:

Xt =

tB1/t , t > 0;

0, t = 0.

Then for any t ≥ 0, we obtain

lim supn→∞

Bt+ 1n− Bt

1n

= lim supn→∞

B 1n− B0

1n

(Exercise 2)

≥ lim supn→∞

√nB 1

n

= lim supn→∞

Xn√n

=∞ (Previous exercise).

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7. Almost surely the paths of B are not differentiable at anypoint t ≥ 0, more precisely, the set

ω ∈ Ω : for each t ∈ [0,∞), either lim sup

h→0+

Bt+h − Bt

h=∞

or lim infh→0+

Bt+h − Bt

h= −∞

contains an event A ∈ F with P(A) = 1.

Proof: Denote

D+Bt = lim suph→0+

Bt+h − Bt

h

andD+Bt = lim inf

h→0+

Bt+h − Bt

h.

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It suffices to show the result on the interval [0,1]. For any fixedintegers j ≥ 1 and k ≥ 1, define

Ajk =⋃

t∈[0,1]

⋂h∈[0,1/k ]

ω ∈ Ω : |Bt+h(ω)− Bt (ω)| ≤ jh.

Note that Ajk is not an event, that is Ajk /∈ F . Note also that

ω ∈ Ω : −∞ < D+Bt (ω) ≤ D+Bt (ω) <∞, for some t ∈ [0,1]

=∞⋃

j=1

∞⋃k=1

Ajk .

Then the proof of the result will be complete if we can find anevent C ∈ F with P(C) = 0 such that Ajk ⊆ C for all j and k .

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For any fixed j and k , we consider a sample path ω ∈ Ajk , thatis, there exists t ∈ [0,1] with

|Bt+h(ω)− Bt (ω)| ≤ jh

for all h ∈ [0,1/k ].

Let n be an integer with n ≥ 4k . Then there exists some i ,1 ≤ i ≤ n such that i−1

n ≤ t < in . Then it is also easy to verify

that, for l = 1,2,3,

i + ln− t ≤ l + 1

n≤ 1

k.

Thus,

|B i+1n

(ω)− B in(ω)| ≤ |B i+1

n(ω)− Bt (ω)|+ |B i

n(ω)− Bt (ω)|

≤ 2jn

+jn

=3jn,

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Note that the random variables√

n(

B i+ln− B i+l−1

n

), l = 1, 2,3,

are iid with law N(0,1). Note also that if a random variable ξhas law N(0,1) then

P(|ξ| ≤ x) ≤ x , for any x ≥ 0.

Therefore, we have

P(C(n)i ) ≤

(3j√

n

)(5j√

n

)(7j√

n

)=

105j3

n32

, i = 1,2, · · · ,n.

Then

P(n⋃

i=1

C(n)i ) ≤ 105j3√

n.

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Take

C =∞⋂

n=4k

n⋃i=1

C(n)i ∈ F .

Then Ajk ⊆ C and it also holds that

P(C) = P(∞⋂

n=4k

n⋃i=1

C(n)i ) ≤ inf

n≥4kP(

n⋃i=1

C(n)i ) = 0,

which completes the proof.

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9. E(E(X |G)) = E(X ).

Proof: In the definition of conditional expectation, let A = Ω,then

E(E(X |G)) =

∫Ω

E(X |G)dP =

∫Ω

XdP = E(X ).

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10. If X and G are independent, then E(X |G) = E(X ).

Proof: It is obvious that E(X ) is G-measurable. For any A ∈ G,∫A

E(X |G)dP =

∫A

XdP = E(1AX )

= E(1A)E(X ) (X and 1A are independent)= P(A)E(X )

=

∫A

E(X )dP.

Thus, it implies that E(X |G) = E(X ).

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11. If X is G-measurable, then E(X |G) = X .

Proof: The proof follows from the fact that X is G-measurableand the definition of conditional expectation: for any A ∈ G∫

AE(X |G)dP =

∫A

XdP.

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Dominated convergence theorem

TheoremLet ξ1, ξ2, ξ3, . . . be a sequence of random variables. If thesequence ξn converges almost surely to a random variable ξand there exists a positive and integrable random variable ηsuch that |ξn| ≤ η for all n, then

limn→∞

E(|ξn − ξ|) = 0,

which also implies

limn→∞

E(ξn) = E(ξ).

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12. If Y is bounded and G-measurable, then

E(XY |G) = YE(X |G).

Proof: If Y = 1B is an indicator function where B ∈ G, then forany A ∈ G∫

AE(YX |G)dP =

∫A

1BXdP =

∫A∩B

XdP

=

∫A∩B

E(X |G)dP

= intA1BE(X |G)dP,

which implies E(XY |G) = YE(X |G).

By the linearity of conditional expectation, we know that theresult is true if Y is a simple function (a linear combination ofindicator functions).

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13. Given two σ-fields B ⊂ G, then

E(E(X |B)|G) = E(E(X |G)|B) = E(X |B).

Proof: Since B ⊂ G, we know that E(X |B) is G-measurable.Then by Exercise 4, we obtain

E(E(X |B)|G) = E(X |B).

For any A ∈ B ⊂ G, we have∫A

E(E(X |G)|B)dP =

∫A

E(X |G)dP =

∫A

XdP =

∫A

E(X |B)dP,

which impliesE(E(X |G)|B) = E(X |B).

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14. Let X and Z be such that(i) Z is G-measurable.(ii) X is independent of G.

Suppose that E(h(X ,Z )|) <∞. Then,

E(h(X ,Z )|G) = E(h(X , z))|z=Z .

Proof: Denote g(z) = E(h(X , z)) and µX (E) = P(X ∈ E) forE ∈ B(R). For any A ∈ G, we denote

Y = 1A;

µY ,Z (F ) = P((Y ,Z ) ∈ F ), F ∈ B(R2).

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Then,∫A

E(h(X ,Z )|G)dP =

∫A

h(X ,Z )dP

= E(Yh(X ,Z ))

=

∫R3

yh(x , z)µX (dx)µY ,Z (dy ,dz)

=

∫R2

yg(z)µY ,Z (dy ,dz)

= E(Yg(Z ))

= E(1Ag(Z ))

=

∫A

g(Z )dP

=

∫A

E(h(X , z))|z=Z dP.

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Properties of stopping times

15. S ∨ T and S ∧ T are stopping times.

Proof: For all t ≥ 0, we know that S ≤ t ∈ Ft andT ≤ t ∈ Ft . Hence,

S ∨ T ≤ t = S ≤ t ∩ T ≤ t ∈ Ft ,

andS ∧ T ≤ t = S ≤ t ∪ T ≤ t ∈ Ft .

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16. Given a stopping time T ,

FT = A : A ∩ T ≤ t ∈ Ft , for all t ≥ 0

is a σ-field.

Proof: It is obvious that Ω is in FT and FT is closed underintersection of countably infinite many subsets. We only needto show that FT is closed under complement. If A ∈ FT thenA ∩ T ≤ t ∈ Ft , and hence

Ac∩T ≤ t = (A∪T > t)c = ((A∩T ≤ t)∪T > t)c ∈ Ft .

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17. S ≤ T ⇒ FS ⊂ FT .

Proof: If A ∈ FS, then for any t ≥ 0 we have A ∩ S ≤ t ∈ Ft .

Since S ≤ T , we also have T ≤ t ⊂ S ≤ t. Thus

A ∩ T ≤ t = A ∩ S ≤ t ∩ T ≤ t ∈ Ft ,

which implies A ∈ FT .

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18. Let Xt , t ≥ 0 be a continuous and adapted process. Thehitting time of a set A ⊂ R is defined by

TA = inft ≥ 0 : Xt ∈ A.

Then, if A is open or closed, TA is a stopping time.

Proof: If A is an open set, then for any t > 0, we have

TA < t =⋃

r∈Q+,r<t

Xr ∈ A ∈ Ft .

By Exercise 1 in Professor Nualart’s lecture notes, we canshow that TA is a stopping time.

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If A is a closed set, then for any t ≥ 0 we have

TA ≥ t =⋂

r∈Q+,r<t

Xr /∈ A ∈ Ft .

Hence TA < t ∈ Ft . By Exercise 1 in Professor Nualart’slecture notes, we can show that TA is a stopping time.

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19. Let Xt be an adapted stochastic process withright-continuous paths and T <∞ is a stopping time. Then therandom variable

XT (ω) = XT (ω)(ω)

is FT -measurable.

Proof: For any n ∈ N, define

Tn =∞∑

i=0

i + 12n 1 i

2n<T≤ i+12n

.

For any t ≥ 0,

Tn ≤ t =⋃

i, i+12n ≤t

T ≤ i + 12n ∈ Ft .

Then Tn is a stopping time for each n, and moreover, Tn ↓ T .

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Then it suffices to show that for any open set A ⊂ R

XTn ∈ A ∩ Tn ≤ t ∈ Ft , t ≥ 0.

In fact,

XTn ∈ A ∩ Tn ≤ t

= ∪ k2n≤t (X k

2n∈ A ∩ k − 1

2n < T ≤ k2n ) ∈ Ft .

Since Tn ↓ T and the process X is right-continuous, we can letn go to∞ and obtain

XT ∈ A ∩ T ≤ t = limn→∞XTn ∈ A ∩ Tn ≤ t ∈ Ft .

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Application to Brownian hitting times

Let Bt be a Brownian motion. Fix a ∈ R and consider the hittingtime

τa = inft ≥ 0 : Bt = a.

PropositionIf a < 0 < b, then

P(τa < τb) =b

b − a.

Proof: From (4) and (5), we can get τa <∞ and τb <∞ almostsurely.

Exercise 18 implies that τa and τb are two stopping times, andhence Exercise 15 implies that τa ∧ τb is also a stopping time.

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For any t ≥ 0, τa ∧ τb ∧ t is a bounded stopping time. Then bythe optional stopping theorem, we have

E(Bτa∧τb∧t ) = E(B0) = 0.

Sincea ≤ Bτa∧τb∧t ≤ b, ∀t ≥ 0

andlim

t→∞Bτa∧τb∧t = Bτa∧τb

almost surely, by the dominated convergence theorem, we have

E(Bτa∧τb ) = E( limt→∞

Bτa∧τb∧t ) = Bτa∧τb∧tE(Bτa∧τb∧t ) = 0.

The random variable Bτa∧τb takes only two values a and b andits distribution is given by

Bτa∧τb =

a, with probability P(τa < τb),

b, with probability P(τa > τb).

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Hence,

E(Bτa∧τb ) = aP(τa < τb) + bP(τa > τb)

= aP(τa < τb) + b(1− P(τa < τb))

= 0.

Solving this equation for P(τa < τb), we obtain

P(τa < τb) =b

b − a.

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Proposition

Let T = inft ≥ 0 : Bt /∈ (a,b), where a < 0 < b. Then

E(T ) = −ab.

Proof: In fact T = τa ∧ τb <∞ almost surely.

Using that B2t − t is a martingale, we get, by the optional

stopping theorem,

E(B2T∧t − (T ∧ t)) = 0,

that is,E(B2

T∧t ) = E(T ∧ t). (7)

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Since T ∧ t ↑ T as t ↑ ∞, by the monotone convergencetheorem we get

E(T ) = E( limt→∞

(T ∧ t)) = limt→∞

E(T ∧ t). (8)

Since B2T∧t is bounded for all t ≥ 0 and B2

T∧t → BT almostsurely as t →∞, using the dominated convergence theoremand (7) and (8) we have

E(B2T ) = E( lim

t→∞B2

T∧t ) = limt→∞

E(B2T∧t ) = E(T ).

From the previous Proposition, we get

E(B2T ) = a2

(b

b − a

)+ b2

(1− b

b − a

)= −ab.

Therefore,E(T ) = −ab.

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Additional exercises

1. Let Z be a Gaussian random variable with law N(0, σ2).From the expression

E(eλZ ) = e12λ

2σ2,

deduce the following formulas for the moments of Z :

E(Z 2k ) =(2k)!

2kK !σ2k , k = 1,2, . . . ,

E(Z 2k−1) = 0, k = 1,2, . . . .

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2. Let Bt , t ∈ [0,1] be a standard Brownian motion. Showthat

ξ =

∫ 1

0Btdt

is a well-defined random variable. Calculate its first and secondmoments.

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