measure, problems and solutions

8/4/2019 Measure, Problems and Solutions

http://slidepdf.com/reader/full/measure-problems-and-solutions 1/31

Measure Theory and

Filtering: Introduction and Applications

Lakhdar Aggoun and Robert J. Elliott

Solutions to SelectedProblems

Chapter 1Problem 1. Let F ii∈I be a family of σ-fields on Ω. Prove that

i∈I F i is

a σ-field.Solution:

(i) Ω belongs to all the F ii∈I and hence it belongs toi∈I F i.

(ii) If A1, A2, . . . is a countable sequence, then

An ∈ F i fori ∈ I . Therefore

An ∈

i∈I F i.(iii) Let A ∈

i∈I F i, then A ∈ F i for all i ∈ I which impliesthat Ac ∈ F i for all i ∈ I and Ac ∈

i∈I F i. Thereforei∈I F i

is a σ-field.

Problem 2. Let A and B be two events. Express by means of the indicatorfunctions of A and B,I A∪B, I A∩B, I A−B, I B−A, I (A−B)∪(B−A), where A − B = A ∩ B.Solution:

The following equalities are easily checked.I B−A = I A + I B − I A∩B,I A∩B = I AI B,I A−B = I A − I A∩B,I B−A = I B − I A∩B,I (A−B)∪(B−A) = I A−B + I B−A = I A − I A∩B + I B − I A∩B =I A + I B

−2I AI B.

Problem 3. Let Ω = Rm and define the sequences

C 2n = [−1, 2 +1

2n) and C 2n+1 = [−2 − 1

2n + 1, 1).

Show that lim sup C n = [−2, +2], lim inf C n = [−1, 1].Solution:

Recall that lim sup C n =n≥1

k≥nC k.

1



2

However,

k≥nC k =

k≥n[−1, 2 +

1

2k)

k≥n[−2 − 1

2k + 1, 1)

= [−1, 2 + 12n

)

[−2 − 12n + 1

, 1) = [−2 − 12n + 1

, 2 + 12n

) and

lim sup C n =n≥1[−2 − 1

2n + 1, 2 +

1

2n) = [−2, +2].

Similarly lim inf C n =n≥1

k≥n C k =

n≥1[−2, 1]

[−1, 2] =

[−1, +1]

Problem 4. Let Ω = (ω1, ω2, ω3, ω4) and P (ω1) =1

12, P (ω2) =

1

6, P (ω3) =

1

3, and P (ω4) =

5

12. Let

An =ω1, ω3 if n is odd,

ω2, ω4 if n is even.

Find P (lim sup An), P (lim inf An), lim sup P (An), and lim inf P (An)and compare.

Solution:

Using the definition, it is easily seen thatlim sup An = Ω and lim inf An = ∅.Also

P (An) = 5/12 if n is odd,

7/12 if n is even.

Hence lim sup P (An) = inf sup P (An) = 7/12 andlim inf P (An) = supinf P (An) = 5/12.Therefore P (lim sup An) ≥ lim sup P (An) andP (lim inf An) ≤ lim inf P (An).

Problem 5. Theorem 1.3.36.Let X n be a uniformly integrable family of random variables.Then E [lim inf X n] ≤ lim inf E [X n].Proof.

First note that for any A > 0: E [X n] = E [X nI (X n <

−A)] +

E [X nI (X n ≥ −A)]. By uniform integrability: for any > 0 wecan take A so large that supn |E [X nI (X n < −A)]| < .By Fatou’s lemma:lim inf E [X nI (X n ≥ −A)] ≥ E [lim inf X nI (X n ≥ −A)].But X nI (X n ≥ −A) ≥ X n, thereforelim inf E [X nI (X n ≥ −A)] ≥ E [lim inf X n].Hence lim inf E [X n] ≥ E [lim inf X n] − .



3

Since > 0 is arbitrary, the result follows.

Problem 7. Show that if X is a random variable, then σ|X | ⊆ σX .Solution:

It suffices to note that |X | = XI (X > 0) + XI (X ≤ 0). Clearlythe RHS is σX -measurable. Hence σ|X | being the smallestσ-field with respect to which |X | is measurable, is included inσX .

Problem 9. Show that the class of finite unions of intervals of the form(−∞, a], (b, c], and (d, ∞) is a field but not a σ-field.

Solution:

For instance the open interval (a, b) = ∞n=1(a, b−1/n] is not in

this collection despite the fact it contains each interval (a, b −1/n].

Problem 10. Show that a sequence of random variables X n converges (a.s)to X if and only if ∀ > 0 limm→∞ P [|X n−X | ≤ ∀ n ≥ m] =1.

Solution:

“⇒” Suppose X nas→ X . Let N = [ω : lim X n(ω) = X (ω)] and

Ω0 = Ω − N . Let Am =n≥m[|X n − X | ≤ ]. Am is monotoni-

cally increasing. Hence lim P (Am) = P (

Am).

If ω0

∈Ω, then there exists m(ω0, ) such that

|X n(ω0) − X (ω0)| ≤ , for all n ≥ m(ω0, ) (*)which implies that Ω0 ⊂

Am and P (

Am) = 1.“⇐” Let A =

m≥1 Am. If ω0 ∈ A then there exists m(ω0, )

such that (*) holds. LetA =

>0 =

n≥1 A1/n.

By assumption P (A1/n) = 1 for all n. Therefore P (A) =lim P (A1/n) = 1.So if ω0 ∈ A, then (*) holdsfor all = 1/n and hence X n(ω0) →X (ω0). Let Ω0 = A and N = Ω − A which implies P (N ) = 0.

Problem 11. Show that if

X k

converges (a.s) to X then

X k

converges

to X in probability but the converse is false.Solution:

Let A = [ω : lim X n(ω) = X (ω)]=

k≥1

N ≥1

n≥N [|X n − X | < 1/k].

Since P (A) = 1 it implies thatP (

N ≥1

n≥N [|X n − X | < ]) = 1 for all > 0.



4

But BN =n≥N [|X n − X | < ] is monotonically increasing.

Therefore lim P (BN ) = 1. However BN ⊂ [|X n − X | < ].

Hence lim P (|X n − X | < ) = 1.In fact the statement in Problem 10 is equivalent to: X n

as→ X if and only if P [supn≥m |X n − X | ≥ 0] → 0, n → ∞.(*)

On the other hand, by definition X nP → X if

P [|X n − X | ≥ 0] → 0, n → ∞.(**)Then it is clear that (*) is stronger than (**).

Problem 13. Let X n be a sequence of random variables with

P [X n = 2n] = P [X n =−

2n] =1

2n,

P [X n = 0] = 1 − 1

2n−1.

Show that X n converges (a.s) to 0 but E |X n| p does not con-verge to 0.

Solution:

To prove the a.s. convergence of X n to 0 it suffices to showthat

∞k=1 P (|X k| ≥ ) < ∞ is satisfied for all > 0. But

P (|X k| ≥ ) =1

2k−1and the result follows.

Now E |X n| p = 2np 12n−1

= 2n( p−1)+1 which does not converge to

0 for any p ≥ 1.

Problem 15. Suppose Q is another probability measure on (Ω, F ) such thatP (A) = 0 implies Q(A) = 0 (Q P ). Show that a.s.P conver-gence implies a.s.Q convergence.

Solution:

Suppose that X nas−P → X . then there exists N such that for all

ω /∈ N : X n(ω) → X (ω), and P (N ) = 0. But P (N ) = 0 im-

plies Q(N ) = 0 and X n(ω)

→X (ω) for all ω /

∈N so X n

as−Q

→X .

Problem 16. Prove that if F 1 and F 2 are independent sub-σ-fields and F 3 iscoarser than F 1, then F 3 and F 2 are independent.

Solution:

Since F 3 is coarser than F 1 then F 3 ⊂ F 1 and the result followsfrom the independence of F 1 and F 2.



5

Problem 17. Let Ω = (ω1, ω2, ω3, ω4, ω5, ω6), P (ωi) = pi = 1

6and the sub-σ-

fields

F 1 = σω1, ω2, ω3, ω4, ω5, ω6,

F 2 = σω1, ω2, ω3, ω4, ω5, ω6.

Show that F 1 and F 2 are not independent. What can be saidabout the sub-σ-fields

F 3 = σω1, ω2, ω3, ω4, ω5, ω6,

and

F 5 = σω1, ω4, ω2, ω5, ω3, ω6?

Solution:

F 1 and F 2 are not independent since if we know, for instance,that ω1, ω2 has occurred in F 1 then we also know that theevent ω1, ω2 in F 2 has occurred. This fact can be checked bydirect calculation using the definition.

Problem 18. Let Ω = (i, j) : i, j = 1, . . . , 6 and P (i, j) =1

36. Define the

quantity

X (ω) =∞k=0

kI (i,j):i+ j=k.

Is X a random variable? Find P X(x) = P (X = x), calculateE [X ] and describe σ(X ), the σ-field generated by X .Solution:

First note that X (ω) =12k=2 kI (i,j):i+ j=k and as the sum of

indicator functions of events in the σ-field generated by Ω, it isa random variable taking values in the set 2, 3, . . . , 12. Forx ∈ 2, 3, . . . , 12, P (X = x) = P (i + j = x).E [X ] =

12k=2 kP (i+ j = k). σ(X ) is the generated by the set of

events (1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (2, 2), (1, 4), (4, 1),(2, 3), (3, 2), . . . , (6, 6), that is, it is generated by 11 atoms.

Problem 19. For the function X defined in the previous exercise, describe therandom variable P (A | X ), where A = (i, j) : i odd, j evenand find its expected value E [P (A | X )].

Solution:

P (A | X )(ω) =12k=2 P (A | X = k)I (X = k)(ω)

=12k=2

P (A

[X = k])

P (X = k)I (X = k)(ω).



6

E [P (A | X )] =

12k=2

P (A

[X = k])

P (X = k)P (X = k)

=12k=2

P (A

[X = k]) = P (A).

Problem 20. Let Ω be the unit interval (0, 1] and on it are given the followingσ-fields:

F 1 = σ(0,1

2], (

1

2,

3

4], (

3

4, 1],

F 2 = σ(0,1

4], (

1

4,

1

2], (

1

2,

3

4], (

3

4, 1],

F 3 = σ(0,

1

8], (

1

8 ,

2

8], . . . , (

7

8 , 1].Consider the mapping

X (ω) = x1I (0, 14](ω) + x2I ( 1

4, 12](ω) + x3I ( 1

2, 34](ω) + x4I ( 3

4,1](ω).

Find E [X | F 1], E [X | F 2], and E [X | F 3].Solution:

E [X | F 1] = E [x1I (0, 14] + x2I ( 1

4, 12] + x3I ( 1

2, 34] + x4I ( 3

4,1] | F 1]

= x1E [I (0, 14] | F 1] + x2E [I ( 1

4, 12] | F 1] + x3E [I ( 1

2, 34] | F 1]

+ x4E [I ( 34,1] | F 1].

For instance the first component in the sum gives:

x1E [I (0, 14 ] | F 1] = x1

P (0,14 ] | (0,

12 ]I ((0,

12 ])

+ P (0, 14

] | (12

, 34

]I ((12

, 34

]) + P (0, 14

] | (34

, 1]I ((34

, 1])

= x1

P (0, 1

4] | (0, 1

2]I ((0, 1

2])

=1

2I ((0,

1

2]).

Problem 21. Let Ω be the unit interval and ((0, 1], P ) be the Lebesgue mea-surable space and consider the following sub-σ-fields:

F 1 = σ(0,1

2], (

1

2,

3

4], (

3

4, 1],

F 2 = σ(0,1

4], (

1

4,

1

2], (

1

2,

3

4], (

3

4, 1].

Consider the mapping

X (ω) = ω.

Find E [E [X | F 1] | F 2], E [E [X | F 2] | F 1] and compare.Solution:

E [X | F 1] must be constant on the atoms of F 1 so thatE [X | F 1](ω) = x1I ((0, 1

2]) + x2I ((1

2, 34

]) + x3I ((34

, 1]). where



7

x1 =E [XI ((0, 1

2])]

P ((0, 12 ])

, x2 =E [XI ((1

2, 34

])]

P ((12 , 34 ])

, x3 =E [XI ((3

4, 1])]

P ((34 , 1]),

or

x1 =

(0, 1

2]

xdx

1/2= 1/4, x2 =

( 12, 34]

xdx

1/4= 5/8, x3 =

( 34,1]

xdx

1/4=

7/8Hence

E [X | F 1](ω) =1

4I ((0,

1

2])(ω) +

5

8I ((

1

2,

3

4])(ω) +

7

8I ((

3

4, 1])(ω),

which is a F 1-measurable random variable.E [E [X | F 1] | F 2] = E [X | F 1].

For the derivation of E [E [X | F 2] | F 1] see the steps in problem20.

Problem 22. Consider the probability measure P on the real line such that:

P (0) = p, P ((0, 1)) = q, p + q = 1,

and the random variables defined on Ω = Rm

X 1(x) = 1 + x, X 2(x) = 0I x≤0 + (1 + x)I 0<x<1 + 2I x≥1,

X 3(x) =+∞

k=−∞(1 + x + k)I k≤x≤k+1.

Is there any a.s.P equality between X 1, X 2 and X 3?Solution:

When k = 0, X 3(x) = 1 + x and hence X 1 = X 3 a.s.P .X 2(0) = 0 = X 1(0) = 1 with probability pand X 2(0) = 0 = X 3(0) = 1 with probability p.

Problem 23. Let X 1, X 2 and X 3 be three independent, identically distributed(i.i.d.) random variables such that P (X i = 1) = p = 1−P (X i =0) = 1 − q. Find P (X 1 + X 2 + X 3 = s | X 1, X 2).Solution:

See Example 1.4.3.

Problem 25. On Ω = [0, 1] and P being Lebesgue measure show thatX = x1I (0, 1

2] + x2I ( 1

2,1] and Y = y1I (0, 1

4]∪( 3

4,1] + y2I ( 1

4], 34]

are independent.Solution:

σ(X ) = σ(0, 12

], (12

, 1] and σ(Y ) = σ(0, 14

] ∪ (34

, 1], (14

], 34

].



8

Hence direct calculations give the result.

Chapter 2

Problem 2. Suppose that at time 0 you have $a and your component has$b. At times 1, 2, . . . you bet a dollar and the game endswhen somebody has $0. Let S n be a random walk on theintegers . . . , −2, −1, 0, +1, +2, . . . with P (X = −1) = q,P (X = +1) = p. Let α = inf n ≥ 1 : S n = −a, +b, i.e. thefirst time you or your component is ruined, then

S n∧α

∞n=0 is

the running total of your profit. Show that if p = q = 12

S n∧αis a bounded martingale with mean 0 and that the probability

of your ruin isb

a + b.

Show that if the game is not fair ( p = q) then S n is not a

martingale but Y n=

q

p

S n is a martingale. Find the probabil-

ity of your ruin and check that if a = b = 500, p = .499 and

q = .501 then P (ruin) = .8806 and it is almost 1 if p =1

3.

Solution:

From Theorem 2.3.11 we know that S n∧α∞n=0 is a martingaleand |S n∧α| is bounded by max(a, b).Also S α = aI (S α = a) − bI (S α = b). Taking expectation onboth side we find0 = E [S α] = aP (S α = a) − bP (S α = b)= aP (S α = a) − b(1 − P (S α = a))

which gives P (S α = a) =b

a + b.

If ( p = q) then it is easy to check that S n is not a martingale.Clearly E [Y n] < ∞.

E [Y n+1| F

n] = E [q

pS n+1 | q

pS n] = q

pS nE [q

pXn+1

] =

Y n

q

p

p +

q

p

−1q

= Y n which shows that Y n is a mar-

tingale.1 = E [S α] = (q/p)aP (S α = a) + (q/p)−b(1 − P (S α = a))

which gives P (S α = a) =(q/p)b − 1

(q/p)a+b − 1.



9

Problem 3. Show that if X n is an integrable, real-valued process, withindependent increments and mean 0 then it is a martingale

with respect to the filtration it generates and if in addition X 2n

is integrable, X 2n − E (X 2n) is a martingale with respect to thesame filtration.

Solution:

The first assertion is obvious.Let Z n = X 2n − E (X 2n). Clearly the Z n are integrable.E [Z n+1 − Z n | F n] = E [X 2n+1 − X 2n | F n] − E [X 2n+1] + E [X 2n].We must show thatE [X 2n+1 − X 2n | F n] = E [X 2n+1] − E [X 2n].Let X n+1 − X n = ∆X n+1.We see that X 2n+1 − X 2n = X n+1∆X n+1 + X n∆X n+1. Therefore

E [X 2n+1 − X

2n | F n] = E [X n+1∆X n+1 | F n] + X nE [∆X n+1 | F n]

= E [X n+1∆X n+1 | F n] using the assumptions on X n.NowX n+1∆X n+1 = (X n+1−X n+X n)∆X n+1 = (∆X n+1)2+X n∆X n+1.ThereforeE [X n+1∆X n+1 | F n] = E [(∆X n+1)2 + X n∆X n+1 | F n]= E [(∆X n+1)2] using again the assumptions on X n.Now E [(∆X n+1)2] = E [X 2n+1 + X 2n − 2X n+1X n]= E [X 2n+1] + E [X 2n] − 2E [X nE [X n+1 | F n]]= E [X 2n+1] + E [X 2n]−2E [X 2n] = E [X 2n+1]−E [X 2n] using the factthat

X n

is a martingale. This finishes the proof.

Problem 4. Let X n be a sequence of i.i.d random variables with E [X n] =0 and E [X 2n] = 1. Show that S 2n− n is an F n = σX 1, . . . , X n-martingale, where S n =

ni=1 X i.

Solution:

Let Y n = S 2n − n. Clearly the Y n are adapted and integrable.E [Y n+1 | F n] = E [S 2n+1 − (n + 1) | F n]= E [Y n + X 2n+1 − 1 + 2X n+1

ni=1 X i | F n]

= Y n + E [X 2n+1 − 1 | F n] + 2E [X n+1ni=1 X i | F n].

Using the independence assumption of X n,E [X 2n+1 − 1 | F n] = E [X 2n+1 − 1] = 1 − 1 = 0and

E [X n+1ni=1 X i | F n] =

ni=1 X iE [X n+1] = 0.

Hence E [Y n+1 | F n] = Y n and the result follows.

Problem 5. Let yn be a sequence of independent random variables withE [yn] = 1. Show that the sequence X n =

nk=0 yn is a martin-

gale with respect to the filtration F n = σy0, . . . , yn.



10

Solution:

Since the yn are integrable so are the X n.

E [X n+1 | F n] =n

k=0 ynE [yn+1 | F n] =n

k=0 yn = X n. There-fore X n is a martingale.

Problem 7. Show that two (square integrable) martingales X and Y areorthogonal if and only if X 0Y 0 = 0 and the process X nY n isa martingale.

Solution:

“⇒” follows from Theorem 3.2.6.“⇐” follows directly from the definition of the:X, Y n = E [X 0Y 0] +

ni=1 E [(X i − X i−1)(Y i − Y i−1) | F i−1] =

ni=1 E [X iY i − X iY i−1 − X i−1Y i + X i−1Y i−1 | F i−1]

=n

i=1[X i−1Y i−1 − X i−1Y i−1 − X i−1Y i−1 + X i−1Y i−1] = 0.Therefore the (square integrable) martingales X and Y are or-thogonal.

Problem 9. Let Bt be a standard Brownian motion process (B0 = 0,a.s., σ2 = 1). Show that the conditional density of Bt fort1 < t < t2

P (Bt ∈ dx | Bt1 = x1, Bt2 = x2)

is a normal density with mean and variance

µ = x1 + x2 − x1

t2 − t1(t − t1), σ2 = (t2 − t)(t − t1)

t2 − t1.

Solution:

The known distribution and independence of the incrementsBt1, Bt − Bt1, Bt2 − Bt lead to the joint distribution

f (x1, x , x2) =1√

2πt1exp

− x2

1

2t1

1

2π(t − t1)exp

−(x − x1)2

2(t − t1)

1

2π(t2 − t)exp

−(x2 − x)2

2(t2 − t)

.

Dividing by the density

f (x1, x2) =1√

2πt1exp

− x2

1

2t1

1

2π(t2 − t1)exp

−(x2 − x1)2

2(t2 − t1)

and after a bit of algebra yields the desired result.



11

Problem 12. Let N t be a standard Poisson process and Z 1, Z 2 . . . a sequenceof i.i.d. random variables such that P (Y i = 1) = P (Y i = −1) =

12

. Show that the process

X t =N ti=1

Z i

is a martingale with respect to the filtration F t = σX s, s ≤t.

Solution:

Using Wald’s equationE [|X t|] = E [|

N ti=1 Z i|] ≤ E [

N ti=1 |Z i|]

= E [N t]E [|Z i|] < ∞. Therefore X t is integrable.Now for s ≤ tE [X t | F s] = E [

N si=1 Z i + Z N t | F s] = X s + E [Z N t | F s].

However E [Z N t | F s] = E [E [Z N t | F s

N t] | F s] = 0 and theresult follows.

Problem 13. Show that the process B2t − t, F Bt is a martingale where B

is the standard Brownian motion process and F Bt its naturalfiltration.

Solution:

Clearly B2t − t is adapted and integrable. Let s ≤ t.

E [B2t − t | F

Bs ] = B

2s − s + E [B

2t − t − B

2s + s | F

Bs ].

But E [B2t − t − B2

s + s | F Bs ] = s − t + E [B2t − B2

s | F Bs ] =s − t + t − s = 0.Hence B2

t − t, F Bt is a martingale.

Problem 14. Show that the process (N t − λt)2 − λt is a martingale whereN t is a Poisson process with parameter λ.

Solution:

Clearly (N t − λt)2 − λt is adapted and integrable.Let Lt = (N t − λt)2 − λt, M t = N t − λt and recall that M t is amartingale.

NowE [Lt − Ls | F N s ] = E [M 2t − M 2s | F N s ] − λt − λs,E [M 2t − M 2s | F N s ] = E [(M t − M s)(M t + M s) | F N s ] =E [M t(M t − M s) | F N s ] because M t is a martingale and M s isF N s -measurable. AlsoE [M t(M t − M s) | F N s ] = E [M 2t | F N s ] − M 2s ,E [M 2t | F N s ] = E [N 2t | F N s ] − 2λtM s − (λt)2,



12

E [N 2t | F N s ] = E [(N t − N s)(N t + N s) | F N s ] + N 2s ,E [(N t − N s)(N t + N s) | F N s ]

= E [(N t − N s + N s)(N t − N s) | F N s ] + N sE [(N t − N s) | F

N s ].Using the independent increment property of N

E [(N t − N s + N s)(N t − N s) | F N s ]= E [(N t − N s)

2] + N sE [(N t − N s)]= λt − λs + (λt − λs)2 + M s(λt − λs) + λs(λt − λs),and N sE [(N t − N s) | F N s ] = N sM s + N sλt − N 2s= M 2s + M sλs + M sλt + λsλt − N 2s .putting all this together yieldsE [Lt − Ls | F N s ] = 0 and the result follows.

Problem 15. Show that the process

I t = t

0

f (ω, s)dM s

is a martingale. Here f (.) is an adapted, bounded, continuoussample paths process and M t = N t − λt is the Poisson martin-gale.

Solution:

First note that the integral is just a finite sum w.p.1 in anyfinite interval and since f is bounded I t is integrable.E [I t | F N s ] = I s + E [

s<u≤t f (u−)∆M u | F N s ], and

E [s<u≤t f (u−)∆M u | F N s ]

=s<u≤t E [f (u−)E [∆M u | F N u−] | F N s ]. Since M is martingale

E [∆M u | F N u−] = 0 which finishes the proof.

Problem 16. Referring to Example 2.4.4 define the processes:

N srn =

nk=1

I (ηk−1=s,ηk=r) =

nk=1

X k−1, esX k, er, (0.1)

and

Orn =nk=1

I (ηk=r) =nk=1

X k, er. (0.2)

Show that 0.1 and 0.2 are increasing processes and give theirDoob decompositions.

Solution:

Being somes of indicator functions, N srn and Orn are clearly in-creasing.Recall the representation X n = ΠX n−1 + M n.Subsituting this form of the Markov chain X in 0.1 and 0.2



13

gives their Doob decompositions.

Problem 18. Let α be a stopping time with respect to the filtration X n, F n.Show that

F α = A ∈ F ∞ : A ∩ ω : α(ω) ≤ n ∈ F n ∀ n ≥ 0.

is a σ-field and that α is F α-measurable.Solution:

The first assertion is easy.To show that α is F α-measurable note that for all k, nα ≤ kα ≤ n = α ≤ min(k, n) ∈ F min(k,n) ⊂ F n fromwhich the result follows.

Problem 19. Let X n be a stochastic process adapted to the filtration F nand B a Borel set. Show that

αB = inf n ≥ 0, X n ∈ Bis a stopping time with respect F n.

Solution:

Since[αB ≤ k] = [X 0 ∈ B]

[[X 0 /∈ B, X 1 ∈ B]

. . .

[[X 0 /∈ B, X 1 /∈ B , . . . , X k−1 /∈ B, X k ∈ B] ∈ F k, therefore αB

is a stopping time with respect F n.

Problem 20. Show that if α1, α2, are two stopping times such that α1 ≤ α2

(a.s.) then F α1 ⊂ F α2.Solution:

Let A ∈ F α1 .A

[α2 ≤ n] = (A

[α1 ≤ n])

[α2 ≤ n] ∈ F n for all n.

Problem 21. Show that if α is a stopping time and a is a positive constant,then α + a is a stopping time.

Solution:

If a > n then [α + a

≤n] = [α

≤n

−a] =

∅ ∈ F n.

If a ≤ n then [α + a ≤ n] = [α ≤ n − a] ∈ F n−a ⊂ F n, thereforeα + a is a stopping time. Note that if a < 0, α + a is a notstopping time.

Problem 22. Show that if αn is a sequence of stopping times and thefiltration F t is right-continuous then inf αn, liminf αn andlim sup αn are stopping times.



14

Solution:

Since [inf αn < t] = [αn < t] ∈ F t,[lim inf αn < t] = [supn≥1 inf k≥n αk < t]=

m≥1

n≥1

k≥n[αk > t + 1/m] ∈ F t,

[lim sup αn < t] = [inf n≥1 supk≥n αk < t]=

m≥1

n≥1

k≥n[αk < t − 1/m] ∈ F t,

therefore by the right continuity of F t inf αn, liminf αn andlim sup αn are stopping times

Chapter 3

Problem 1. Let X 1, X 2, . . . be a sequence of i.i.d N (0, 1) random variables

and consider the process Z 0 = 0 and Z n =nk=1 X k.

Show that

[Z, Z ]n =nk=1

X 2k ,

Z, Z n = n,

E ([Z, Z ]n) = E (nk=1

X 2k) = n.

Solution:

Direct substitution in the formula [Z, Z ]n =nk=1(Z k − Z k−1)

2

gives the result.Z, Z n =

nk=1 E [(Z k−Z k−1)2 | F k−1] =

nk=1 E [X 2k | F k−1] =n

k=1 E [X 2k ] = n using the independence and ditribution as-sumptions of X 1, X 2, . . . .

Problem 2. Show that if X and Y are (square integrable) martingales, thenXY − X, Y is a martingale.

Solution:

Clearly XY − X, Y is adapted and integrable.E [X nY n

− X, Y

n

| F n−1] =

− X, Y n−1 − E [(X n − X n−1)(Y n − Y n−1) | F n−1]+ E [X nY n | F n−1].But X nY n = (X n − X n−1 + X n−1)(Y n − Y n−1 + Y n−1)= (X n − X n−1)(Y n − Y n−1) + X nY n−1 − X n−1Y n−1+ Y nX n−1 − X n−1Y n−1 + X n−1Y n−1.HenceE [X nY n | F n−1] = E [(X n − X n−1)(Y n − Y n−1) | F n−1]



15

+ X n−1Y n−1 which yields the result.

Problem 3. Establish the identity:

[X, Y ]n =1

2([X + Y, X + Y ]n − [X, X ]n − [Y, Y ]n).

Solution:

[X + Y, X + Y ]n = (X 0 + Y 0)2 +ni=1(X i + Y i − X i−1 − Y i−1)2

= (X 0 + Y 0)2 +ni=1(X i − X i−1)2 +

ni=1(Y i − Y i−1)2

+ 2ni=1(X i − X i−1)(Y i − Y i−1)

= [X, X ]n + [Y, Y ]n + 2[X, Y ]n from which the result follows.

Problem 6. Show that if X n is a square integrable martingale then X 2 −X, X is a martingale.Solution:

See the solution to problem 2.

Problem 7. Find [B + N, B + N ]t and B + N, B + N t for a Brownianmotion process Bt and a Poisson process N t.

Solution:

From the identities[B +N, B +N ]t = [B, B]t+[N, N ]t+2[B, N ] = [B, B]t+[N, N ]tandB + N, B + N t = B, Bt + N, N t + 2B, N = B, Bt +

N, N tRecall that [X, X ]t = X c, X ct +s≤t(∆X s)

2,

[N, N ]t =

0≤s≤t(∆N s)2 = N t, N, N t = λt, and

[B, B]t = B, Bt = t.Therefore[B + N, B + N ]t = t + N t andB + N, B + N t = t + λt.

Problem 9. Let f be a deterministic square integrable function and Bt aBrownian motion. Show that the stochastic integral

t0

f (s)dBs

is a normally distributed random variable with distributionN (0,

t0 f 2(s)ds).

Solution:

Since f is square integrable there exists a sequence of simplefunctions f n such that



16

t

0|f (s) − f n(s)|2ds → 0, (n → ∞), and

t0

f n(s)dBsL2

→ t0

f (s)dBs. (*)Clearly each of the random variables t0

f n(s)dBs =k f n(tk−1)(Btk − Btk−1)

has distribution N (0, t0

f 2n(s)ds).

From (*) t0

f 2n(s)ds → t0

f 2(s)ds and the result follows.

Problem 10. Show that if t0

E [f (s)]2ds < ∞,

the Ito process

I t = t0 f (s)dBs

has orthogonal increments, i.e., for 0 ≤ r ≤ s ≤ t ≤ u

E [(I u − I t)(I s − I r)] = 0.

Solution:

Using the fact that I is a martingale we obtainE [(I u − I t)(I s − I r)] = E [(I s − I r)E [(I u − I t) | F s]]= E [(I s − I r)(I s − I s)] = 0.

Problem 11. Show that t0

(B2s − s)dBs = B

3

t

3− tBt.

Solution:

Using the Ito formulaB3t = 3

t0

B2sdBs + 3

t0

Bsds or using integration by part

B3t /3 =

t0

B2sdBs + tBt −

t0

sdBs and the result follows.

Problem 16. If N is a standard Poisson process show that the stochasticintegral

t

0

N sd(N s−

s)

is not a martingale. However, show that t0

N s−d(N s − s)

is a martingale. Here N t is a Poisson process. (Note that atany jump time s, N s− = N s − 1.)



17

Solution:

t

0 N sd(N s − s) = t

0 N sdN s − t

0 N sds.

Since dN s = ∆N s is either 0 or 1, we have t0

N sdN s = N T 1 + N T 2 + · · · + N t

= 1 + 2 + · · · + N t =N t(N t + 1)

2(*),

where T 1, T 2, . . . are the jump times of N .Let s ≤ t, then in view of (*)

E [ t0

N u−d(N u − u) − [ s0

N u−d(N u − u) | F s]= E [

ts

N u−d(N u − u) | F s] = E [ ts

N u−dN u − ts

N udu | F s].In view of (*)

E [ t

sN udN u | F s] = E [

N t(N t + 1)

2− N s(N s + 1)

2| F s]

= (1/2)E [(N t − N s)2

+ 2N s(N t − N s) + N t − N s | F s]= (1/2)[E (N t − N s)

2 + 2N sE (N t − N s) + E (N t − N s)]= (1/2)[t − s + (t − s)2 + (2N s + 1)(t − s)]= (t − s)/2[2N s + t − s + 2].

E [ ts N udu | F s] = E [

ts

(N u − N s + N s)du | F s]=

ts

E (N u − N s)du + N s(t − s)

= ts

(u − s)du + N s(t − s) = (t − s)/2[2N s + t − s].SinceE [

ts N udN u | F s] − E [

ts N udu | F s] = 0

it follows that t

0N ud(N u − u) is not a martingale.

However t0 N s−dN s = 0 + N T 1 + N T 2 + · · · + N t − 1

=N t(N t − 1)

2,

(or using the product rule t0

N s−dN s = (1/2)(N 2t − [N, N ]t)).

E [ ts

N u−du | F s] = (t − s)/2[2N s + t − s].

Hence E [ ts

N u−dN u | F s] − E [ ts

N u−du | F s] = 0.

It follows that t0 N ud(N u − u) is a martingale.

Problem 17. Prove that

t0

2N s−dN s = 2N t − 1.

Here N t is a Poisson process.Solution: t

02N s−dN s = 20 + 21 + · · · + 2N t−1 = 2N t − 1.



18

Problem 19. Show that the unique solution of

xt = 1 + t0 xs−dN s

is given by xt = 2N t. Here N t is a Poisson process.Solution:

The result is obtained by using the stochastic exponential for-mulaxt = eN t−

1

2N c,V ct

s≤t(1 + ∆N s)e−∆N s

= eN ts≤t e−∆N s

s≤t(1 + ∆N s)

= eN t−N t2N t = 2N t .

Problem 21. Show that the linear stochastic differential equation

dX t = F (t)X tdt + G(t)dt + H (t)dBt, (0.3)

with X 0 = ξ has the solution

X t = Φ(t)

ξ +

t0

Φ−1(s)G(s)ds +

t0

Φ−1(s)H (s)dBs

. (0.4)

Here F (t) is an n × n bounded measurable matrix, H (t) is ann × m bounded measurable matrix, Bt is an m-dimensionalBrownian motion and G(t) is an Rmn-valued bounded measur-able function. Φ(t) is the fundamental matrix solution of thedeterministic equation

dX t = F (t)X tdt.Solution:

Differentiating (0.4) shows that X is solution of (0.3).

Problem 24. Show that the linear stochastic differential equation

dX t = −αX tdt + σdBt,

with E |X 0|2 = E |ξ|2 < ∞ has the solution

X t = e−αtξ + σ

t0

e−α(t−s)dBs,

and

µt = E [X t] = e−αtE [ξ],

P (t) = V ar(X t) = e−2αtV ar(ξ) +σ2(1 − e−2αt)

2α.

Solution:

Direct calculation yield



19

d(e−αtξ + σ

t

0e−α(t−s)dBs)

=

−αe−αtξdt +

−ασ t0 e−α(t−s)dBs + σe−αteαtdBt

= −αX tdt + σdBt.Clearly E [X t] = e−αtE [ξ] because E

t0

e−α(t−s)dBs

= 0.

We also have E t

0e−α(t−s)dBs

2=

t0

e−2α(t−s)ds

from which the variance of X follows.

Problem 26. Suppose for θ ∈ Rm

X θt = eθM t−

1

2θ2At

is a martingale, and suppose there is an open neighborhood I of θ = 0 such that for all θ ∈ I and all t (P a.s.),

(i) |X θt | ≤ a,

(ii) |dX θtdθ

| ≤ b,

(iii) |d2X θtdθ2

| ≤ c.

Here a, b, c are nonrandom constants which depend on I , butnot on t. Show that then the processes M t and M 2t − Atare martingales.

Solution:

Take s ≤ t and A ∈ F s. Then using (i) and (ii) A

E

dX θtdθ

θ=0

| F s

dP = E

A

dX θtdθ

θ=0

dP | F s

= E

d

dθ

A

X θt dP

θ=0

| F s

=

d

dθ

A

X θsdP

θ=0

= A

dX θsdθ

θ=0

dP.

That is,

E

dX θtdθ

θ=0

| F s

=

dX θsdθ

θ=0

P a.s.

so E [M t| F

s] = M s a.s.The second assertion follows similarly using (iii), because

d2X θtdθ2

θ=0

= M 2t − At.

Chapter 4



20

Problem 1. Consider the probability space ([0, 1], B([0, 1]), λ), where λ isthe Lebesgue measure on the Borel sigma-field B([0, 1]). Let P

be another probability measure carried by the singleton 0, i.eP (0) = 1. Let

π1 = [0,1

2], (

1

2, 1],

π2 = [0,1

4], (

1

4,

3

4], (

3

4, 1], . . . ,

πn = [0,1

2n], . . . , (1 − 1

2n, 1].

Define the random variable

Λn([0,1

2n]) =

P ([0, 2−n])

λ([0, 2−n])=

2n,

0 elswhere in [0, 1].

Show that the sequence Λn is a positive martingale (with re-spect to the filtration generated by the partitions πn) such thatE λ[Λn] = 1 for all n but lim Λn = 0 λ-almost surely.

Solution:

Clearly Λn is adapted, λ-integrable and positive for all n .MoreoverE [Λn] = 2nλ([0, 2−n]) = 2n2−n = 1.E [Λn+1 | F n] = 2n+1E [I ([0, 2−n−1]) | F n]

= 2n+1λ([0, 2−n−1]

[0, 2−n])

λ([0, 2−n])I ([0, 2−n]) = 2nI ([0, 2−n]).

Therefore Λ is a martingale.To prove the λ-a.s. convergence of Λn to 0 it suffices toshow that

∞k=1 λ(Λn ≥ ) < ∞ is satisfied for all > 0. But

λ(Λn ≥ ) =1

2nand the result follows.

Problem 2. Prove Lemma 4.2.14.Solution:

Using Bayes’ Theorem we haveE [Y k+1 | Gk] = E [Y k+1 | Gk]= E [Y k+1 | Gk]/E [Λk+1 | Gk]

= ΛkE [λk+1Y k+1 | Gk]/ΛkE [λk+1 | Gk]= E [λk+1Y k+1 | Gk]

= E M

i=1 Mcik+1Y k+1, f iY k+1 | Gk

=M i=1 Mcik+1f iE

Y k+1, f i | Gk

=

M i=1 Mcik+1f i(1/M ) =

M i=1 cik+1f i = ck+1 = CX k,

which finishes the proof.



21

Problem 3. Consider the order-2 Markov chain X n, 1 ≤ n ≤ N dis-cussed in Example 2.4.6. Define a new probability measure P on

(Ω,F n such that P (X n = ek | X n−2 = ei, X n−1 = e j) = pk,ij.

Solution:

Let Λ0 = 1. Define

λn(ω) =kij

pk,ij pk,ij

I (Xn=ek,Xn−1=ej,Xn−2=ei)(ω), ΛN =N n=1 λn.

It is easy to show that Λn is an F n, P -martingale.We have to show that and under P the order-2 Markov chainX has transitions probabilities pk,ij.

Using Bayes’ Theorem write

P [X n = e | F n−1] = E [I (Xn=e) | F n−1] =E [I (Xn=e)Λn

| F n−1]

E [Λn | F n−1]

=Λn−1E [I (Xn=e)λn | F n−1]

Λn−1E [λn | F n−1]

=E [I (Xn=e)λn | F n−1]

E [λn | F n−1]

= E [I (Xn=e)λn | F n−1],

since E [λn] = 1]. Therefore

P [X n = e| F

n−1] = ij

p,ij

p,ijI (Xn−1=ej ,Xn−2=ei)P [X n = e

|X n−2 = ei, X n−1 = e j]

=ij

p,ij p,ij

I (Xn−1=ej ,Xn−2=ei)P [X n = e | X n−2 = ei, X n−1 = e j]

=ij

p,ij p,ij

I (Xn−1=ej ,Xn−2=ei) p,ij

= pe,Xn−1,Xn−1,

which gives the result.

Problem 6. Show that the exponential martingale

Λt

given by (4.7.5) isthe unique solution of

Λt = 1 +i,j

t0

Λs−(λijs )−1(λij

s − λijs )(dJ ijs − λijs ds).

Solution:

Equation (4.7.5) is



22

Λt = i= j exp

t

0 logλij

s

λijsdJ ijr −

t

0

(λij

s − λijs )dr

.

Let’s simply the jumping part of Λti= j exp

t0

logλij

s

λijsdJ ijr

=

i= j exp

s log

λij

s

λijs∆J ijr

=i= j

0<s≤t

λij

s

λijs

∆J ijr

Write Λt = e

− t0(λ

ijs −λ

ijs

ds)

Y t, where

Y t =

i= j0<s≤t

λij

s

λijs

∆J ijs

, and Λt = f (t, Y t).

Using Ito rule

f (t, Y t) = 1 +

t0

∂f (s, Y s−)

∂sds +

t0

∂f (s, Y s−)

∂Y dY s

+0<s≤t

f (s, Y s) − f (s, Y s−) − ∂f (s, Y s−)

∂Y ∆Y s

. (0.5)

The first integral in (0.5) is simply

t

0

Λs(λij

s

−λijs )ds.

Because Y t is a purely discontinuous process and of boundedvariation the second integral in (0.5) is equal to

0<s≤t e t0(λijs −λ

ijs )ds∆Y s.

In expression (0.5) we havef (s, Y s) − f (s, Y s−) = Λs − Λs−

= e t0(λijs −λ

ijs )ds

i= j

r≤s−

λij

r

λijr

∆J ijr

λij

s

λijs

∆J ijs

− r≤s−λij

r

λijr

∆J ijr

= Λs−i= j λij

s

λijs

∆J ijs

−1

=i= j Λs−

λij

s

λijs− 1

∆J ijs .

Putting all these results together gives (0.5). For the uniquenesssee the reference in Example 3.6.11.

Problem 7. Prove Lemma 4.7.3.



23

Solution:

By the characterization theorem of Poisson processes (see Theo-

rem 3.6.14) we must show that M ijt

= J ijt −λij

t and (M ijt )2−λij

tare (P , F t)-martingales.

By Bayes’ Theorem, for t ≥ s ≥ 0

E [M ijt | F s] = E [ΛtM ijt | F s]E [Λt | F s] =

E [ΛtM ijt | F s]Λs

.

Therefore, M ijt is a (P , F t) martingale if and only if ΛtM ijt is a(P, F t)-martingale. Now

ΛtM ijt = t0

Λs−dM ijs + t0

M ijs−dΛs + [Λ, M ij ]t.

Recall

[Λ, M ij]t =o<s≤t

∆Λs∆M ijs

=o<s≤t

∆

t0

Λs−(λij)−1(λij − λij

)dJ ijs

∆J ijs

=− t

0

Λs−

(λij)−1(λij

−λij

)d[J ij ,

J ij ]s

= − t0


)dJ ijs .

Therefore

ΛtM ijt =

t0

Λs−(dJ ijs −λij

ds)+

t0

M ijs−dΛs− t0

Λs−(λij)−1(λij−λij

)dJ ijs .

(0.6)The second integral on the right of (0.6) is a (P, F t) martingale.(Recall that

J ij

t −λijt is a (P,

F t) martingale). The other two

integrals are written as

t0

Λs−(dJ ijs − λij

ds) =

t0

Λs−(dJ ijs − λijds + λijds − λij

ds)

=

t0

Λs−(dJ ijs − λijds) +

t0

Λs−λijds − t0

Λs−λij

ds, (0.7)



24

and

t0 Λs−(λ

ij

)

−1

(λ

ij

− λ

ij

)dJ ij

s = t0 Λs−(λ

ij

)

−1

(λ

ij

− λ

ij

)(dJ ij

s − λ

ij

ds + λ

ij

ds)

=

t0

Λs−(dJ ijs − λijds) +

t0


)λijds. (0.8)

Substituting (0.7) and (0.8) in (0.6) yields the desired result and

it remains to show (M ijt )2 − λij

t is also a (P , F t) martingale.Now

(M ij)2t = 2

t0

M ijs−dM ijs + [M ij , M ij ]t = 2

t0

M ijs−dM ijs + J ijt . (0.9)

Subtracting λij

t from both sides of (0.9) makes the last term

on the right of (0.9) a (P , F t)-martingale and since the dM ijintegral is a (P , F t)-martingale the result follows.

Chapter 5

Problem 1. Assume that the state and observation processes of a systemare given by the vector dynamics (5.4.1) and (5.4.2). For m,k ∈ IN, m < k , write the unnormalized conditional density suchthat

E [ΛkI (X m ∈ dx) | Y k] = γ m,k(x)dx.

Using the change of measure techniques described in Section5.3, show that

γ m,k(x) = αm(x)β m,k(x),

where αm(x) is given recursively by (5.4.6). Show that

β m,k(x) = E [Λm+1,k | X m = x, Y k]

=1

φ(ym+1)

Rm

φm+1(Y m+1 − C m+1z)

×ψm+1(z − Am+1x)β m+1,k(z)dz. (0.10)

Solution:

For an arbitrary integrable function f : Rm

→ R writeE [Λkf (X m) | Y k] =

Rm

f (x)γ m,k(x)dx.

However,E [Λkf (X m) | Y k]= E [Λ1,mf (X m)E [Λm+1,k | X 0, . . . , X m, Y k] | Y k].

Let E [Λm+1,k | X m = x, Y k]= β m,k(x).



25

Consequently,E [Λkf (X m) | Y k] = E [Λ1,mf (X m)β m,k(xm) | Y k].

Therefore Rm

f (x)γ m,k(x)dx =

Rm

f (x)αm(x)β m,k(x)dx

from which the result follows.Nowβ m,k(x) = E [Λm+1,k | X m = x, Y k]= E [λm+1Λm+2,k | X m = x, Y k]= E [λm+1E [Λm+2,k | X m = x, X m+1Y k] | X m = x, Y k]

= E [φ(D−1

m+1(Y m+1 − C m+1X m+1))

|Dm+1|φ(Y m+1)

ψ(B−1m+1(X m+1 − Am+1x))

|Bm+1|ψ((X m+1)β m+1,k(X m+1) | X m = x, Y k].

Now recall that under P , X m+1 is normally distributed andindependent of the Y process. Therefore

β m,k(x) =

Rm

φ(D−1m+1(Y m+1 − C m+1z)

|Dm+1|φ(Y m+1)

ψ(B−1m+1(z − Am+1x))

|Bm+1|ψ((z)β m+1,k(z)ψ(z)dz,which is the result.

Problem 3. Assume that the state and observation processes are given bythe vector dynamics

X k+1 = Ak+1X k + V k+1 + W k+1 ∈ Rm,

Y k = C kX k + W k ∈ Rm.

Ak, C k are matrices of appropriate dimensions, V k and W k arenormally distributed with means 0 and respective covariancematrices Qk and Rk, assumed non singular. Using measurechange techniques derive recursions for the conditional meanand covariance matrix of the state X given the observations Y .

Solution:

Notice the noise W appears in both signal and observationsprocesses. To circumvent this write:

X k+1 = (Ak+1

−C k+1)X k + V k+1 + Y k+1, (0.11)

Y k = C kX k + W k. (0.12)

Then, we assume that we start with a reference measure P un-der which X k and Y k are two independent sequences of randomvariables normally distributed with means 0 and covariance ma-trices the identity matrix I m×m. With ψ and φ denoting stan-dard normal densities, define the positive mean one martingale



26

sequence :

Λk = k

m=0

φ(R−1m+1(Y m+1 − C m+1X m+1))

|Rm+1|φ(Y m+1)

× ψ(Q−1m+1(X m+1 − Am+1X m − Y m+1))

|Qm+1|ψ(X m+1).

Now define the ‘real world’ measure P by setting the restriction

of dP

dP to Gk = σX 0, . . . , X k, Y 0, . . . , Y k equal to Λk.

It can be shown that under the measure P the dynamics (0.11)and (0.12) hold.

The remaining steps are standard and are left to the reader.Problem 4. Let m = n = 1 in (5.8.1) and (5.8.2). The notation in Section

5.8 and Section 5.9 is used here.

Let Γt be the process defined as

Γt =

t0

x psds, p = 1, 2, . . . .

Write

E [ΛtI (Γt∈dx) | Y t] = µt(x)dx.

Show that at time t, the density µt(x) is completely described bythe p + 3 statistics st(0), st(1), . . . , st( p), Σt, and mt as follows:

µt(x) = pi=1

st(i)

q(x, t), (0.13)

where s0(i) = 0, i = 1, . . . , p, and

dst( p)

dt= − p(At + Σ−1

t B2t )st( p) + 1,

dst( p − 1)

dt= −( p − 1)(At + Σ−1

t B2t )st( p − 1) + pst( p)Σ−1

t B2tmt,

dst(i)

dt= −i(At + Σ−1

t B2t )st(i) +

1

2(i + 1)(i + 2)st(i + 2)

+ (i + 1)st(i + 1)Σ−1t B2t , i = 1, . . . , p − 2,dst(0)

dt= B2

t st(2) + Σ−1t st(1)mt.

(0.14)

Solution:



27

ΛtΓtg(xt) =

0

t

ΛsΓsg(xs)Asxsds +

0

t

ΛsΓsg(xs)Bsdws

+ (1/2)

0

t

ΛsΓs 2 g(xs)B2sds +

0

t

ΛsΓs g(xs)xsC sD−2s dys

+

0t

Λsg(xs)x psds.

Conditioning on Y t, t0

µt(x)g(x)dx =

t0

R

µs(x) g(xs)Asxdxds

+ (1/2)

0t

R

µs(x) 2 g(xs)B2sdxds

+ t

0 R

µs(x)g(x)xC sD−2s dxdys

+

t0

R

x pg(x)αs(x)dxds.

Integrating by part in x, we that µt(.) satisfies the stochasticpartial differential equation

µt(x) = − t0

(d/dx)(µs(x)Asx)ds+(1/2)

t0

(d2/dx2)µs(x)B2sds

+

t0

µs(x)xC sD−2s dys

+ t

0

x pαs(x)ds.

It can be verified that (0.13) is a solution to the above equa-tion if the time-varying coefficients at(0), . . . , at( p) satisfy theordinary differential equations (0.14).

Problem 5. Give a detailed proof of Lemma 5.7.1.

Solution:

Let us recall

λk+1 =φ(D−1

k+1(Y k+1 − C k+1X k+1))

|Dk+1|φ((Y k+1)

ψ(B−1l (X k+1 − Ak+1X k))

|Bk+1|ψ((X k+1).

Suppose f, g : Rm → R are “test” functions (i.e. measur-able functions with compact support). Then with E (resp. E )

denoting expectation under P (resp. P ) and using Bayes’ The-orem

E [f (V k+1)g(W k+1) | Gk] =E [Λk+1f (V k+1)g(W k+1) | Gk]

E [Λk+1 | Gk]= E [λk+1f (V k+1)g(W k+1) | Gk],



28

Consequently

E [f (V k+1

)g(W k+1

)| Gk

] = E [λk+1

f (V k+1

)g(W k+1

)| Gk

]

= E

φ(D−1

k+1(Y k+1 − C k+1X k+1))

|Dk+1|φ((Y k+1)

ψ(B−1k+1(X k+1 − Ak+1X k))

|Bk+1|ψ((X k+1)

× f (B−1k+1(X k+1 − Ak+1X k))g(D−1

k+1(Y k+1 − C k+1X k+1)) | Gk

= E

ψ(B−1

k+1(X k+1 − Ak+1X k))

|Bk+1|ψ((X k+1)f (B−1

k+1(X k+1 − Ak+1X k))

× E

φ(D−1

k+1(Y k+1 − C k+1X k+1))

|Dk+1|φ((Y k+1)g(D−1

k+1(Y k+1 − C k+1X k+1)) | Gk, X k+1

| Gk

.

Now

E

φ(D−1

k+1(Y k+1 − C k+1X k+1))

|Dk+1|φ((Y k+1)g(D−1

k+1(Y k+1 − C k+1X k+1)) | Gk, X k+1

=

Rm

φ(D−1k+1(y − C k+1X k+1))

|Dk+1|φ(y)g(D−1

k+1(y − C k+1X k+1))φ(y)dy

=

Rm

φ(u)g(u)du,

after an appropriate change of variable. Similar calculationsshow that

E [f (V k+1)g(W k+1) | Gk] = Rm

φ(u)f (u)du Rm

ψ(u)g(u)du,

which finishes the proof.

Problem 6. Prove (5.7.5), (5.7.6), (5.7.7) and (5.7.3).Solution:

We have to show that

αk+1(x) = Ψk+1(x, Y k+1)

Rm

Φ(x, z)αk(z)dz.

(0.15)

Here

Ψk+1(x, Y k+1) =ψ(D−1

k+1(Y k+1 − C k+1x))

|Bk+1||Dk+1|ψ(Y k+1)(0.16)

and

Φ(x, z) = φ(B−1k+1(x − Ak+1z)). (0.17)



29

For any “test” function g

Rm g(x)αk+1(x)dx = E [Λk+1g(X k+1) | Y k+1]

= E [Λkλk+1g(X k+1) | Y k+1]

= E

Λk

φ(B−1k+1(X k+1 − Ak+1X k))ψ(D−1

k+1(Y k+1 − C k+1X k+1))

|Bk+1||Dk+1|φ(X k+1)ψ(Y k+1)

× g(X k+1) | Y k+1]

=1

|Bk+1||Dk+1|ψ(Y k+1)

×E Λk Rmφ(B−1

k+1(x − Ak+1X k))ψ(D−1k+1(Y k+1 − C k+1x))

φ(x)

× φ(x)g(x)dx | Y k+1] .

The last equality follows from the fact that X k+1 has distribu-tion φ and is independent of everything else under P . Also,note that given Y k+1 we condition only on Y k to get

Rm

g(x)αk+1(x)dx =1

|Bk+1||Dk+1|ψ(Y k+1)

× Rm

Rm

φ(B−1k+1(x − Ak+1z))ψ(D−1

k+1(Y k+1 − C k+1x))

×g(x)αk(z)dxdz.

This holds for all “test” functions g so we can conclude that(0.15) holds.

Let S k =km=1

X m−1 ⊗ X m−1,

S ijk =km=1

X m−1, eiX m−1, e j and write

β ijk (x) = E [ΛkS ijk I (X k ∈ dx) | Y k].

E [ΛkS ijk g(X k)

| Y k] = Rm β ijk (x)g(x)dx

β k+1(x) = Ψk+1(x, Y k+1)

RmΦ(x, z)β k(z)dz

+ Rm

z, eiz, e jαk(z)Φ(x, z)dz

We have to show that.β k+1(x) = Ψk+1(x, Y k+1)

Rm

Φ(x, z)β k(z)dz



30

+

Rm

z, eiz, e jαk(z)Φ(x, z)dz

.

Note that S ijk+1

= S ijk

+X k

, ei

X k

, e j

, therefore Rm

β ijk+1(x)g(x)dx

= E [Λk+1S ijk g(X k+1) | Y k+1]+ E [Λk+1X k, eiX k, e jg(X k+1) | Y k] =

= E [λk+1ΛkS ijk g(X k+1) | Y k+1]+ E [λk+1ΛkX k, eiX k, e jg(X k+1) | Y k].The remaining steps are the by now familiar calculation so theyare skipped.

Problem 11. Give the proof of theorem 5.11.4.Solution:

Here the dynamics are given by

dxt = g(xt)dt + s(xt)dBt,

dyt = h(xt)dt + α(yt)dW t.

We have to show that if φ ∈ C 2(Rd) is a real-valued functionwith compact support. Then

σ(φ)t = σ(φ)0 + t

0

σ(Aφ)sds + t

0

σ[(

φ(xs))s(xs)ρα(ys)

−2h(xs)]ds

+

t0

(α−1(ys)σ(φh))α−1(ys)dys,

where Aφ(x) =1

2

di,j=1

(ss)ij(xs)∂ 2φ(xs)

∂xi∂x j+

di=1

gi(xs)∂φ(xs)

∂xi.

Using the vector Ito rule we establish

φ(xt) = φ(x0) + t

0

Aφ(xs)ds + t

0

(

φ(xs))s(xs)dws (0.18)

Recall that

Λt = exp

t0

(α(ys)−1h(xs))α(ys)−

1dys − 1

2

t0

|α(ys)−1h(xs)|2ds

.

andΛt = 1 +

t0

Λs(α(ys)−1h(xs))α(ys)−

1dys



31

using the Ito product rule

Λtφ(xt) = φ(x0) + t0 ΛsAφ(xs)ds +

t0 Λs(φ(xs))

Bsdws

+

t0

Λsφ(xs)(α(ys)−1h(xs))α(ys)

−1dys

+ [Λ, φ]t (0.19)

. (0.20)

But[Λ, φ]t = [

t0

Λs(α(ys)−1h(xs))α(ys)

−1dys, t0

(φ(xs))s(xs)dws]

= t0

Λs(φ(xs))s(xs)ρα(ys)−2h(xs)ds.

Conditioning both sides of (0.20) on Y t gives the result.

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