Martingales Exercises

Martingales Exercises

Exercise 4.1. Show that a sum of martingales is a martingale.

Exercise 4.2.

a)Is any Markovian process is a martingale? If yes, prove it. Otherwise, construct a counter- example.

b)Is any martingale is Markovian? If yes, prove it. Otherwise, construct a counter-example.

Exercise 4.3. LetM =fM_t:t2 f0;1;2; :::ggbe a square-integrable martingale¹existing on a …ltered probability space( ;F;F;P), whereFis the …ltration …ltrationfF^t :t2 f0;1;2; :::gg. The predictable process =f t:t 2 f0;1;2; :::gg is constructed on the same space, that is, for all t2 f1;2; :::g, _t isF^{t 1} measurable, ₀ beingF⁰ measurable. Moreover, we assume that for all t2 f0;1;2; :::g, the random variable _t is square-integrable,that is, E [j ²tj]<1: Show that the process N =fN_t :t 2 f0;1;2; :::gg de…ned as

N_t=N₀+ Xt k=1

k(M_k M_{k 1}) is a martingale provided thatN₀ is F⁰ measurable.

Exercise 4.4. LetX be a square-integrable random variable E^P[jXj]<1 constructed on the …ltered probability space( ;F;F;P). Prove that the stochastic processfM_t :t2 f0;1;2; :::gg de…ned as

M_t = E^P[XjF^t], t 0 is a martingale.

Exercise 4.5. Consider the probability space ( ;F;P) on which two …ltrations are con- structed, fF^t:t 0g and fG^t:t 0g; satisfying

F^t G^t

a) Let M = fM_t:t 0g be a fF^t:t 0g martingale and let N = fN_t :t 0g be a fG^t:t 0g martingale. IsM afG^t:t 0g martingale ? IsN afF^t:t 0g martingale ? Justify.

b) Let be a fF^t :t 0g stopping time and be a fG^t:t 0g stopping time. Is a fF^t:t 0g stopping time ? Is a fG^t :t 0g stopping time ? Justify.

18t 0; E M_t² <1

Exercise 4.6. Let "1; "2; :::be a sequence of independent random variables with zero mean and variance Var ["_i] = ²_i. Let

Sn= Xn

i=1

"i and T_n² = Xn

i=1 2 i:

Prove that fS_n² T_n²gn2N is a martingale.

Exercise 4.7. Let fX_t :t 0g be a fG^t :t 0g martingale and F^t= (X_s; s t). Prove that fX_t:t 0g is also afF^t:t 0g martingale.

Solutions

1 Exercise 4.1

LetX =fXt:t2 f0;1; :::gg and Y =fYt:t2 f0;1; :::gg, two martingales on ( ;F;F;P): Since 8t2 f0;1; :::g,

E^P[jX_t+Y_tj] E^P[jX_tj]

| {z }

<1

sinceX is a martingale

+ E^P[jY_tj]

| {z }

<1

sinceY is a martingale

<1;

the(M1)condition is veri…ed.

Since twoF^t measurable random variables isF^t measurable, then8t2 f0;1; :::g,X_t+Y_t is F^t measurable. Therefore, the stochastic processX+Y is adapted to the …ltration F:

Condition (M3) :8s; t 2 f0;1; :::gsuch that s < t

E^P[X_t+Y_tjF^s] = E^P[X_tjF^s] + E^P[Y_tjF^s] =X_s+Y_s:

The proof can be generalized to the sum of more than 2 martingales using induction.

2 Exercise 4.2

a) It is not all Markovian processes that are martingales.

Counter-example. On the probability space ( ;F;P),consider the random walk X de…ne as

X₀ = 0 and 8t2 f0;1;2; :::g, X_t= Xt n=1

n

where the sequence of independent and identically distributed random variablesf t:t 2 f1;2; :::gg has a positive expectation E^P[ _t] = >0. SinceX is a random walk, it is Markovian. But X cannot be a martingale since its expectation varies over time. Indeed,

E^P[X_t] = E^P

" _t X

n=1 n

#

= Xt n=1

E^P[ _n] = Xt n=1

=t :

b) It is not all martingales that are Markovian. Counter-example :

! X₁(!) X₂(!) X₃(!) P

!₁ 1 2 4 ¹₈

!₂ 1 2 0 ¹₈

!₃ 1 0 0 ¹₈

!₄ 1 0 0 ¹₈

!₅ 1 0 2 ¹₈

!₆ 1 0 2 ¹₈

!7 1 2 2 ¹₈

!₈ 1 2 2 ¹₈

Indeed,

fX₁g = ff!₁; !₂; !₃; !₄g;f!₅; !₆; !₇; !₈gg fX1; X2g = ff!1; !2g;f!3; !4g;f!5; !6g;f!7; !8gg fX₁; X₂; X₃g = ff!₁g;f!₂g;f!₃; !₄g;f!₅g;f!₆g;f!₇; !₈gg

The processX =fX_t :t2 f1;2;3ggis not Markovian on the space( ;F;P)whereF is the algebra generated by all elements of : Indeed,

P(X₃ = 2jX₂ = 0 ) = P(X₃ = 2 and X₂ = 0) P(X₂ = 0)

= Pf!₅g

Pf!₃; !₄; !₅; !₆g =

1 8 1 2

= 1 4 but

P(X₃ = 2jX₂ = 0 and X₁ = 1 ) = P(X₃ = 2, X₂ = 0 and X₁ = 1) P(X₂ = 0 and X₁ = 1) P(?) 0

(M2)is also satis…ed from the construction of the …ltration. Need to verify the last condition (M3 ). But

E^P[X₃j fX₁; X₂g] =X₂ since

8! 2 f!₁; !₂g, E^P[X₃j fX₁; X₂g] (!) = 4 ¹₈ + 0 ¹₈

1 4

= 2 =X₂(!) 8! 2 f!3; !4g, E^P[X3j fX1; X2g] (!) = 0 ¹₈ + 0 ¹₈

1 4

= 0 =X2(!) 8! 2 f!₅; !₆g, E^P[X₃j fX₁; X₂g] (!) = 2 ¹₈ 2 ¹₈

1 4

= 0 =X₂(!) 8! 2 f!₇; !₈g, E^P[X₃j fX₁; X₂g] (!) = 2 ¹₈ + 2 ¹₈

1 4

= 2 =X₂(!) and

E^P[X₂j fX₁g] =X₁ since

8! 2 f!₁; !₂; !₃; !₄g

E^P[X₂j fX₁g] (!) = 2 ¹₈ 2 ¹₈ + 0 ¹₈ + 0 ¹₈

1 2

= 1 = X₁(!) 8! 2 f!₅; !₆; !₇; !₈g

E^P[X₂j fX₁g] (!) = 0 ¹₈ + 0 ¹₈ + 2 ¹₈ + 2 ¹₈

1 2

= 1 =X₁(!):

3 Exercise 4.3

Veri…cation of (M1) : E [jN_tj] = E

"

N₀+ Xt k=1

k(M_k M_{k 1})

#

E [jN₀j] + Xt

k=1

E [j kj jM_k M_{k 1}j]

E [jN₀j] + Xt

k=1

E j kj^{2 1=2} E jM_k M_{k 1}j^{2 1=2} from Cauchy-Schwarz inequality

= E [jN₀j] + Xt

k=1

E j kj^{2 1=2} E M_k² 2M_kM_{k 1}+M_k² ₁ ¹⁼²

= E [jN₀j] + Xt

k=1

E j kj^{2 1=2} E M_k² 2E [M_kM_{k 1}] + E M_k² ₁

= E [jN₀j] + Xt

k=1

E j kj^{2 1=2} E M_k² 2E [M_kM_{k 1}] + E M_k² ₁ since E [M_kM_{k 1}] = E [E [M_kM_{k 1}jF^{k 1}]]

= E [M_{k 1}E [M_kjF^{k 1}]]

= E M_k² ₁

= E [jN₀j]

| {z }

<1

+ Xt

k=1

E j kj^{2 1=2}

| {z }

<1

E M_k² E M_k² ₁ ¹⁼²

| {z }

<1

< 1

Veri…cation de (M2) : N_t=N₀+

Xt

|{z}k (M_k M_{k 1})

| {z } estF^t mesurable.

Veri…cation de (M3) : …rstly, note that for all 0 s < t <1; N_t=N_s+

Xt k=s+1

k(M_k M_{k 1}): Indeed,

Nt = N0+ Xt k=1

k(Mk Mk 1)

= N₀+ Xs k=1

k(M_k M_{k 1}) + Xt k=s+1

k(M_k M_{k 1})

= N_s+ Xt k=s+1

k(M_k M_{k 1}): Therefore,

E [N_tj F^s] = E

"

N_s+ Xt k=s+1

k(M_k M_{k 1}) F^s

#

= N_s+ Xt k=s+1

E [ _k(M_k M_{k 1})j F^s]

= Ns+ Xt k=s+1

E [ E [ _k(Mk Mk 1)j F^{k 1}]j F^s] from (EC3)

= Ns+ Xt k=s+1

E 2

4 _kE [ (Mk Mk 1)j F^{k 1}]

| {z }

=0

F^s 3 5

= N_s:

4 Exercise 4.4

Veri…cation of (M1) :

E^P[jM_tj] = E^P E^P[XjF^t]

E^P E^P[jXj jF^t] since X jXj

= E^P E^P[jXj jF^t] since E^P[jXj jF^t] 0

= E^P[jXj] since (EC3)

< 1by hypothesis

Veri…cation of (M2) : M_t= E^P[XjF^t] isF^t measurable because constant on the atoms F^t.

Veri…cation of (M3) : Let0 s t.

E^P[M_tjF^s] = E^P E^P[XjF^t]jF^s from the defn. of M_t

= E^P[XjF^s] from(EC3)since F^s F^t

= M_s from the de…nition of M_s.