80-646-08 Stochastic Calculus I

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De…nition Example Stopped process Optional Stopping Theorem Markovian process

Martingales

80-646-08 Stochastic Calculus I

Geneviève Gauthier

HEC Montr éal

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De…nition Lemma 1 Example lemma 2

Example Stopped process Optional Stopping Theorem Markovian process

De…nition

On the …ltered probability space ( Ω , F ^, ^F ^, ^P ) , where F is the

…ltration fF ^t ^: ^t 2 f ^0, ^1, ^2, ^{. . .} gg , the stochastic process M = f ^M ^t ^: ^t 2 f ^0, ^1, ^2, ^{. . .} gg

is a discrete-time martingale if

(M1) 8 ^t 2 f ^0, ^1, ^2, ^{. . .} g ^, ^E

^P

[ j ^M ^t j ] < _∞ ; (M2) 8 ^t 2 f ^0, ^1, ^2, ^{. . .} g ^, ^M ^t ^is F ^t measurable;

(M3) 8 ^s ^, ^t 2 f ^0, ^1, ^2, ^{. . .} g ^{such that} ^s < t, E

^P

[ M _t jF ^s ] = M _s .

(3)

Martingale

Constant expectation process

Lemma. Let M = f ^M ^t ^: ^t 2 f ^0, ^1, ^2, ^{. . .} gg be a martingale built on the …ltered probability space ( Ω , F ^, F , P ) . Then

8 ^t 2 f ^1, ^2, ^{. . .} g ^, ^E

^P

[ M _t ] = E

^P

[ M ₀ ] . Proof of the lemma. 8 ^t 2 f ^1, ^2, ^{. . .} g ^,

E

^P

[ M _t ] = E

^P

h

E

^P

[ M _t jF ⁰ ] ⁱ by ( EC 3 )

= E

^P

[ M ₀ ] by ( M 3 ) .

Interpretation. A martingale is a stochastic process that,

on average, is constant. This doesn’t mean however that

such a process varies little, since the variance Var

^P

[ M _t ] ,

at every time, can be in…nite.

(4)

Example I

Example. Let f ^ξ t : t 2 f ^1, ^2, ^{. . .} gg be a sequence of ( _Ω , F ) independent and identically distributed random variables with respect to the measure P and such that

E

^P

[ ξ _t ] = 0 and E

^P

ξ ² _t < _∞ . Let’s de…ne

F ⁰ = f? ^, ^Ω g ^;

8 ^t 2 f ^1, ^2, ^{. . .} g ^, F ^t = σ f ^ξ s : s 2 f ^1, ^{. . .} ^, ^t gg ^; and

M ₀ = 0, M _t =

∑ t s = 1

ξ _s .

The stochastic process M is a martingale on the space

( Ω , F ^, F , P ) .

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Example II

Indeed,

E

^P

[ j ^M ^t j ] = E

^P

"

∑ t s = 1

ξ _s

# t

s ∑ = 1

E

^P

[ j ^ξ s j ]

∑ t s = 1

q

E

^P

ξ ² _s < ∞

where the second inequality comes from the fact that, for any random variable,

0 Var [ j ^X j ] = E h

j ^X j ² ⁱ ( E [ j ^X j ]) ² ) ^E [ j ^X j ] r

E h

j ^X j ² ⁱ ^.

Given the selected …ltration, M is adapted (which is to say that

8 ^t 2 f ^0, ^1, ^2, ^{. . .} g ^, ^M ^t ^is F ^t mesurable).

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Example III

Lastly, 8 ^s ^, ^t 2 f ^0, ^1, ^2, ^{. . .} g ^{such that} ^s < t, E

^P

[ M _t jF ^s ]

= E

^P

"

M s +

∑ t u = s + 1

ξ _u jF ^s

#

= E

^P

[ M _s jF ^s ] +

∑ t u = s + 1

E

^P

[ ξ _u jF ^s ]

= M s +

∑ t u = s + 1

E

^P

[ ξ _u jF ^s ]

| {z }

=

_E^P

[

ξ_u

]

from ( EC 1 ) since M _s is F ^s measurable, and from ( EC 7 ) since ξ _u is independent from ξ ₁ , . . . , ξ _s .

= M s .

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Martingale I

Lemma

In the de…nition of a martingale, the condition ( M 3 ) is equivalent to

M3* 8 ^t 2 f ^1, ^2, ^{. . .} g ^, ^E

^P

[ M _t jF ^t ¹ ] = M _t ₁ .

Proof of the lemma. Clearly, ( M 3 ) ) ( M3 ) since ( M 3 ) is only a special case of ( M 3 ) . Indeed, it is su¢ cient to de…ne s = t 1.

So we must show that ( M 3 ) ) ( M 3 ) . This can be proved by induction. Intuitively, if s < t then

E

^P

[ M t jF ^s ] = E

^P

h

E

^P

[ M t jF ^t ¹ ] jF ^s ⁱ ^from ( EC 3 ) ,

= E

^P

[ M _t ₁ jF ^s ] from ( M 3 )

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Martingale II

But if s < t 1, then we can use the same logic again, and we get

E

^P

[ M _t ₁ jF ^s ] = E

^P

h

E

^P

[ M _t ₁ jF ^t ² ] jF ^s ⁱ ^from ( EC 3 ) ,

= E

^P

[ M _t ₂ jF ^s ] from ( M 3 ) . Now just substitute this result into the …rst equation:

E

^P

[ M t jF ^s ] = E

^P

[ M t 2 jF ^s ] .

By iterating such an algorithm, we will eventually obtain E

^P

[ M _t jF ^s ] = E

^P

[ M _s ₊ ₁ jF ^s ]

= M _s from ( M 3 ) .

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Example I

ω ξ

₁

ξ

₂

ξ

₃

P Q

ω

1

1 1 1

¹₈ ¹₂

ω

2

1 1 1

¹₈ ₁₄¹

ω

3

1 1 1

¹₈ ₁₄¹

ω

₄

1 1 1

¹₈ ₁₄¹

ω ξ

₁

ξ

₂

ξ

₃

P Q

ω

5

1 1 1

¹₈ ₁₄¹

ω

6

1 1 1

¹₈ ₁₄¹

ω

7

1 1 1

¹₈ ₁₄¹

ω

₈

1 1 1

¹₈ ₁₄¹

On the sample space Ω = f ^ω ¹ ^, ^{. . .} ^, ^ω ⁸ g , we will use the σ-algebra F = the set of all events in Ω . The …ltration F is made up of the σ-subalgebras

F

0

= f? ^, ^Ω g ^,

F

1

= σ f ^ξ

1

g = σ ff ^ω

1

, ω

₂

, ω

₃

, ω

₄

g ^, f ^ω

5

, ω

₆

, ω

₇

, ω

₈

gg ^,

F

²

= σ f ^ξ

1

, ξ

₂

g = σ ff ^ω

¹

^, ^ω

²

g ^, f ^ω

³

^, ^ω

⁴

g ^, f ^ω

⁵

^, ^ω

⁶

g ^, f ^ω

⁷

^, ^ω

⁸

gg ^,

F

³

= σ f ^ξ

1

, ξ

₂

, ξ

₃

g = F ^.

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Example II

The stochastic process M, built on the …ltered measurable space ( _Ω , F ^, F ) is de…ned as follows :

M ₀ = 0, M ₁ = ξ ₁

M ₂ = ξ ₁ + ξ ₂ and M 3 = ξ ₁ + ξ ₂ + ξ ₃ .

By construction, M is adapted to the …ltration F .

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Example III

M = f ^M ^t ^: ^t 2 f ^0, ^1, ^2, ³ gg is a martingale on ( Ω , F ^, ^F ^, ^P ) . Indeed, condition ( M2 ) is already veri…ed since M is F adapted.

Condition ( M1 ) is also satis…ed since 8 ^t 2 f ^0, ^1, ^2, ³ g ^, E

^P

[ j ^M ^t j ] E

^P

[ j ^ξ 1 j ] + E

^P

[ j ^ξ 2 j ] + E

^P

[ j ^ξ 3 j ] = 3.

Let’s verify condition ( M 3 ) .

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Example IV

E

^P

[ M

₁

jF

0

] = E

^P

[ M

₁

] from ( EC 4 )

= 0

= M

0

8 ^ω 2 f ^ω

¹

^, ^ω

²

^, ^ω

³

^, ^ω

⁴

g ^, E

^P

[ M

2

jF

¹

] ( ω ) = ¹

₁

2

2 1

8 2 1

8 + 0 1 8 + 0 1

8 = 1

M

1

( ω ) = 1;

8 ^ω 2 f ^ω

5

, ω

6

, ω

7

, ω

8

g ^, E

^P

[ M

2

jF

1

] ( ω ) = ¹

₁

2

0 1

8 + 0 1

8 + 2 1 8 + 2 1

8 = 1

M

₁

( ω ) = 1

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Example V

8 ^ω 2 f ^ω

1

, ω

2

g ^, E

^P

[ M

3

jF

2

] ( ω ) = ¹

₁

4

3 1

8 1 1

8 = 2 and M

2

( ω ) = 2;

8 ^ω 2 f ^ω

3

, ω

4

g ^, E

^P

[ M

3

jF

2

] ( ω ) = ¹

₁

4

1 1

8 + 1 1

8 = 0 and M

2

( ω ) = 0;

8 ^ω 2 f ^ω

5

, ω

₆

g ^, E

^P

[ M

₃

jF

2

] ( ω ) = ¹

₁

4

1 1

8 + 1 1

8 = 0 and M

₂

( ω ) = 0;

8 ^ω 2 f ^ω

7

, ω

₈

g ^, E

^P

[ M

₃

jF

2

] ( ω ) = ¹

₁

4

1 1

8 + 3 1

8 = 2 and M

₂

( ω ) = 2.

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Example VI

By contrast M = f ^M ^t ^: ^t 2 f ^0, ^1, ^2, ³ gg is not a martingale on ( Ω , F ^, ^F ^, ^Q ) . Indeed,

E

^Q

[ M ₁ jF ⁰ ]

= E

^Q

[ M 1 ] from ( EC 4 )

= ¹

2 + ¹

14 + ¹ 14 + ¹

14 = ⁶

14

6 = 0

= M 0 .

(15)

Example VII

Conclusion. For a stochastic process, the property of

being a martingale depends on the …ltration and on the

measure. That’s why the notation ( F , P ) martingale

may sometimes be seen.

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De…nition Example Stopped process

De…nition Example Theorem Optional Stopping Theorem Markovian process

De…nition

The stochastic process X and the stopping time τ are built on the same …ltered measurable space ( Ω , F ^, ^F ) . The stochastic process X

^τ

de…ned by

X _t

^τ

( ω ) = X _t _^

_τ

₍

_ω

₎ ( ω ) (1)

is called a stopped process with stopping time τ.

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Example I

ω X 0 X 1 X 2 X 3 τ X ₀

^τ

X ₁

^τ

X ₂

^τ

X ₃

^τ

ω 1 1 ¹ ₂ 1 ¹ ₂ 0 1 1 1 1 ω 2 1 ¹ ₂ 1 ¹ ₂ 3 1 ¹ ₂ 1 ¹ ₂

ω 3 1 2 1 1 1 1 2 2 2

ω 4 1 2 2 1 1 1 2 2 2

(18)

Stopped martingale I

Theorem

If the martingale M and the stopping time τ are built on the

same …ltered probability space ( Ω , F ^, ^F ^, ^P ) then the stopped

process M

^τ

is also a martingale on that space.

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Stopped martingale II

Theorem

Proof of the theorem. The key to the proof is to express M _t

^τ

in terms of the components of the process M .

M ₀

^τ

= M ₀ and 8 ^t 2 f ^1, ^2, ^{. . .} g ^, ^M t

^τ

= M _t

^τ

t 1 k ∑ = 0

I _f

τ

= k g + _I _f

_τ

_t _g

!

=

t 1 k ∑ = 0

I _f

_τ

= k g M _t

^τ

+ I _f

_τ

t g M _t

^τ

=

t 1 k ∑ = 0

I _f

_τ

= k g M _k + I _f

_τ

t g M _t .

(20)

Stopped martingale III

Theorem

Verifying condition ( M 1 ) :

E

^P

[ j ^M 0

^τ

j ] = E

^P

[ j ^M ⁰ j ] < ∞ and 8 ^t 2 f ^1, ^2, ^{. . .} g ^,

E

^P

[ j ^M t

^τ

j ] = E

^P

"

t 1 k ∑ = 0

I _f

τ

= k g M _k + _I _f

_τ

_t _g M _t

#

t 1 k ∑ = 0

E

^P

I _f

τ

= k g M _k + E

^P

I _f

τ

t g M _t

t 1 k ∑ = 0

E

^P

[ j ^M ^k j ] + E

^P

[ j ^M ^t j ] < ∞

since, M being a martingale, we have that 8 ^t 2 f ^0, ^1, ^2, ^{. . .} g ^,

E

^P

[ j ^M ^t j ] < ∞ .

(21)

Stopped martingale IV

Theorem

Verifying condition ( M 2 ) :

M ₀

^τ

= M ₀ is F ⁰ measurable. (2) Now, 8 ^t 2 f ^1, ^2, ^{. . .} g ^,

M _t

^τ

=

t 1 k ∑ = 0

I _f

τ

= k g

| {z }

F

k

measurable since f

^τ

= k g2F

^k

M _k

|{z}

F

k

measurable since M is adapted.

| {z }

F

^t

measurable since k < t )F

^k

F

^t

+ _I _f

_τ

_t _g

| {z }

F

^t ¹

^measurable since

f

^τ

^t g = f

^τ

^t ¹ g

^c

2F

t 1

M _t

|{z}

F

^t

^measurable since M is adapted.

is F ^t measurable.

(22)

Stopped martingale V

Theorem

Verifying condition ( M 3 ) : 8 ^t 2 f ^1, ^2, ^{. . .} g ^, M _t

^τ

M _t

^τ

₁

=

t 1 k ∑ = 0

I _f

_τ

= k g M _k + I _f

_τ

t g M _t

!

t 2 k ∑ = 0

I _f

_τ

= k g M _k + I _f

_τ

t 1 g M _t ₁

!

= I _f

τ

= t 1 g M _t ₁ + I _f

τ

t g M _t I _f

τ

t 1 g M _t ₁

= _I _f

_τ

_t _g M t I _f

τ

t 1 g I _f

τ

= t 1 g M t 1

= _I _f

_τ

_t _g M _t I _f

τ

t g M _t ₁

since, f ^τ = t 1 g ^and f ^τ ^t g being disjoint, I _f

_τ

= t 1 g + I _f

_τ

t g = I _f

_τ

= t 1 g[f

^τ

^t g = I _f

_τ

t 1 g .

= I _f

τ

t g ( M t M t 1 ) .

(23)

Stopped martingale VI

Theorem

As a consequence, since I _f

_τ

t g is F ^t ¹ ^measurable E

^P

[ M _t

^τ

jF ^t ¹ ] M _t

^τ

₁

= E

^P

[ M _t

^τ

M _t

^τ

₁ jF ^t ¹ ]

= E

^P

I _f

_τ

t g ( M _t M _t ₁ ) jF ^t ¹

= I _f

_τ

t g E

^P

[ M _t M _t ₁ jF ^t ¹ ]

= _I _f

_τ

_t _g E

^P

[ M t jF ^t ¹ ] E

^P

[ M t 1 jF ^t ¹ ]

= I _f

τ

t g ( M t 1 M t 1 ) = 0 hence

E

^P

[ M _t

^τ

jF ^t ¹ ] = M _t

^τ

₁ .

(24)

Optional Stopping Theorem

Theorem

(Optional Stopping Theorem). Let

X = f ^X ^t ^: ^t 2 f ^0, ^1, ^2, ^{. . .} gg be a process built on the …ltered probability space

( Ω , F ^, ^F ^, ^P ) , where F is the …ltration fF ^t ^: ^t 2 f ^0, ^1, ^2, ^{. . .} gg ^. Let’s assume that the stochastic process X is F adapted and that it is integrable, i.e. E

^P

[ j ^X ^t j ] < _∞ . Then X is a

martingale if and only if

E

^P

[ X

_τ

] = E

^P

[ X 0 ]

for any bounded stopping time τ, i.e for any given stopping time τ, there exists a constant b such that

8 ^ω 2 Ω , 0 τ ( ω ) b.

(25)

Optional Stopping Theorem I

Proof of the theorem

First part. Let’s assume that X is a martingale and let’s show that, in such a case, E

^P

[ X

_τ

] = E

^P

[ X ₀ ] for any bounded stopping time.

Let τ be any bounded stopping time. Then, there exists a

constant b such that 8 ^ω 2 Ω , 0 τ ( ω ) b. As a

(26)

Optional Stopping Theorem II

Proof of the theorem

consequence,

E

^P

[ X

τ

] = E

^P

"

_b

k=0

∑

X

_k

I

_fτ=k_g

#

= E

^P

"

∑

b k=0

X

_k

I

_f_τ _k_g

I

_f_τ k+1g

#

=

∑

b k=0

E

^P

h

X

_k

I

_f_τ _k_g

i

^b

k

∑

=0

E

^P

h

X

_k

I

_f_τ k+1g

i

= E

^P

[ X

₀

] +

∑

b k=1

E

^P

h

X

_k

I

_f_τ _k_g

i

^b ¹

k=0

∑

E

^P

h

X

_k

I

_f_τ _k+1g

i

since I

_f_τ ₀_g

= I

_Ω

= 1 and I

_f_τ _b+1_g

= I

_?

= 0.

= E

^P

[ X

0

] +

∑

b k=1

E

^P

h

X

_k

I

_fτ kg

i

^b

k=1

∑

E

^P

[ X

_k ₁

I

_fτ kg

(27)

Optional Stopping Theorem III

Proof of the theorem

= E

^P

[ X

0

] +

∑

b k=1

E

^P

h

( X

_k

X

_k ₁

) I

_fτ _kg

i

= E

^P

[ X

0

] +

∑

b k=1

E

^P

h E

^P

h

( X

_k

X

_k ₁

) I

_f_τ _k_g

jF

k 1

ii

from ( EC3 ) ,

= E

^P

[ X

₀

] +

∑

b k=1

E

^P

h

I

_f_τ _k_g

E

^P

[ X

_k

X

_k ₁

jF

k 1

] ⁱ from ( EC 6 ) ,

= E

^P

[ X

0

]

since, X being a martingale,

E

^P

[ X _k X _k ₁ jF ^k ¹ ] = E

^P

[ X _k jF ^k ¹ ] E

^P

[ X _k ₁ jF ^k ¹ ]

= X _k ₁ X _k ₁ = 0.

(28)

Optional Stopping Theorem IV

Proof of the theorem

Second part. Let’s now assume that, for any bounded stopping time τ, E

^P

[ X

_τ

] = E

^P

[ X ₀ ] and let’s show that, in such a case, the adapted and integrable stochastic process is a martingale.

By hypothesis, X already satis…es conditions ( M1 ) and ( M 2 ) . The only thing left to verify is that 8 ^s ^, ^t 2 f ^0, ^1, ^2, ^{. . .} g ^such that s < t, E

^P

[ X _t jF ^s ] = X _s .

So, let’s set s and t 2 f ^0, ^1, ^2, ^{. . .} g ^{such that} ^s < t.

We denote by _P s = ⁿ A ⁽ ₁ ^s ⁾ , . . . , A ⁽ n ^s

s

⁾

o

the …nite

partition generated by F ^s ^.

(29)

Optional Stopping Theorem V

Proof of the theorem

For any i 2 f ^1, ^{. . .} ^, ⁿ ^s g we build a random time :

S _i ( ω ) = 8 >

<

> :

s if ω 2 ^A ⁽ i ^s ⁾

t if ω 2 / ^A i ⁽ ^s ⁾ . S i is a stopping time (obviously bounded) since 8 ^u 2 f ^0, ^1, ^2, ^{. . .} g ^,

f ^ω 2 Ω : S _i ( ω ) = u g = 8 >

> >

<

> >

> :

A ⁽ _i ^s ⁾ if u = s 2 F ^s A ⁽ _i ^s ⁾ ^c if u = t 2 F ^s F ^t

? ^otherwise 2 F ⁰ F ^u

.

(30)

Optional Stopping Theorem VI

Proof of the theorem

So, by hypothesis, we have that

E

^P

[ X _S

_i

] = E

^P

[ X ₀ ] .

Besides, since the random time τ t de…ned as 8 ^ω 2 Ω ,

τ t ( ω ) = t is also a bounded stopping time, we have, again by hypothesis, that

E

^P

[ X _t ] = E

^P

[ X

_τ_t

] = E

^P

[ X ₀ ] hence

E

^P

[ X _S

_i

] = E

^P

[ X _t ] .

(31)

Optional Stopping Theorem VII

Proof of the theorem

As a consequence, 8 ⁱ 2 f ^1, ^{. . .} ^, ⁿ ^s g ^, 0 = E

^P

[ X _t ] E

^P

[ X _S

_i

]

= E

^P

[ X _t X _S

_i

]

= E

^P

( X _t X _S

_i

) _I _A

(s) i

+ ( X _t X _S

_i

) _I

A

_i⁽^s⁾ ^c

= E

^P

( X t X s ) _I

A

_i⁽^s⁾

+ ( X t X t ) _I

A

⁽_i^s⁾ ^c

= E

^P

h

( X t X s ) I _A

⁽^s⁾

i

= ∑

ω

2 ^A

⁽i^s⁾

( X _t ( ω ) X _s ( ω )) P ( ω ) hence

∑

ω

2 ^A

⁽i^s⁾

X t ( ω ) _P ( ω ) = ∑

ω

2 ^A

⁽i^s⁾

X s ( ω ) _P ( ω ) .

(32)

Optional Stopping Theorem VIII

Proof of the theorem

Now we can conclude the proof, since E

^P

[ X _t jF ^s ] =

n

s

i ∑ = 1

I _A

⁽^s⁾

i

P A _i ⁽ ^s ⁾ ∑

ω

2 ^A

⁽i^s⁾

X _t ( ω ) P ( ω )

=

n

s

i ∑ = 1

I _A

(s) i

P A _i ⁽ ^s ⁾ ∑

ω

2 ^A

⁽i^s⁾

X _s ( ω ) _P ( ω )

= E

^P

[ X _s jF ^s ]

= X _s .

(33)

Markovian process I

De…nition

A stochastic process X = f ^X ^t ^: ^t 2 T g ^{, where} T is a set of indices ^a , is said to be markovian if, for any

t ₁ < t ₂ < . . . < t _n 2 T , the conditional distribution of X _t

_n

given X _t

₁

, . . . , X _t

_n ₁

is equal to the conditional distribution of X t

n

given X t

n 1

, i.e. for any x 1 , . . . , x n 2 R ,

P [ X _t

_n

x _n j ^X ^t

1

= x ₁ , . . . , X _t

_n ₁

= x _n ₁ ]

= P [ X _t

_n

x _n j ^X ^t

n 1

= x _n ₁ ] .

a

Examples: T = f ^0, ^1, ^2, ^{. . .} g ^, T = f ^0, ^1, ^2, ^{. . .} ^, ^T g ^where ^T ^{is a}

positive integer, T = [ 0, T ] where T is a positive real number, T = [ 0, ∞ ) ,

etc.

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Markovian process II

De…nition

Intuitively, if we assume that t are temporal indices, the process X is markovian if its distribution in the future, given the present and the past, only depends on the present.

”A Markov chain is then a memoryless random phenomenon: the distribution of an observation to come, given our present knowledge of the system and its whole history, is the same when only its present state is known.” ¹

The set of values that the process may take is called the state space of X and we denote it by E ^X ^.

1

Jean Vaillancourt.

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Markovian process I

Example

Example. We throw a dice repeatedly.

The random variable ξ _n represents the number of points obtained on the n th throw.

The stochastic process X represents the total cumulative number of points obtained at any time, i.e. for any natural integer t,

X _t =

∑ t n = 1

ξ _n .

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Markovian process II

Example

X is a Markovian process. Indeed, X _t =

∑ t n = 1

ξ _n =

t 1 n ∑ = 1

ξ _n + ξ _t = X _t ₁ + ξ _t .

But the outcome of the t th throw of dice, ξ _t , is independent from the results obtained on the …rst t 1 th throws,

σ f ^ξ n : n 2 f ^1, ^{. . .} ^, ^t ¹ gg . As a consequence, the distribution of X t depends on the past of the stochastic process,

σ f ^ξ n : n 2 f ^1, ^{. . .} ^, ^t ¹ gg ^{, through} ^σ f ^X ^t ¹ g ^only.

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Markovian process

Remark

Question. Why do we need a probability space? Wouldn’t a measurable space have been su¢ cient?

Answer. A probability measure is required to ensure that

the independence property is satis…ed.

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Markovian process

Random walk

The stochastic process X , built on the probability space ( Ω , F ^, ^P ) , is a random walk if it admits the representation

X 0 = 0 et 8 ^t 2 f ^1, ^2, ^{. . .} g ^, ^X ^t =

∑ t n = 1

ξ _n

where the sequence f ^ξ t : t 2 f ^1, ^2, ^{. . .} gg is made up of independent and identically distributed random variables.

Random walks are Markovian processes.

80-646-08 Stochastic Calculus I

Martingales