proof of Doob’s inequalities

proof of Doob’s inequalities

Let

(

Ω

,

ℱ

,

(

ℱ

t

)

t

∈

𝕋

,

ℙ

)

be a

filtered probability space

with countable index set

𝕋

.
If

(

X

t

)

t

∈

𝕋

is a

submartingale

, we show that

ℙ

(

sup

s

≤

t

X

s

≥

K

)

≤

K

-

1

𝔼

[

(

X

t

)

+

]

(1)

and if

X

is a

martingale

or nonnegative submartingale then,

ℙ

(

X

t

*

≥

K

)

≤

K

-

1

𝔼

[

|

X

t

|

]

,

(2)

∥

X

t

*

∥

p

≤

p

p

-

1

⁢

∥

X

t

∥

p

.

(3)

for every

K

>

0

and

p

>

1

.

First, let us consider the case where

𝕋

is finite. The first time at which

X

t

≥

K

,

τ

=

inf

⁡

{

t

∈

𝕋

:

X

t

≥

K

}

is a

stopping time

(as

hitting times are stopping times

). By Doob’s

optional

sampling theorem for submartingales

X

τ

∧

t

≤

𝔼

[

X

t

∣

ℱ

τ

∧

t

]

and therefore,

K

ℙ

(

τ

≤

t

)

≤

𝔼

[

1

{

τ

≤

t

}

X

τ

∧

t

]

≤

𝔼

[

1

{

τ

≤

t

}

X

t

]

However,

τ

≤

t

if and only if

sup

s

≤

t

⁡

X

s

≥

K

giving,

ℙ

(

sup

s

≤

t

X

s

≥

K

)

≤

K

-

1

𝔼

[

1

{

sup

s

≤

t

X

s

≥

K

}

X

t

]

,

(4)

where the supremum is understood to be over

s

∈

𝕋

.
Now suppose that

𝕋

is countable. Then choose finite subsets

𝕋

n

⊆

𝕋

which increase to

𝕋

as

n

goes to

infinity

. Replacing

𝕋

by

𝕋

n

in

inequality

(

4

) and using the

monotone convergence theorem

to take the limit

n

→

∞

extends (

4

) to arbitrary uncountable index sets. Then, inequality (

1

) follows immediately from (

4

).

Now, suppose that

X

is a martingale. Jensen’s inequality gives

𝔼

[

|

X

t

|

∣

ℱ

s

]

≥

|

𝔼

[

X

t

∣

ℱ

s

]

|

=

|

X

s

|

for any

s

<

t

, so

|

X

|

is a nonnegative submartingale. Therefore, it is enough to prove inequalities (

2

) and (

3

) for

X

a nonnegative submartingale, and the martingale case follows by replacing

X

by

|

X

|

.

So, we take

X

to be a nonnegative submartingale in the following. In this case, (

2

) just reduces to (

1

) and it only remains to prove inequality (

3

).

For

p

>

1

, multiply (

4

) by

K

p

-

1

and integrate up to some limit

L

>

0

,

∫

0

L

K

p

-

1

ℙ

(

X

t

*

≥

K

)

d

K

≤

∫

0

L

K

p

-

2

𝔼

[

1

{

X

t

*

≥

K

}

X

t

]

d

K

.

(5)

The

left hand side

of this inequality can be computed by commuting the order of integration with respect to

ℙ

and

d

⁢

K

(Fubini’s theorem),

∫

0

L

K

p

-

1

ℙ

(

X

t

*

≥

K

)

d

K

=

𝔼

⁢

[

∫

0

L

K

p

-

1

⁢

1

{

X

t

*

≥

K

}

⁢

𝑑

K

]

=

1

p

⁢

𝔼

⁢

[

(

L

∧

X

*

)

p

]

.

The right hand side of (

5

) can be computed similarly,

∫

0

L

K

p

-

2

⁢

𝔼

⁢

[

1

{

X

t

*

≥

K

}

⁢

X

t

]

⁢

𝑑

K

=

𝔼

⁢

[

X

t

⁢

∫

0

L

K

p

-

2

⁢

1

{

X

t

*

≥

K

}

⁢

𝑑

K

]

=

1

p

-

1

⁢

𝔼

⁢

[

X

t

⁢

(

L

∧

X

t

*

)

p

-

1

]

.

Putting these back into (

5

),

∥

L

∧

X

t

*

∥

p

p

≤

p

p

-

1

⁢

𝔼

⁢

[

X

t

⁢

(

L

∧

X

t

*

)

p

-

1

]

.

(6)

Now let

q

=

p

/

(

p

-

1

)

, so that

p

,

q

are

conjugate

(

http://planetmath.org/ConjugateIndex

) and the Hölder inequality gives

𝔼

⁢

[

X

t

⁢

(

L

∧

X

t

*

)

p

-

1

]

≤

∥

X

t

∥

p

⁢

∥

(

L

∧

X

t

*

)

p

-

1

∥

q

=

∥

X

t

∥

p

⁢

∥

L

∧

X

t

*

∥

p

p

-

1

.

Substituting into (

6

), the finite term

∥

L

∧

X

t

*

∥

p

p

-

1

cancels to get

∥

L

∧

X

t

*

∥

p

≤

p

p

-

1

⁢

∥

X

t

∥

p

,

and the result follows by letting

L

increase to infinity.

Title

proof of Doob’s inequalities

Canonical name

ProofOfDoobsInequalities

Date of creation

2013-03-22 18:39:55

Last modified on

2013-03-22 18:39:55

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

5

Author

gel (22282)

Entry type

Proof

Classification

msc 60G46

Classification

msc 60G44

Classification

msc 60G42

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