proof of Prohorov inequality

proof of Prohorov inequality

Starting from the basic inequality

exp

⁡

(

-

x

)

≥

1

-

x

, it’s
easy to derive by elementary algebraic manipulations the two inequalities

exp

⁡

(

x

)

-

x

-

1

≤

2

⁢

(

cosh

⁡

(

x

)

-

1

)

2

⁢

(

cosh

⁡

(

x

)

-

1

)

≤

x

⁢

sinh

⁡

(

x

)

By the Chernoff-Cramèr bound (

http://planetmath.org/ChernoffCramerBound

), we have:

Pr

⁡

{

∑

i

=

1

n

(

X

i

-

E

⁢

[

X

i

]

)

>

ε

}

≤

exp

⁡

[

-

sup

t

>

0

⁡

(

t

⁢

ε

-

ψ

⁢

(

t

)

)

]

where

ψ

⁢

(

t

)

=

∑

i

=

1

n

(

ln

⁡

E

⁢

[

e

t

⁢

X

i

]

-

t

⁢

E

⁢

[

X

i

]

)

Keeping in mind that the condition

Pr

⁡

{

|

X

i

|

≤

M

}

=

1

⁢

⁢

∀

i

implies that, for all

i

,

E

⁢

[

|

X

i

|

k

]

≤

M

k

⁢

⁢

∀

k

≥

0

(see here (

http://planetmath.org/RelationBetweenAlmostSurelyAbsolutelyBoundedRandomVariablesAndTheirAbsoluteMoments

) for a proof) and since

ln

⁡

x

≤

x

-

1

∀

x

>

0

, and

E

⁢

[

|

X

|

k

]

≤

M

k

⁢

⟹

⁢

E

⁢

[

|

X

|

k

]

≤

E

⁢

[

X

2

]

⁢

M

k

-

2

⁢

⁢

∀

k

≥

2

,

k

∈

N

(see here (

http://planetmath.org/AbsoluteMomentsBoundingNecessaryAndSufficientCondition

) for a proof), one has:

ψ

⁢

(

t

)

=

∑

i

=

1

n

(

ln

⁡

E

⁢

[

e

t

⁢

X

i

]

-

t

⁢

E

⁢

[

X

i

]

)

≤

∑

i

=

1

n

E

⁢

[

e

t

⁢

X

i

]

-

t

⁢

E

⁢

[

X

i

]

-

1

=

∑

i

=

1

n

E

⁢

[

e

t

⁢

X

i

-

t

⁢

X

i

-

1

]

≤

∑

i

=

1

n

2

⁢

E

⁢

[

cosh

⁡

(

t

⁢

X

i

)

-

1

]

≤

∑

i

=

1

n

E

⁢

[

t

⁢

X

i

⁢

sinh

⁡

(

t

⁢

X

i

)

]

≤

∑

i

=

1

n

E

⁢

[

|

t

⁢

X

i

⁢

sinh

⁡

(

t

⁢

X

i

)

|

]

=

∑

i

=

1

n

t

⁢

E

⁢

[

|

X

i

|

⁢

sinh

⁡

(

t

⁢

|

X

i

|

)

]

=

∑

i

=

1

n

t

⁢

E

⁢

[

∑

k

=

0

∞

t

2

⁢

k

+

1

⁢

|

X

i

|

2

⁢

k

+

2

(

2

⁢

k

+

1

)

!

]

=

∑

i

=

1

n

t

⁢

∑

k

=

0

∞

t

2

⁢

k

+

1

⁢

E

⁢

[

|

X

i

|

2

⁢

k

+

2

]

(

2

⁢

k

+

1

)

!

≤

∑

i

=

1

n

t

⁢

∑

k

=

0

∞

t

2

⁢

k

+

1

⁢

E

⁢

[

X

i

2

]

⁢

M

2

⁢

k

(

2

⁢

k

+

1

)

!

=

t

M

⁢

∑

k

=

0

∞

t

2

⁢

k

+

1

⁢

M

2

⁢

k

+

1

⁢

∑

i

=

1

n

E

⁢

[

X

i

2

]

(

2

⁢

k

+

1

)

!

=

t

⁢

v

2

M

⁢

∑

k

=

0

∞

(

t

⁢

M

)

2

⁢

k

+

1

(

2

⁢

k

+

1

)

!

=

t

⁢

v

2

M

⁢

sinh

⁡

(

t

⁢

M

)

.

One can now write

sup

t

>

0

⁡

(

t

⁢

ε

-

ψ

⁢

(

t

)

)

≥

sup

t

>

0

⁡

(

t

⁢

ε

-

t

⁢

v

2

M

⁢

sinh

⁡

(

t

⁢

M

)

)

=

sup

t

>

0

⁡

[

v

2

M

2

⁢

(

M

2

⁢

ε

v

2

⁢

t

-

t

⁢

M

⁢

sinh

⁡

(

t

⁢

M

)

)

]

Optimizing this

expression

with respect to

t

would lead to solving the

transcendental equation

:

M

⁢

ε

v

2

=

M

⁢

t

o

⁢

p

⁢

t

⁢

cosh

⁡

(

M

⁢

t

o

⁢

p

⁢

t

)

+

sinh

⁡

(

M

⁢

t

o

⁢

p

⁢

t

)

which is analytically infeasible. So, one can choose the sup-optimal yet
manageable solution

t

~

=

1

M

⁢

arsinh

⁡

(

M

⁢

ε

2

⁢

v

2

)

which, once plugged into the bound, yields

Pr

⁡

{

∑

i

=

1

n

(

X

i

-

E

⁢

[

X

i

]

)

>

ε

}

≤

exp

⁡

[

-

v

2

M

2

⁢

(

M

⁢

ε

2

⁢

v

2

⁢

arsinh

⁡

(

M

⁢

ε

2

⁢

v

2

)

)

]

Title

proof of Prohorov inequality

Canonical

name

ProofOfProhorovInequality

Date of creation

2013-03-22 16:12:58

Last modified on

2013-03-22 16:12:58

Owner

Andrea Ambrosio (7332)

Last modified by

Andrea Ambrosio (7332)

Numerical id

14

Author

Andrea Ambrosio (7332)

Entry type

Proof

Classification

msc 60E15

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