proof of hitting times are stopping times for right-continuous processes

proof of hitting times are stopping times for right-continuous processes

Let

(

ℱ

)

t

∈

𝕋

be a

filtration

(

http://planetmath.org/FiltrationOfSigmaAlgebras

) on the

measurable space

(

Ω

,

ℱ

)

, It is assumed that

𝕋

is a

closed subset

of

ℝ

and that

ℱ

t

is

universally complete

for each

t

∈

𝕋

.

Let

X

be a right-continuous and

adapted process

taking values in a

metric space

E

and

S

⊆

E

closed.
We show that

τ

=

inf

⁡

{

t

∈

𝕋

:

X

t

∈

S

}

is a

stopping time

. Assuming

S

is nonempty and defining the

continuous function

d

S

⁢

(

x

)

≡

inf

⁡

{

d

⁢

(

x

,

y

)

:

y

∈

S

}

, then

τ

is the first time at which the right-continuous process

Y

t

=

d

S

⁢

(

X

t

)

hits

0

.

Let us start by supposing that

𝕋

has a

minimum

element

t

0

.

If

ℙ

is a

probability measure

on

(

Ω

,

ℱ

)

and

ℱ

t

ℙ

represents the

completion

(

http://planetmath.org/CompleteMeasure

) of the

σ

-algebra

ℱ

t

with respect to

ℙ

, then it is enough to show that

τ

is an

(

ℱ

t

ℙ

)

-stopping time. By the

universal

completeness of

ℱ

t

it would then follow that

{

τ

≤

t

}

∈

⋂

ℙ

ℱ

t

ℙ

=

ℱ

t

for every

t

∈

𝕋

and, therefore, that

τ

is a stopping time.
So, by replacing

ℱ

t

by

ℱ

t

ℙ

if necessary, we may assume without loss of generality that

ℱ

t

is

complete

with respect to the probability measure

ℙ

for each

t

.

Let

𝒯

consist of the set of

measurable

times

σ

:

Ω

→

𝕋

∪

{

∞

}

such that

{

σ

<

t

}

∈

ℱ

t

for every

t

and that

σ

≤

τ

. Then let

σ

*

be the

essential supremum

of

𝒯

.
That is,

σ

*

is the smallest (up to sets of zero

probability

)

random variable

taking values in

ℝ

∪

{

±

∞

}

such that

σ

*

≥

σ

(almost surely) for all

σ

∈

𝒯

.

Then, by the properties of the essential supremum, there is a

countable

sequence

σ

n

∈

𝒯

such that

σ

*

=

sup

n

⁡

σ

n

. It follows that

σ

*

∈

𝒯

.

For any

n

=

1

,

2

,

…

set

σ

1

=

inf

⁡

{

t

∈

𝕋

:

t

≥

σ

*

,

Y

t

<

1

/

n

}

.

Clearly,

σ

1

≤

τ

and, choosing any countable

dense subset

A

of

𝕋

, the right-continuity of

Y

gives

{

σ

1

<

t

}

=

⋃

s

<

t

,

s

∈

A

{

σ

*

<

s

,

Y

s

<

1

/

n

}

∈

ℱ

t

.

So,

σ

1

∈

𝒯

, which implies that

σ

1

≤

σ

*

with probability one.
However, by the right-continuity of

Y

,

σ

1

>

σ

*

whenever

σ

*

is finite and

Y

σ

*

>

1

/

n

, so

ℙ

(

σ

*

<

∞

,

Y

σ

*

>

0

)

≤

∑

n

ℙ

(

σ

*

<

∞

,

Y

σ

*

>

1

/

n

)

=

0

.

This shows that

Y

σ

*

=

0

and therefore

σ

*

≥

τ

whenever

σ

*

<

∞

. So,

σ

*

=

τ

almost surely and

τ

∈

𝒯

giving,

{

τ

≤

t

}

=

{

τ

<

t

}

∪

{

Y

t

=

0

}

∈

ℱ

t

.

So,

τ

is a stopping time.

Finally, suppose that

𝕋

does not have a minimum element. Choosing a sequence

t

n

→

-

∞

in

𝕋

then the above

argument

shows that

τ

n

=

inf

⁡

{

t

∈

𝕋

:

t

≥

t

n

,

Y

t

=

0

}

are stopping times so,

{

τ

≤

t

}

=

⋃

n

{

τ

n

≤

t

}

∈

ℱ

t

as required.

Title

proof of hitting times are stopping times for right-continuous processes

Canonical name

ProofOfHittingTimesAreStoppingTimesForRightcontinuousProcesses

Date of creation

2013-03-22 18:39:12

Last modified on

2013-03-22 18:39:12

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

5

Author

gel (22282)

Entry type

Proof

Classification

msc 60G05

Classification

msc 60G40

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