On Borell's Theorem (Gaussian processes)

Nobody1111 · **Joined:** Mon, 28 Dec 2009 16:23:32 UTC **Posts:** 10

Let ${X(t):t \geq 0}$ be a Gaussian process with mean

and bounded (with probability 1) sample paths. Borell's Theorem states then that for all u>0

we have
$P(sup_{t \geq 0} X(t)>u) \leq 2 \Psi (\frac{u-m}{\sigma_T})$ ,
where

is the median of supX(t)

, $\sigma_T$ is the supremum of VarX(t)

and $\Psi = 1 - \Phi$ is the tail of a standard normal distribution.
I need to show that under the assumptions of this theorem (we can use the theorem as well) supX(t)

has finite all moments, i.e. E(supX(t))^k

exists and is finite fot all $k \geq 1$ .

Thank you for any help.

outermeasure · **Posted:** Sun, 5 Jun 2011 07:01:43 UTC

Nobody1111 wrote:

Let ${X(t):t \geq 0}$ be a Gaussian process with mean

and bounded (with probability 1) sample paths. Borell's Theorem states then that for all u>0

we have
$P(sup_{t \geq 0} X(t)>u) \leq 2 \Psi (\frac{u-m}{\sigma_T})$ ,
where

is the median of supX(t)

, $\sigma_T$ is the supremum of VarX(t)

and $\Psi = 1 - \Phi$ is the tail of a standard normal distribution.
I need to show that under the assumptions of this theorem (we can use the theorem as well) supX(t)

has finite all moments, i.e. E(supX(t))^k

exists and is finite fot all $k \geq 1$ .

Thank you for any help.

You forgot a finiteness assumption of $m,\sigma_T$ , e.g. Brownian motion on $[0,\infty)$ .

Once you have finiteness, it is just $\Psi(\cdot)$ is rapidly decreasing.

Nobody1111 · **Joined:** Mon, 28 Dec 2009 16:23:32 UTC **Posts:** 10

Assuming that

and $\sigma_T$ are finite, I have still some problems with proving that E(supX(t))^k

is finite.
My first problem is how toexploit the knowledge about the tail distribution of supremum to calculate the

-th moment. I know the formula
$EX=\int_0 ^\infty (1-F(x))dx - \int_{-\infty} ^0 F(x)dx$ if

is the cdf of X. But is there any similar formula valid for higher moments?
The second problem is that the bound in Borell's theorem holds for u>0

and we don't know what happens if u<0

.
I have been trying to solve the problem without Borell's theorem but I gave it up.

outermeasure · **Posted:** Tue, 7 Jun 2011 11:43:30 UTC

Nobody1111 wrote:

Assuming that

and $\sigma_T$ are finite, I have still some problems with proving that E(supX(t))^k

is finite.
My first problem is how toexploit the knowledge about the tail distribution of supremum to calculate the

-th moment. I know the formula
$EX=\int_0 ^\infty (1-F(x))dx - \int_{-\infty} ^0 F(x)dx$ if

is the cdf of X. But is there any similar formula valid for higher moments?
The second problem is that the bound in Borell's theorem holds for u>0

and we don't know what happens if u<0

.
I have been trying to solve the problem without Borell's theorem but I gave it up.

Answer to your first question: $\mathbb{E}[Y^p:Y>0]=p\int_0^\infty t^{p-1}\mathbb{P}(\{Y>t\})\,\mathrm{d}t$ follows from a simple change of variable in Cavalieri's formula (or Tonelli's theorem).

Answer to your second question: Since the process has mean 0 (and assumed to have continuous paths?), you don't really have to worry about u<0 --- remember $\sup A=-\inf(-A)$ and reflect the process will give you an immediate bound.

Nobody1111 · **Joined:** Mon, 28 Dec 2009 16:23:32 UTC **Posts:** 10

Thank you very much. I have been thinking about your post and now everything is clear for me.
Cheers :-)

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