Ganesh Ujwal
New member
- Joined
- Aug 10, 2014
- Messages
- 32
Let \(\displaystyle {X(t):t \geq 0}\) be a Gaussian process with mean \(\displaystyle 0\) and bounded (with probability \(\displaystyle 1\)) sample paths. Borell's Theorem states then that for all \(\displaystyle u>0\) we have
\(\displaystyle P(\sup_{t \geq 0} X(t)>u) \leq 2 \Psi \left(\frac{u-m}{\sigma_T}\right)\),
where \(\displaystyle m\) is the median of \(\displaystyle \sup X(t)\), \(\displaystyle \sigma_T\) is the supremum of \(\displaystyle Var(X(t))\) and \(\displaystyle \Psi = 1 - \Phi\) 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) \(\displaystyle \sup X(t)\) has finite all moments, i.e. \(\displaystyle E(\sup X(t))^k\) exists and is finite \(\displaystyle \forall k \geq 1\).
\(\displaystyle P(\sup_{t \geq 0} X(t)>u) \leq 2 \Psi \left(\frac{u-m}{\sigma_T}\right)\),
where \(\displaystyle m\) is the median of \(\displaystyle \sup X(t)\), \(\displaystyle \sigma_T\) is the supremum of \(\displaystyle Var(X(t))\) and \(\displaystyle \Psi = 1 - \Phi\) 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) \(\displaystyle \sup X(t)\) has finite all moments, i.e. \(\displaystyle E(\sup X(t))^k\) exists and is finite \(\displaystyle \forall k \geq 1\).