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Given a filtered probability space $\left ( \mathbb P, \mathcal F, \left(\mathcal F_t \right)_{t \geq 0}, \Omega\right)$, consider a process $\phi = \left( \phi_t \right)_{t\geq 0}$ $\mathcal F_t$- progressively measurable such that $ \mathbb E \left\{ \int _0 ^t \left |\phi_s \right |^2 ds\right\}< \infty$ for all $t \geq 0$. I would like to show that the stopping time sequence $\left(\tau_p \right)_{p\geq0}$ defined by

$$\tau_p := \inf \{ t\geq 0: \left |\phi_s \right | \geq p \} $$

is such that $\tau_p \nearrow \infty $ $ \mathbb P$- a.e as $ p \rightarrow \infty$

Edit: As "pgassiat" notice, the definition of $\tau_p$ is mistaked. But, I believe it can be changed in order to work. Could you help me on it please ?

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Something's missing. As written, take e.g. $\phi_s=s^{-1/4}$ deterministic, then $\tau_p=0$ for all $p$. –  pgassiat Jan 15 '13 at 20:10

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up vote 0 down vote accepted

I belive I've found the solution for my own question. Changing the definition of $\tau_p$ for

$$\tau_p := \inf \{ t\geq 0: \int_0 ^t \left |\phi_s \right |^2 ds\geq p \} $$

and supposing by contratiction there is a constant $M >0$ such that $$\mathbb P \left\{ \tau_p \leq M, \forall p \geq 0 \right\} >0 \ , $$ we have for all $p \geq 0$

\begin{align}\mathbb E \left \{ \int_0 ^M \left| \phi _s\right|^2 ds\right\} &= \mathbb E \left \{ \int_0 ^M \left| \phi _s\right|^2 ds \mathbf{1}_{\{\tau_p \leq M\} }+\int_0 ^M \left| \phi _s\right|^2 ds \mathbf{1}_{\{\tau_p > M\} }\right\}\\&\geq \mathbb E \left \{ \int_0 ^M \left| \phi _s\right|^2 ds \mathbf{1}_{\{\tau_p \leq M\} } \right\}\\&\geq \mathbb E \left \{ \int_0 ^{\tau_p} \left| \phi _s\right|^2 ds \mathbf{1}_{\{\tau_p \leq M\} } \right\}\\&\geq p \ \mathbb P \left\{ \tau_p \leq M, \forall p \geq 0 \right\} \underset {p\rightarrow +\infty}{\longrightarrow } + \infty\end{align}

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Why the last inequality? –  Did Jan 15 '13 at 22:06
    
My mistake, did. I've put $p$ where it was $M$ almost everywhere (not in the mathematical sense! =) ). Please, see the edited version and say me if you agree now. –  Paul Jan 15 '13 at 22:20

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