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Consider $\left( M^\alpha _t\right)_{t \geq 0} \in \mathcal M_c ^2$ such that

$$ \mathbb E \left\{ \sup_{0\leq s \leq t} \left |M^\alpha _s \right| ^2\right \} \leq C_\alpha t^{1-\alpha} $$

How to propelly show that for all $\beta > 1- \frac{\alpha}{2}$

$$ \sup_{0\leq s \leq t} \frac{\left |M^\alpha _s \right|}{t^\beta} \underset{t\rightarrow \infty}{ \longrightarrow } 0 \ a.e.$$

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Are you sure that it is not $\beta > 1 - \frac{\alpha}{2}$ ? –  Siméon Jan 18 '13 at 18:58
    
You are right! $\beta > 1 -\alpha / 2$ I'll fixe it in the text. –  Paul Jan 18 '13 at 19:26
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1 Answer 1

up vote 4 down vote accepted

We note $M^*_t = C_\alpha^{-1/2}\sup_{0 \leq s \leq t}|M^\alpha_s|$ so that the hypothesis writes $E\left[(M^*_t)^2\right] \leq t^{1-\alpha}$.

Fix $\epsilon > 0$. Chebyshev's inequality shows that, for all $n \geq 1$, $$ P\left(\frac{M_n^*}{n^\beta} \geq \epsilon\right) \leq \epsilon^{-2}\frac{1}{n^{2\beta-1+\alpha}}. $$

Suppose that $\beta > 1 - \frac{\alpha}{2}$. We have $2\beta - 1 + \alpha > 1$, hence Borel-Cantelli lemma ensures that $$ P\left(\frac{M_n^*}{n^\beta} \geq \epsilon\;;\;i.o\right) = 0. $$ In other words: $$ P\left(\limsup_{n\to\infty} \frac{M_n^*}{n^\beta} \leq \epsilon\right) = 1. $$

Taking a countable intersection, we first obtain a weak form a the result: $$ P\left(\limsup_{n\to\infty} \frac{M_n^*}{n^\beta} = 0\right) = P\left(\bigcap_{k\geq 1}\left\{\limsup_{n\to\infty} \frac{M_n^*}{n^\beta} \leq 1/k\right\}\right) = 1. $$

We finally show that the convergence $\lim_{n\to\infty} \dfrac{M_n^*}{n^\beta}=0$ naturally extends to a convergence for $t\to\infty$ due to monotonicity properties. Notice that the function $t \mapsto M_{t}^*$ is non-decreasing. Thus, if $n = [t]$ we have $$ \frac{M_n^*}{(n+1)^\beta} \leq \frac{M_t^*}{t^\beta} \leq \frac{M_{n+1}^*}{n^\beta} $$ and we conclude with the squeeze theorem (also know as « théorème des keufs »).

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Thank you for your answer. What do you mean by "OP" and "i.o."? –  Paul Jan 18 '13 at 19:32
    
@Paul: OP refers to the Original Poster and that would be you, and i.o. is short for infinitely often. –  Stefan Hansen Jan 18 '13 at 20:23
    
Now it is complete. –  Siméon Jan 18 '13 at 22:22
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