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How to show that for all $t\geq 0$ $$ \int _0^t \frac{\left|B_u \right|}{u}du < \infty \ a.e.,$$ where $ \left( B_t \right)_{t\geq 0}$ is the real standard brownian motion starting from zero ?

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Using Fubini-Tonelli theorem and Cauchy-Schwarz's inequality, we have $$ E\left[\int_0^t \frac{|B_u|}{u}\,du\right] = \int_0^t\frac{E(|B_u|)}{u}\,du \leq \int_0^t \frac{E(B_u^2)^{1/2}}{u}\,du = \int_0^t \frac{du}{\sqrt{u}} =2\sqrt{t} <\infty. $$

Edit: As noted by did, the expectation can be computed exactly:

$$ E\left[\int_0^t \frac{|B_u|}{u}\,du\right] = \sqrt{\frac{8t}{\pi}}. $$

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    $\begingroup$ By scaling, $E(|B_u|)=\sqrt{u}E(|B_1|)$ hence Cauchy-Schwarz is not required. $\endgroup$ – Did Jan 16 '13 at 13:08
  • $\begingroup$ You're right of course, but it gives the bound $E(|B_1|) \leq 1$. $\endgroup$ – Siméon Jan 16 '13 at 13:24
  • $\begingroup$ Which is perfect, no? $\endgroup$ – Stefan Hansen Jan 16 '13 at 13:25
  • $\begingroup$ @StefanHansen: It's not perfect since $E(|B_1|) = \sqrt{2/\pi} < 1$. $\endgroup$ – Siméon Jan 16 '13 at 13:34
  • $\begingroup$ Which changes nothing to show the wanted result. Enven if it's refined estimation. Thank you, everybody! $\endgroup$ – Paul Jan 16 '13 at 13:38
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Hint: Show that $$ E\left[\int_0^t \frac{|B_u|}{u}\,\mathrm du\right]<\infty $$ by the use of Tonelli's theorem.

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