Let $W_t$ be a Wiener process and for $a\geq0$

$$\tau_a:=\inf \left\{ t\geq0: |W_t|=\sqrt{at+7} \right\}.$$

Is $\tau_a<\infty$ almost everywhere? What about $E(\tau_a)$ then?

  • 3
    $\begingroup$ By the law of the iterated logarithm, you should have $\tau_a<\infty$. $\endgroup$ – ShawnD Feb 17 '12 at 0:45
  • $\begingroup$ So I could use ${\limsup_{t\rightarrow \infty} \frac{W_t}{\sqrt{2t\cdot ln(ln(t))}}=1}$ almost surely. But still it is $\sqrt{at+7} \geq \sqrt{2tln(ln(t))}$, for $a>2$, isn't it? $\endgroup$ – user25070 Feb 18 '12 at 12:39
  • $\begingroup$ Got something out of the answer? $\endgroup$ – Did Mar 8 '12 at 8:02

As explained by Shawn, the law of the iterated logarithm shows that $\tau_a$ is almost surely finite.

Since $\tau_a=\inf\{t\geqslant0\mid W_t^2=at+7\}$ and $\tau_a$ is almost surely finite, $W_{\tau_a}^2=a\tau_a+7$. Likewise, for every $t\geqslant0$, $\tau_a\wedge t\leqslant\tau_a$ almost surely hence $W_{\tau_a\wedge t}^2\leqslant a(\tau_a\wedge t)+7$ almost surely and in particular $\mathrm E(W_{\tau_a\wedge t}^2)\leqslant a\mathrm E(\tau_a\wedge t)+7$. On the other hand, $(W_t^2-t)_{t\geqslant0}$ is a martingale hence $\mathrm E(W_{\tau_a\wedge t}^2)=\mathrm E(\tau_a\wedge t)$ for every $t\geqslant0$. One sees that $(1-a)\mathrm E(\tau_a\wedge t)\leqslant7$.

If $a\lt1$, one gets $(1-a)\mathrm E(\tau_a)\leqslant7$ hence $\tau_a$ is integrable. Since $W_{\tau_a}^2=a\tau_a+7$, $W_{\tau_a}^2$ is integrable as well and $\mathrm E(W_{\tau_a}^2)=a\mathrm E(\tau_a)+7=\mathrm E(\tau_a)$, which shows that $$ \mathrm E(\tau_a)=\frac{7}{1-a}. $$ If $a\geqslant1$, $\tau_a\geqslant\tau_b$ for every $b\lt1$, hence $\mathrm E(\tau_a)\geqslant\mathrm E(\tau_b)=7/(1-b)$ for every $b\lt1$, in particular $\mathrm E(\tau_a)$ is infinite.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.