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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?

vonbrand
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user25070
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    By the law of the iterated logarithm, you should have $\tau_a<\infty$. – ShawnD Feb 17 '12 at 00:45
  • 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? – user25070 Feb 18 '12 at 12:39
  • Got something out of the answer? – Did Mar 08 '12 at 08:02

1 Answers1

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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.

Did
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