0
$\begingroup$

Define $\tau_a = \inf \left\lbrace t \geq 0 | B(t) \geq a \right\rbrace $ for some $a>0$. The problem is to show that $ \tau_a \stackrel{d}{=} \sqrt a\tau_1 $. What I've done so far:

$$P(\tau_a \leq t) = 2P(B(t) \geq a) = 2P \left(Z \geq \frac{a}{\sqrt t}\right)$$

The first step is from the reflection principle, the second is just normalizing the Brownian Motion. But now doing something similar:

$$P(\sqrt a\tau_1 \leq t) = 2P\left(B \left(\frac{t}{\sqrt a} \right) \geq 1 \right) = 2P \left(Z \geq \left(\frac{\sqrt a}{t}\right)^{\frac{1}{2}} \right)$$

Clearly these are different so I'm not sure where the mistake I'm making is coming in. I've already calculated a density function for $\tau_a$ and showed that $\tau_a$ has stationary, independent increments, but I can't see where that fact would be helpful, if anywhere.

$\endgroup$
1
$\begingroup$

Your calculations are correct, but the claim is not. Instead of

$\tau_a \stackrel{d}{=} \sqrt{a} \tau_1$

it should read

$\tau_a \stackrel{d}{=} a^2 \tau_1$.

References:

  • Revuz/Yor: Continuous martingales and Brownian motion, Proposition III.3.10
  • Schilling/Partzsch: Brownian motion - An introduction to stochastic processes, Problem 6.6
$\endgroup$
  • $\begingroup$ Thank you for this. I will bring this up to the professor $\endgroup$ – Brenton May 17 '15 at 20:51

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.