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.


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

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

it should read

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


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

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