# Brownian Motion Hitting Time Distribution

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.

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