Can a Brownian Motion be scaled by a random time still be a brownian motion?

Question

I was reading the proof that if $\tau=\sup\{s\in [0,1]:B_{s}=0\}\wedge 1$. Then $\bigg(\frac{1}{\sqrt{\tau}}W_{\tau t}\bigg)_{t\in [0,1]}$ is a Standard Brownian Bridge in $[0,1]$ from here.

In the proof (Lemma 15 and Lemma 16), they are using that for a standard Brownian Motion $Y_{t}$ , we have that $(\sqrt{\tau}\cdot Y_{t/\tau})_{t\in [0,1]}$ is again a Standard Brownian Motion.

However, I have not encountered such a scaling by a Random Time before. $\tau$ is also not a stopping time.

So my question is whether $(\sqrt{\tau}\cdot Y_{t/\tau})_{t\in [0,1]}$ is a Brownian Motion or not and if the above is indeed true, then how should I go about proving this?

Firstly, I don't see how this is a Gaussian process. To be honest, I don't even see even for a fixed $t$, how is $\sqrt{\tau} Y_{t/\tau}$ has normal distribution. So I cannot go about computing the covariances. However, I can easily see that conditioned on $\tau$, we have that it is a Brownian Motion by usual scaling invariance.

Can anyone provide a proof or a reference or more details for this ?

Answer 1

You want to prove the following result:

$\left(W_t\right)_{t\ge 0}$ is a standard Brownian motion independent from the $\sigma-$algebra $\mathcal F_{\tau}$ where $\tau$ is a random nonnegative variable, then $\left(\frac1{\sqrt{\tau}}W_{t\tau}\right)_{t\ge 0}$ is a standard Brownian motion.

Let $\tau_n = 2^{-n} \left(\left\lfloor 2^{n}\tau\right\rfloor + 1\right)$. It is clear that $\left[\tau_n = k2^{-n}\right]\in \mathcal F_{\tau}$ and

\begin{align} \mathbb E\left[e^{iu \frac1{\sqrt{\tau_n}} W_{\tau_n t}}\right] &= \sum_{k=1}^{\infty} \mathbb E\left[e^{iu\frac1{\sqrt{\tau_n}} W_{\tau_n t}}\mathbf 1_{\left\{\tau_n=k2^{-n}\right\}}\right] \\ &= \sum_{k=1}^{\infty} \mathbb E\left[\mathbb E\left[e^{iu\frac1{\sqrt{\tau_n}} W_{\tau_n t}}\mathbf 1_{\left\{\tau_n=k2^{-n}\right\}}\Big | \mathcal F_{\tau}\right]\right] \\ &= \sum_{k=1}^{\infty} \mathbb E\left[\mathbb E\left[e^{iu\frac1{\sqrt{k2^{-n}}} W_{k2^{-n}t}}\Big | \mathcal F_{\tau}\right]\mathbf 1_{\left\{\tau_n=k2^{-n}\right\}}\right] \\ &= \sum_{k=1}^{\infty} e^{-\frac12tu^2} \mathbb P\left[\tau_n = k2^{-n}\right]\\ &= e^{-\frac12tu^2}. \end{align}

Since $\tau_n\to \tau$ a.s. then you have the result you are looking for.