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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 ?

Dovahkiin
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    What is wrong with relying on Lemma 16 from the ref. you shared? – Math-fun Jan 23 '24 at 17:51
  • @Math-fun In proving Lemma 16, the fact that $(\sqrt{\tau}\cdot Y_{t/\tau})_{t\in [0,1]}$ is a Standard Brownian Motion is used, where $Y$ is a Standard Brownian Motion also. – Dovahkiin Jan 23 '24 at 18:07
  • right, but then "measurability" is assumed which is essential for the validity of the arguments. – Math-fun Jan 23 '24 at 18:13
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    Lemma 16 assumes that for every $t>0$, the event $[\tau \ge t]$ depends only on $(X_s)_{s \ge t}$, and the prood uses heavily the independence of $Y$ and $\tau$, which follows from this assumption. – Christophe Leuridan Jan 28 '24 at 13:19

1 Answers1

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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.

Kroki
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  • Unless I'm mistaken, this proof uses the independence lemma to compute the conditional expectation, which would require that $\tau$ is independent of $Y$. Do you get the same result if $\tau$ is not independent of $Y$? – user6247850 Jan 23 '24 at 18:44
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    Sorry, it requires the measurability of what? Regardless, could you please elaborate on the line $\mathbb{E}[e^{\sqrt{\tau} u (Y_{t/\tau} - Y_{s/\tau})}|\tau] = e^{-\frac 12 \tau u^2 \times \left(\frac t\tau - \frac s\tau \right)}$? Or am I misunderstanding that proof, and the claim is only that they are equal in expectation? – user6247850 Jan 23 '24 at 19:14
  • Ah ofcourse. I had conditioned on $\tau$ but forgot that in the expression for the mgf or characteristic function, we would get $\sqrt{\tau}^{2}$ and it nicely cancels out. Thanks – Dovahkiin Jan 23 '24 at 21:06
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    Sorry for the late response but akin to user6247850's question, how are you saying $\mathbb{E}[e^{\sqrt{\tau} u (Y_{t/\tau} - Y_{s/\tau})}|\tau] = e^{-\frac 12 \tau u^2 \times \left(\frac t\tau - \frac s\tau \right)}$ ? And how does the measurability of $\tau$ help here? For all I know, $\tau$ is such that ${\tau\geq t}$ is measurable wrt the tail $\sigma(X_{s}:s\geq t)$. @Kroki – Dovahkiin Jan 28 '24 at 12:37
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    The independence of $Y$ and $\tau$ is actually important and used in the proof of Lemma 16. It follows from the assumption (forgotten in this post) that for every $t>0$, the event $[\tau \ge t]$ depends only on $(X_s)_{s \ge t}$. – Christophe Leuridan Jan 28 '24 at 13:22
  • As @user6247850 mentioned the independence of $Y$ and $\tau$ is crucial to this proof. The independence of them is assured by the result on $\tau$. I will update my answer to include that. – Kroki Jan 30 '24 at 23:44
  • @Dovahkiin I edited my answer you can look at it now and you will have proof of what you are looking for. – Kroki Jan 31 '24 at 00:48
  • @Kroki What is wrong with just using $W_{\tau t}/\sqrt{\tau}$ for a stopping time $\tau$ and then repeating your proof. I think this will prove what I am trying to show as I want to show that for $1/\tau$ (as in question), which is a stopping time, we have $W_{\tau t}/\sqrt{\tau}$ is an SBM where $W$ is an SBM . The proof you are using is by following how Strong Markov Property is proved. Also I don't think the scaling $\tau(1+t)$ really helps me as we do have at $t=0$ that we get $\tau$ but at $t=1$, we get $2\tau$ which seems irrelevant to me. – Dovahkiin Jan 31 '24 at 08:44
  • @Kroki The above is providing that $W$ is independent of $\mathcal{F}{\tau}$. i.e. in the post, they first consider $Y{t}=\bar{X_{t+\overline{\tau}}}-\bar{X_\overline{{\tau}}}$ which is independent of $\mathcal{F}{\tau}$ and then they have said that $\sqrt{\tau}Y{t/\tau}$ is an SBM. Thereafter, can I directly use your proof to conclude that $\sqrt{\tau}Y_{t/\tau}=W_{t\sqrt{\bar{\tau}}}/\sqrt{\bar{\tau}}$ is a Brownian motion? – Dovahkiin Jan 31 '24 at 09:12
  • I edited the answer to prove the general result. – Kroki Jan 31 '24 at 18:15
  • What is $\mathcal F_\tau$? Just $\sigma$-algebra generated by $\tau$? 2. How does this answer relate to the question?
    • – zhoraster Feb 07 '24 at 05:20
  • Yes $\mathcal F_\tau$ is any $\sigma-$algebra that $\tau$ is mesurable inside it. 2. It is related to the question perfectly. – Kroki Feb 07 '24 at 12:58
  • No it's not. $W_t$ does not – zhoraster Feb 09 '24 at 15:46
  • I am not sure I understand your comment. Can you please explain? – Kroki Feb 09 '24 at 16:09