# Why the time inversion of a Brownian Motion a martingale

Given $$X(t) = t W(1/t)$$, I want to show that $$X(t)$$ is a Brownian Motion. Specifically, I want to show it using Levy's Characterization of Brownian Motion. Levy's Characterization of Brownian Motion states that if a process is almost surely a continuous martingale with $$X(0) = 0$$ and quadratic variation $$t$$, then the process is a Brownian Motion. I know how to show the continuity of $$X(t)$$ and that the quadratic variation is equal to $$t$$, but it is not clear to me why this process is a martingale. In my attempt to show this I have done the following:

\begin{align} E(X(t) | \mathcal{F}_s) &= E(t W(1/t) | \mathcal{F}_s)\\\\ &= tE(W(1/t) | \mathcal{F}_s)\\\\ &= t W(1/s). \end{align} My understanding is that the martingale property states that $$E(X(t) | \mathcal{F}_s) = X(s)$$, so, in this case, we would need $$E(tW(1/t) | \mathcal{F}_s) = sW(1/s)$$ which we do not have. Is this correct or am I mistaken?

• What is $\mathcal F_s$ here? Apr 21 at 23:46
• $\mathcal{F}_s$ is the filtration of sigma algebras defined as $\mathcal{F}_t = \sigma(sW(1/s) | s < t)$ Apr 24 at 14:32
• Okay, so why is $E[W(1/t)|\mathcal F_s] = W(1/s)$? Apr 24 at 14:35
• I think originally I may have defined $\mathcal{F}$ as $\sigma(W(1/s) | s<t)$. If I define the filtration as I did in my previous comment, would it follow that the process is a martingale? Apr 24 at 14:40
• I think those filtrations are the same - scaling the random variables $W(1/s)$ doesn't really affect the information they give. The property you seem to be trying to use is that $E[W(t)|W(s)] = W(s)$ when $s < t$, but here you have something that looks more like $E[W(1/t)|W(1/s)]$ when $s < t$. Apr 24 at 15:13

We need to compute $$E[W(1/t)|W(1/s)]$$ for $$s < t$$, or equivalently we need to compute $$E[W(t)|W(s)]$$ for $$t < s$$. We make a guess that $$E[W(t)|W(s)] = \beta W(s)$$ for some constant (possibly depending on $$t$$ and $$s$$) $$\beta \in \mathbb{R}$$. Now, note that we must have $$0 = E[W(s)(W(t)-E[W(t)|W(s)])] = E[W(s)(W(t)-\beta W(s))],$$ so solving for $$\beta$$ we have $$\beta = \frac{E[W(s)W(t)]}{E[W(s)^2]} = \frac ts$$. Furthermore, since $$W(s)$$ and $$W(t)-\beta W(s)$$ are jointly normal, the fact that they are uncorrelated implies that they are independent, so we confirm that our guess $$E[W(t)|W(s)] = \beta W(s)$$ is correct.
Applying this to the problem we had originally, $$E[W(1/t)|W(1/s)] = \frac{1/t}{1/s} W(1/s) = \frac st W(1/s)$$, so \begin{align*} E[X(t)|\mathcal F_s] &= t E[W(1/t)|W(1/s)] \\ &= t \left( \frac st W(1/s) \right) \\ &= s W(1/s) = X(s). \end{align*}