Why the time inversion of a Brownian Motion a martingale

Question

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?

Answer 1

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*}