I was wondering with respect to what filtration the process $X_t=\begin{cases}\frac{1}{t}B_{1/t}\text{ if }t>0 \\ 0\text{ if }t=0
\end{cases}$ $\;\;\;$($B_t$ is a Brownian motion) is a martingale. Is it with respect to $\mathcal{F_{1/t}}$ where $\mathcal{F}_t$ is the natural filtration of the Brownian motion? Is it what we call a reverse martingale?
Filtration with respect to which reverse brownian motion is martingale
brownian motionfiltrationsprobability theory
Best Answer
$X$ is not a martingale with respect to any filtration. If $X$ is a martingale, then $X^2$ is a sub-martingale, and in particular $t \mapsto \mathbb{E}[X_t^2]$ is increasing. For this $X$, we have $$\mathbb{E}[X_t^2] = \frac 1{t^2} \mathbb{E}[B_{1/t}^2] = \frac 1{t^3},$$ which is decreasing. Hence $X^2$ cannot be a sub-martingale, and therefore $X$ cannot be a martingale with respect to any filtration.