Let $W$ be a standard Brownian motion on a probability space $(X, \mathcal F, \mathbb P)$ let and $\mathcal F\_t$ its natural filtration.


For $\varepsilon > 0, T \in [0, \infty)$ let $A\_{\varepsilon, t}$ be the event $\{\sup\_{s \in [0, T]} |W\_s|\ < \varepsilon\}$.


Let $X\_t$ be the solution to the SDE


$$dX\_t = \sigma(t, X\_t) dW\_t$$


where $\sigma$ is nice enough for a solution to exist.


For each $\varepsilon, T$ define the probability measure $\mathbb Q\_{\epsilon, T}$ by


$$\mathbb Q\_{\varepsilon, t} (E) = \frac{\mathbb P(E \cap A\_{\varepsilon, T})}{\mathbb P(A\_{\varepsilon, T)}}.$$



> 
> **Question:** Is $X$ a local $\mathcal F\_t$ martingale under $\mathbb Q\_{\varepsilon, T}$?
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**Remark:** We note that $\mathbb Q\_{\varepsilon, T}$ is *not* equivalent to $\mathbb P$, so that we cannot apply Girsanov’s theorem.



Is the integral against a Brownian motion conditioned to stay bounded a local martingale?
In case $\sigma = 1$ the claim is that the conditioned process itself is a martingale, however, as it paths are the paths of a brownian motion, it will have quadratic variation t, and therefore, it is an ordinary brownian motion.


