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$B$ is a one-dimensional Brownian motion and $X_t$ is defined as$\\$ $X_t:=f_{1-t}(B_t)$, $0\le t<1$ and $0$, $1\le t<\infty$ where $f_s(x)=\frac{1}{\sqrt{2\pi s}}e^{-\frac{x^2}{2s}}$. I have to show that $X$ is a continuous local martingale.

$f$ is continuous as a combination of continuous functions. I tried to use Ito's Formula, but i cannot see how it helps me. I also tried to show that it is a martingale by using stopping times. I would be thankful for any help.

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Ito's lemma is one way to prove it. Using the definition of $f_{1-t}$, we have $$X_t = f_{1-t}(B_t) = \frac{1}{\sqrt{2\pi (1-t)}} \exp\left\{\frac{-B_t^2}{2(1-t)}\right \}:=F(t,B_t),$$ and according to Ito's lemma: $$dX_t = dF(t,B_t) = \left(\frac{\partial F}{\partial t}+\frac{1}{2}\frac{\partial^2 F}{\partial B_t^2}\right)dt+ \frac{\partial F}{\partial B_t}dW_t.$$ Computing the partial derivatives: \begin{align} \frac{\partial F}{\partial t} &= \frac{\exp\left\{\frac{-B_t^2}{2(1-t)}\right\}}{2\sqrt{2\pi(1-t)}}\left(\frac{1}{1-t}-\frac{B_t^2}{(1-t)^2}\right) \\ \frac{\partial F}{\partial B_t} &= \frac{\exp\left\{\frac{-B_t^2}{2(1-t)}\right\}}{\sqrt{2\pi(1-t)}} \left(\frac{-B_t}{1-t}\right) \\ \frac{\partial^2 F}{\partial B_t^2} &= \frac{\exp\left\{\frac{-B_t^2}{2(1-t)}\right\}}{\sqrt{2\pi(1-t)}} \left(\frac{-1}{1-t}+\frac{B_t^2}{(1-t)^2}\right). \end{align} It follows that $$\frac{\partial F}{\partial t}+\frac{1}{2}\frac{\partial^2 F}{\partial B_t} = 0.$$ Hence, we obtain $$dX_t = -\frac{B_t}{1-t}X_tdW_t,$$ which implies that $X_t$ is a (continuous) local martingale ($X_t$ is a stochastic integral w.r.t Brownian motion).

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