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Let $(B_t)_{t\geq 0}$ be a standard Brownian motion. Let $\phi: [0,1)\to [0,\infty)$ be defined by $ \phi(t):=t/(1-t)$. Then $(M_t)_{0\le t<1}$ is a continuous Markov martingale with $M_t:=B_{\phi(t)}$. Do we have its martingale representation? Namely, there exist measurable function $\sigma: [0,1)\times \mathbb R\to \mathbb R_+$ and some Brownian motion $W$ (w.r.t. its natural filtration) s.t.

$$M_t=\int_0^t \sigma(s,M_s)dW_t,\quad \forall t\in [0,1).$$

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  • $\begingroup$ The quadratic variation of $M_t$ is $\phi(t)$, so $dW_t = (\phi(t))^{-1/2} dM_t$ is the Brownian motion you are looking for. $\endgroup$ Commented Oct 21, 2021 at 18:21
  • $\begingroup$ Or rather, $W_t=\int_0^t (1-s) dM_s$. The $(1-s)$ is $[\phi'(s)]^{-1/2}$. $\endgroup$ Commented Oct 23, 2021 at 15:28

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Indeed as mentioned in the comments and in the blog here https://almostsuremath.com/2010/04/20/time-changed-brownian-motion/

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for $\theta(t)=\frac{t}{1-t}$, we get that $B_{\theta(t)}\stackrel{d}{=}\int_{0}^{t}\frac{1}{|1-s|}dB_{s}.$

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