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Given a non decreasing continuous function $f:\mathbb{R}_+\rightarrow\mathbb{R}_+$ with $f(0)=0$.

I want to find a continuous martingale $(X_t)_{t\ge 0}$ , such that it's quadratic variation is equal to $f$, i.e. $\langle X,X\rangle_t=f(t)$ for all $t\ge 0$.

Let $(W_t)_{t\ge 0}$ be standard brownian motion. I find $E[W_{f(t)}^2]=f(t)$. So I was considering to define $(X_t)_{t\ge0}$ with $X_t:=W_{f(t)}$.

Which just looks too simple to me. Is this actually right?

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    $\begingroup$ $X_t = W_{f(t)}$ is correct. However, in order to show that $\langle X,X\rangle_t = f(t)$, it is not enough to prove that $E[X_t^2] = f(t)$. You have to check that $X_t^2 - f(t)$ is a martingale. Also this might be of interest as it shows that any continuous martingale $X$ with quadratic variation $f$ admits the representation $X_t = W_{f(t)}$ for some Brownian motion $W$. $\endgroup$
    – Michh
    May 29, 2020 at 21:09
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    $\begingroup$ If you assume $f$ is continuously differentiable another option could be to define $X_t=\int_0^t (f'(s))^{(1/2)} dW(s)$ then its quadratic variation is $\int_0^t f'(s)ds=f(t)$. $\endgroup$
    – Chaos
    May 29, 2020 at 21:14
  • $\begingroup$ Thanks already, for your fast answers! I will think about this more, because, as you mention: It has to be a martingale with respect to a filtration. I may post an answer on this myself afterwards. $\endgroup$
    – user408858
    May 29, 2020 at 21:20
  • $\begingroup$ Well, I am still struggling finding an idea why $(W_{f(t)})_{t\ge 0}$ is a martingale for some brownian motion $(W_t)_{t\ge 0}$. Do you have any hints on this? Is it just possible to define this process on the Filtration $(\mathcal{F}_{f(t)})_{t\ge 0}$ where $(\mathcal{F}_t)_{t\ge 0}$ denotes the filtration of $W$? $\endgroup$
    – user408858
    May 30, 2020 at 21:00

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