# Find continuous martingale, such that it's quadratic variation is a deterministic continuous function

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?

• $X_t = W_{f(t)}$ 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 $W$. May 29, 2020 at 21:09
• 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 $\int_0^t f'(s)ds=f(t)$. May 29, 2020 at 21:14
• 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. May 29, 2020 at 21:20
• 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 $W$? May 30, 2020 at 21:00