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Let $M$ be a (true) martingale with continuous sample paths, such that $M_0 = 0$. We assume that $(M_t)_{t \geq 0}$ is also a Gaussian process. Show that there exists a continuous monotone nondecreasing function $f: \mathbb{R}_+ \to \mathbb{R}_+$ such that $\langle M, M \rangle_t = f(t)$ for every $t \geq 0$.

Recall that $\langle M, M \rangle_t$ is the quadratic variation of $M$ and that $\langle M, M \rangle_t = \lim_{n \to \infty} \sum_{i = 1}^{p_n} (M_{t_i^n} - M_{t_{i-1}^n})^2$ for a sequence of partitions with mesh going to zero. We know this is an increasing process.

In a previous part of this question I showed that a consequence of the martingale property is that the increments are independent of the past. I don't know how to show that $\langle M, M \rangle_t$ is continuous.

cgmil
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Define $f(t) = \mathbb{E}[M_t^2]$. It remains to check that $M_t^2 - f(t)$ is a (local) martingale since then $f$ will be the quadratic variation of a continuous martingale and hence will be continuous and non-decreasing.

Recall that a continuous martingale that is also a Gaussian process has independent increments. Hence \begin{align*} \mathbb{E}[M_t^2 - M_s^2 \mid \mathcal{F}_s] = \mathbb{E}[(M_t - M_s)^2 \mid \mathcal{F}_s] = \mathbb{E}[(M_t - M_s)^2] = \mathbb{E}[M_t^2 - M_s^2] \end{align*} where we used independence of increments in the third equality. In particular, it follows that $M_t^2 - \mathbb{E}[M_t^2]$ is a martingale.

Rhys Steele
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I figured it out on my own.

Continuous martingales are continuous local martingales, and $\langle M, M\rangle_t$ is unique up to indistinguishability, an increasing (thus monotone and nondecreasing) process, and $N_t = M_t^2 - \langle M, M \rangle_t$ is a continuous local martingale. $M_t^2$ is also continuous, so

$$f(t) = \langle M, M\rangle_t = M_t^2 - N_t$$

must also be continuous.

cgmil
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    This answer doesn't work (especially since you don't use the assumption that $M$ is Gaussian anywhere). The main challenge in the problem you state is to show that the quadratic variation is equal to a deterministic function almost surely. It will then be automatically true that it is continuous and non-decreasing since the quadratic variation always is. This answer does nothing to address that main challenge. – Rhys Steele Mar 24 '18 at 21:18