Showing a Quadratic Variation is increasing and continuous with respect to time Given $M$ a martingale starting from $0$ with continuous sample paths and $(M_t)_{t\ge 0}$ is a Gasussian process. I want to show $\langle M,M\rangle_t$ (the quadratic variation of $M_t$) is a monotone nondecreasing continuous function of $t$).
I was considering using Doob-Meyer. By Doob-Meyer there exists a nondecreasing stochastic process $(A_t)$ with continuous sample paths such that $A_t=<M_t,M_t>_t.$ 
Is there a different way to proceed by using the fact that $M_t$ is Gaussian?
 A: In fact (and maybe this is what you are saying you want to show), the quadratic variation of $M$ is deterministic; more precisely, $\langle M\rangle_t = E[M^2_t]$ a.s. Here is the outline of a proof. The function $q(s):=E[M_s^2]$ is continuous and non-decreasing. A Gaussian martingale necessarily has independent increments. Fix $t>0$. For a given positive integer $n$ choose $0=t_0<t_1<t_2<\cdots<t_{n-1}<t_n=t$ so that $q(t_{k+1})-q(t_k)=q(t)/n$ for $k=0,1,2,\ldots,n-1$. The  increments $M_{t_1}, M_{t_2}-M_{t_2},\ldots,M_{t_n}-M_{t_{n-1}}$ are independent Gaussian random variables with mean $0$ and variance $q(t)/n$. The sum of squares
$\sum_{k=1}^n[M_{t_k}-M_{t_{k-1}}]^2$ therefore has the same distribution as $(q(t)/n)\sum_{k=1}^n Z_k^2$, where $Z_1,Z_2,\ldots$ are iid standard normal; by the law of large numbers, this latter sum converges a.s. to $q(t)$ as $n\to\infty$. On the other hand, $\sum_{k=1}^n[M_{t_k}-M_{t_{k-1}}]^2$ converges in probability to $\langle M\rangle_t$. It follows that $\langle M\rangle_t=q(t)$ a.s.
