I'm reading a remark at page 4 of these lecture notes.
Let $\left(\mathcal{F}_t, t \in \mathbb{R}_{+}\right)$be a filtration and $M=\left(M_t, t \in \mathbb{R}_{+}\right)$be a continuous square-integrable martingale with respect to $\left(\mathcal{F}_t, t \in \mathbb{R}_{+}\right)$.
Reminder. The quadratic variation of $M$ is the unique process $\left(\langle M\rangle_t, t \in \mathbb{R}_{+}\right)$which is increasing, continuous and adapted to $\left(\mathcal{F}_t, t \in \mathbb{R}_{+}\right)$, such that $\langle M\rangle_0=0$ a.s. and $\left(M_t^2-\langle M\rangle_t, t \in \mathbb{R}_{+}\right)$is a martingale with respect to $\left(\mathcal{F}_t, t \in \mathbb{R}_{+}\right)$.
Lemma 1.1. For all $t>s \geq 0$,
$$
\mathbb{E}\left(\left(M_t-M_s\right)^2 | \mathcal{F}_s\right)=\mathbb{E}\left(\langle M\rangle_t-\langle M\rangle_s | \mathcal{F}_s\right).
$$
Remark. In general, $\langle M\rangle_t$ is not deterministic, but when $M$ has independent increments, then $\langle M\rangle_t=\mathbb{E}\left(M_t^2\right)-\mathbb{E}\left(M_0^2\right)$ (and is therefore deterministic).
My understanding In other threads (1, 2, 3, 4, 5), the remark holds if in addition $M$ is a Gaussian process, i.e., if $M$ is Gaussian then $t \mapsto \mathbb{E}\left(M_t^2\right)$ is continuous.
Could you confirm that the remark is not necessarily correct?
Best Answer
The following is an elementary solution and does not involve concepts like quadratic variation.
Let $f:[0,\infty)\rightarrow[0,\infty)$ be defined by $f(t)=E[M_{t}^{2}]$. Let $t_{0}\in[0,\infty)$ be arbitrary. We go to show that $f$ is continuous at $t_{0}$. Let $(t_{n})$ be an arbitrary sequence in $[0,\infty)$ such that $t_{n}\rightarrow t_{0}$. Choose $T\in[0,\infty)$ such that $t_{n}\leq T$ for all $n$.
Claim 1: $\{M_{t}^{2}\mid t\in[0,T]\}$ is uniformly integrable. Note that $\{E\left[M_{T}^{2}\mid\mathcal{F}_{t}\right]\mid t\in[0,T]\}$ is uniformly integrable because it is a family of random variables arising from taking conditional expectation of the integrable random variable $M_{T}^{2}$. For each $t\in[0,T]$, by Jensen inequality, $M_{t}^{2}=\left(E\left[M_{T}\mid\mathcal{F}_{t}\right]\right)^{2}\leq E\left[M_{T}^{2}\mid\mathcal{F}_{t}\right]$. Therefore, $\{M_{t}^{2}\mid t\in[0,T]\}$ is uniformly integrable too.
In particular, $\{M_{t_{n}}^{2}\mid n\in\mathbb{N}\}$ is uniformly integrable. Note that $M_{t_{n}}^{2}\rightarrow M_{t_{0}}^{2}$ pointwisely, so $\int M_{t_{n}}^{2}\,dP\rightarrow\int M_{t_{0}}^{2}\,dP$. That is, $f(t_{n})\rightarrow f(t_{0})$. This shows that $f$ is continuous at an arbitrary point and hence is a continuous function.