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?

  • $\begingroup$ Maybe this will help you: math.stackexchange.com/questions/3945806/… $\endgroup$
    – greyls
    Commented Feb 6, 2023 at 16:37
  • $\begingroup$ @greyls The author of the question in that thread uses the fact that $(M_t^2, t\ge 0)$ is a sub-martingale and thus $M_t^2 \le \mathbb E [M_T^2 | \mathcal F_t] = \eta$ for all $t \in [0, T]$. It seems to me $\eta$ depends on $\mathcal F_t$ and thus on $t$. Could you elaborate on my confusion? $\endgroup$
    – Analyst
    Commented Feb 6, 2023 at 17:33
  • $\begingroup$ @greyls Please have a look at this question which formulates my above confusion. $\endgroup$
    – Analyst
    Commented Feb 6, 2023 at 17:43

2 Answers 2


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.

  • $\begingroup$ Remark: The condition "independent increment" is irrelevant and is not needed. $\endgroup$ Commented Feb 7, 2023 at 3:30
  • $\begingroup$ But the assumption "independent increments" is crucial for the fact that $\langle M\rangle_t=\mathbb{E}\left(M_t^2\right)-\mathbb{E}\left(M_0^2\right)$, right? $\endgroup$
    – Analyst
    Commented Feb 7, 2023 at 10:26
  • 1
    $\begingroup$ @Analyst The task is to prove that $t\mapsto E(M_t^2)$ is continuous. We do not need anything related to quadratic variation. You may trace my proof. $\endgroup$ Commented Feb 7, 2023 at 13:57

I refer to this source, Theorem 2. Every $\mathbb{R}$-valued continuous process $X$ with independent increments is s.t. $X_t-X_s\sim \mathcal{N}(b_t-b_s,\Sigma_t-\Sigma_s)$ for unique continuous functions $b_t,\Sigma_t$ with $b_0=0,\Sigma_0=0$. Now consider a filtration $(\mathscr{F}_t)_{t \geq 0}$ s.t. $X_t-X_s$ is independent of $\mathscr{F}_s$ for all $s<t$ and $\sigma(X_s,s\leq t)\subseteq \mathscr{F}_t,\,\forall t$. Then $$E[X_t-X_s|\mathscr{F}_s]=E[X_t-X_s]=b_t-b_s$$ Therefore $X$ is a $\mathscr{F}_t$-martingale iff $b_t=0,\forall t\geq 0$. Suppose $X$ is also a $\mathscr{F}_t$-martingale. Then $E[X_t^2]=E[X_0^2]+\Sigma_t$, which is continuous.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .