# Let $(M_t)$ be a continuous square-integrable martingale with independent increments. Is $t \mapsto \mathbb E[M_t^2]$ continuous?

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?

• Maybe this will help you: math.stackexchange.com/questions/3945806/… Commented Feb 6, 2023 at 16:37
• @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? Commented Feb 6, 2023 at 17:33
• @greyls Please have a look at this question which formulates my above confusion. Commented Feb 6, 2023 at 17:43

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.

• Remark: The condition "independent increment" is irrelevant and is not needed. Commented Feb 7, 2023 at 3:30
• 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? Commented Feb 7, 2023 at 10:26
• @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. 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 $$s and $$\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.