# If the moving average of a process is a martingale, is the process a martingale?

Problem set up:

Let $$\mathcal F_t$$ be a filtration satisfying the usual conditions. Let $$T > 0$$ be a fixed real number, and define the filtration $$\mathcal H_t := \mathcal F_{T + t}$$.

Suppose a cadlag adapted process $$X$$ is almost surely locally integrable, i.e. for any compact set $$C \subset \mathbb R_+$$we have $$\int_C X_t \ dt < \infty$$ a.s.

Define the moving average process $$M$$ by $$M_t := \int_{[t, t + T]} X_s ds$$.

Suppose that the following conditions hold:

• Almost surely, $$X_0 = X_s$$ for all $$s \leq T$$.

• $$M_t$$ is a $$\mathcal H_t$$-martingale.

Question: Is it true that $$X$$ is an $$\mathcal F_t$$-martingale?

• While I do not know the answer, note that if $X_t$ is an $\mathcal F_t$-martingale, then typically $M_t$ is not an $\mathcal H_t$-martingale. Indeed, $\mathbb E(M_{t+T}|\mathcal H_t) = \mathbb E(M_{t+T}|\mathcal F_{t+T}) = T X_{t+T}$ rather than $M_t$. Commented May 27, 2021 at 7:03
• In discrete time the answer is yes, isn't it, and the H fields are not different from the F fields ?
– mike
Commented May 27, 2021 at 7:26
• I was thinking of a discretization argument too @mike. But contnuity issues are tricky.. Commented May 27, 2021 at 7:27
• Wait a minute: $M_t$ has (locally) finite variation, so if it is a martingale, it is constant, right? But then $0=\frac d{dt}M_t =X_{t+T}-X_t$, and so $X_t$ is periodic with period $T$. Commented Jul 5, 2021 at 20:02
• Oh wow, you are right of course. Damn. Commented Jul 5, 2021 at 21:24

Let $$X(t)$$ be any deterministic $$T$$ periodic function, then
$$M(t)=\int_0^T X(s) ds=\operatorname{Const},$$
but $$X(t)$$ is not a martingale.
• Wait, actually because of the condition on the starting values of $X$, would not any such $T$-periodic function be constant? Commented May 27, 2021 at 5:54