Let $(X_t)$ be a semimartingale with $\mu_t= E[X_t]$. Is $\mu_t$ of bounded variation? Equivalently, is $X_t-\mu_t$ a semimartingale?

My intuition says yes (EDIT: my intuition is wrong, see comments below, so you don't have to read it, you can focus on the question), since $X_t=M_t +A_t$ where $(M_t)$ is a local martingale and $(A_t)$ is of bounded variation. Then $E[X_t]= E[M_t]+E[A_t]= c+E[A_t]=\mu_t$ for some $c \in \mathbb R$. Hence, $\mu_t$ is the mean of a bounded variation process. So the question boils down to is the mean of a bounded variation process of bounded variation?

If not, does adding sample-continuity of $(X_t)$ change the result?

  • $\begingroup$ You should know this result as it is very useful: a function is of bounded variation if and only if it is the difference of two increasing functions. Then $A_t = B_t - C_t,$ for two increasing processes, and therefore $\mu_t$ is also the difference of two increasing functions, which means of bounded variation. $\endgroup$
    – William M.
    Dec 2, 2022 at 19:55
  • 1
    $\begingroup$ It is not true that $E[M_t] = c$ for some $c \in \mathbb{R}$ if $(M_t)$ is only a local martingale rather than a martingale. $\endgroup$ Dec 2, 2022 at 20:03
  • $\begingroup$ @WilliamM. I know of that result for functions. I do not know how you apply it to processes. What is the equality you use? Almost surely? How can you guarantee integrability of $B_t$ and $C_t$? Anyway it seems that I cannot use that argument here following the comment of user6247850 $\endgroup$
    – W. Volante
    Dec 2, 2022 at 22:12
  • $\begingroup$ @user6247850 I didn't know. What is the mean of a local martingale then? What properties does it have? $\endgroup$
    – W. Volante
    Dec 2, 2022 at 22:12
  • $\begingroup$ Local martingale really should've been defined before semi-martingale, but it means that there exists a sequence of stopping times $(\tau_n) \rightarrow \infty$ such that $M_t^{\tau_n}$ is a martingale for each $n$. The stopping theorem guarantees that every martingale is a local martingale, but the reverse is not true. $\endgroup$ Dec 2, 2022 at 22:17

1 Answer 1


First, we will show that if $A_t$ is an integrable bounded variation process, then $\mathbb{E}[A_t]$ is also bounded variation. This follows from the comment by William M. Since $A_t$ is bounded variation, there exist increasing processes $B_t,C_t$ such that $A_t = B_t-C_t$. Then $\mathbb{E}[A_t] = \mathbb{E}[B_t]-\mathbb{E}[C_t]$ is also bounded variation because $\mathbb{E}[B_t]$ and $\mathbb{E}[C_t]$ are both increasing.

For the local martingale part $M$, we need some integrability conditions. Anything that guarantees that $M$ is a true martingale (e.g. $\mathbb{E}[\langle M,M\rangle_t] < \infty$ for all $t$ or $\{M_\tau : \tau \le a \text{ and }\tau\text{ is a stopping time}\}$ is uniformly integrable for all $a$) would work with the exact proof you gave. Anything assumption like $M_t \ge a$ (or $M_t \le a$) for all $t$ and some $a \in \mathbb{R}$ would also work by making $M$ a sub- or supermartingale.

I will make the slightly weaker assumption $\sup_{\tau} \mathbb{E}[|M_\tau|] < \infty$, where the supremum is taken over all finitely valued stopping times $\tau$. This allows us to use the Krickeberg decomposition: there exists a pair of non-negative local martingales $(M^+,M^-)$ such that $M = M^+-M^-$. Now, since a non-negative local martingale is a supermartingale, we know that $\mathbb{E}[M_t^+]$ and $\mathbb{E}[M_t^-]$ are both decreasing. Since $\mathbb{E}[M_t] = \mathbb{E}[M_t^+] - \mathbb{E}[M_t^-]$ is the difference of two monotone processes, it is bounded variation.

In that case, $\mathbb{E}[X_t] = \mathbb{E}[M_t] + \mathbb{E}[A_t]$ is the sum of bounded variation processes and hence bounded variation. Although we needed to make some potentially unnatural looking assumptions about the local martingale part $M$, we could have just imposed the slightly stronger (but easier to check) condition that $\mathbb{E}[\langle X,X\rangle_t] < \infty$ for all $t$.


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