# Is the mean of an integrable semimartingale of bounded variation?

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?

• 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. Dec 2, 2022 at 19:55
• 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. Dec 2, 2022 at 20:03
• @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 Dec 2, 2022 at 22:12
• @user6247850 I didn't know. What is the mean of a local martingale then? What properties does it have? Dec 2, 2022 at 22:12
• 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. Dec 2, 2022 at 22:17

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 $$t$$.