# Can a process with a stochastic drift be a martingale?

In general, a process with a deterministic non-zero drift cannot be a martingale. What if the process has a stochastic drift, that has an expectation of zero? Could such a process ever be a martingale?

I am aware that the zero expectation of the stochastic drift is not a sufficient condition for the process to be a martingale, but is there a general mathematical theorem or a result that would prove that a process with any non-zero drift (stochastic or deterministic) cannot be a martingale?

I provide a "counterexample" below (process with a zero-expectation stochastic drift, that is NOT a martingale).

Question: Could anyone either provide an example of a process with a stochastic drift that is a martingale or alternatively reference a theorem or a result that shows that all processes with a drift can't be martingales?

Let: $$X_t:=\int_{h=0}^{h=t}W_hdh+\int_{h=0}^{h=t}1dW_h=\int_{h=0}^{h=t}W_hdh+W_t$$

We clearly have a zero-expectation drift term with:

$$\mathbb{E}\left[\int_{h=0}^{h=t}W_hdh+\right]=\int_{h=0}^{h=t}\mathbb{E}[W_h]dh=0$$

And in fact we have that:

$$\mathbb{E}[X_t|\mathcal{F}_0]=X_0$$

But, $$\forall 0< s :

$$\mathbb{E}[X_t | \mathcal{F}_s]=\mathbb{E}\left[\int_{h=0}^{h=s}W_hdh+\int_{h=s}^{h=t}W_hdh+W(t-s)+W(s) \mid \mathcal{F}_s\right]=\\=\int_{h=0}^{h=s}W_hdh+W_s+\mathbb{E}\left[\int_{h=s}^{h=t}W_hdh+W(t-s) \mid \mathcal{F}_s\right]=\\=X_s+\int_{h=s}^{h=t}\mathbb{E}[W_h|\mathcal{F}_s]dh=\\=X_s+W_s\int_{h=s}^{h=t}dh=\\=X_s+W_s(t-s)\neq X_s$$

So $$X_t$$ is not a martingale.

• If $X=Y+Z$ is martingale with $Z$ a martingale and $Y$ a drift, then $X-Z=Y$ is also martingale. But martingale that's of finite variation is? Do you know something telling you that such martingales are of specific type? Commented Dec 5, 2020 at 19:25
• @JanStuller: Do you want a proof in an answer or is Shashi's comment enough? (Although his comment is practically a proof)
– UBM
Commented Dec 5, 2020 at 19:42
• @UBM: if you can post an answer with a proof, I will happily up vote & accept :) Commented Dec 5, 2020 at 19:47
• @user408858 no instead bound it by the full integral of the absolute value. That is finite by definition of being integrable. Commented Oct 22, 2021 at 13:59
• An integrable function need not to be bounded. What you are using does not always work, while the integral bound does work. Commented Oct 22, 2021 at 14:18

I will prove it for an Ito process of the type $$X_t = X_0 + \int_0^ta_udu + \int_0^tb_udW_u, \quad 0 \leq t \leq T. \tag{1}$$
Theorem 1. Let M and N be local martingales. Then the sum $$M+N$$ is also a local martingale.
Theorem 2. Let $$c \mathscr{M}_{0,loc}$$ denote the space of continuous local martingale that starts at zero. Let $$M \in c \mathscr{M}_{0,loc}$$ be of finite variation. Then $$M \equiv 0$$ a.s.
Answer: Let $$X$$ be a continuous local martingale of the type SDE(1). Then the drift must be zero.
Proof. We know the process $$\int_0^t b_udW_u, 0 \leq t \leq T,$$ is a continuous local martingale. Then the process $$A_t := \int_0^ta_udu= X_t - X_0 - \int_0^tb_udW_u, \quad 0 \leq t \leq T$$ is a sum of continous local martingales and, by Theorem 1, is also a continuous local martingale. Also, note that $$A_0 = 0,$$ so $$A \in c \mathscr M_{0,loc}^2.$$ But the process $$A$$ is of finite variation, so by Theorem 2, must be zero. This is the same as saying that $$\int_0^ta_udu=0$$ for all $$0 \leq t \leq T.$$ Since this integral is defined path by path, by elementary calculus we have that $$a_t = 0$$ for all $$0 \leq t \leq T.$$