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:


And in fact we have that:


But, $\forall 0< s <t$:

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

So $X_t$ is not a martingale.

  • 4
    $\begingroup$ 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? $\endgroup$
    – Shashi
    Commented Dec 5, 2020 at 19:25
  • $\begingroup$ @JanStuller: Do you want a proof in an answer or is Shashi's comment enough? (Although his comment is practically a proof) $\endgroup$
    – UBM
    Commented Dec 5, 2020 at 19:42
  • $\begingroup$ @UBM: if you can post an answer with a proof, I will happily up vote & accept :) $\endgroup$ Commented Dec 5, 2020 at 19:47
  • 1
    $\begingroup$ @user408858 no instead bound it by the full integral of the absolute value. That is finite by definition of being integrable. $\endgroup$
    – Shashi
    Commented Oct 22, 2021 at 13:59
  • 1
    $\begingroup$ An integrable function need not to be bounded. What you are using does not always work, while the integral bound does work. $\endgroup$
    – Shashi
    Commented Oct 22, 2021 at 14:18

1 Answer 1


I will prove it for an Ito process of the type $$X_t = X_0 + \int_0^ta_udu + \int_0^tb_udW_u, \quad 0 \leq t \leq T. \tag{1}$$

We will need the following two theorems.

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 $0 \leq t \leq T.$


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