I have repeatedly come across the statement that "a process with a drift cannot be a martingale". Is this true also for stochastic drifts?

Suppose I have a process with a stochastic drift:


Above, $f_1$ and $f_2$ are some functions of $W_t$.

What if the expected value of the stochastic drift is zero? i.e.:


I know the above is not a sufficient condition for $X_t$ to be a martingale, but my intuition tells me it should at least ensure that $X_t$ "has a shot" at being a martingale, right?

Edit: the example below serves as a "counterexample" (the drift has zero expectation, but the process is not a martingale) (thank you to @AntoineConze). So I wonder if its possible after all to have a process with a stochastic drift that is a martingale?

Example 1:

Let $X_t$ be defined as:

$$X_t=X_0+\int_{h=0}^{h=t}W_hdh + \int_{h=0}^{h=t}1dW_h$$

NTS: $\forall 0\leq s < t$: $\mathbb{E}\left[X_t | \mathcal{F}_s \right]=X_s$


$$\mathbb{E}\left[X_0+\int_{h=0}^{h=t}W_hdh + \int_{h=0}^{h=t}1dW_h|\mathcal{F}_s \right]=\\=X_0+\mathbb{E}\left[\int_{h=0}^{h=s}W_hdh + \int_{h=s}^{h=t}W_hdh +W_s +W(t-s)|\mathcal{F}_s \right]=\\=X_0+\int_{h=0}^{h=s}W_hdh + \int_{h=0}^{h=s}1dW_h+\int_{h=s}^{h=t}\mathbb{E}[W_h|\mathcal{F}_s]dh + \mathbb{E}[W(t-s)]_{=0}=\\=X_s+\int_s^tW_sdh\neq X_s$$


2 Answers 2


Check your calculation it's wrong. In the first example $$ E[X_t | {\cal F}_s] = X_s + E[\int_s^t W_h dh | {\cal F}_s] = X_s + \int_s^t E[W_h | {\cal F}_s] dh \\= X_s + \int_s^t W_s dh = X_s + W_s (t - s) \neq X_s $$

  • $\begingroup$ So that would make the second example also wrong. But I am not 100% sure I agree, because the condition "given what we know at time "$s$" had already been used by the time the integral $$\int_{h=0}^{h=s}W_hdh$$ had been pulled out of the expectation. So we should then just have an unconditional expectation remaining, which would make the expectation of the integral $$\int_{h=s}^{h=t}W_hdh$$ equal to zero? $\endgroup$ Dec 4, 2020 at 8:29
  • $\begingroup$ Upon reflection, I think you are correct and I am wrong! $\endgroup$ Dec 4, 2020 at 8:32
  • 1
    $\begingroup$ you can't replace $W_s$ (which you know at time $s$) with its expectation viewed from time 0 ! $\endgroup$ Dec 4, 2020 at 8:33

Well, let's just add a proof that it is never a (local) martingale.

Let's consider $(X_t)_t$, a (local) martingale, and $dX_t = A_tdt+H_tdW_t$. We know that $(\int_0^t H_sdW_s)_t$ is a local martingale. Therefore $(V_t)_t:=(\int_0^tA_sds)_t$ must be a local martingale too (let's say with family of stopping times $(\tau_n)_n$ such that we also have $|V_{t\wedge\tau_n}|\leq n$).

It's well known that for a martingale $(M_t)_t$, we have $\mathbb{E}[(M_t-M_s)^2] = \mathbb{E}[M_t^2-M_s^2]$ for any $t>s$, so $$\mathbb{E}[(V_{t \wedge \tau_n}-V_{s \wedge \tau_n})^2] = \mathbb{E}[V_{t \wedge \tau_n}^2 - V_{s \wedge \tau_n}^2]$$

Now, since $(V_{t \wedge \tau_n})_t$ is a (true) bounded martingale, also $(V_{t\wedge\tau_n}^2-\langle V \rangle_{t\wedge\tau_n})_t$ is a local martingale.

However, $\langle V \rangle = 0$, thus $(V_{t\wedge\tau_n}^2)_t$ is a local martingale, it's bounded, so it's a true martingale. This means that $$\mathbb{E}[(V_{t \wedge \tau_n}-V_{s \wedge \tau_n})^2] = \mathbb{E}[V_{t \wedge \tau_n}^2 - V_{s \wedge \tau_n}^2] = 0$$

From here, using the continuity of $V$, we easily show that $V \equiv 0$ almost surely, and thus there can't be any drift, not even a stochastic one.


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