I have two questions.

1) I was reading a presentation by Prof. Protter about some technicalities in stochastic calculus and, at one point, in the slides, I see stated that

"P. A. Meyer (1973) showed that there are no local martingales in discrete time; they are a continuous time phenomenon."

I wonder if anyone could either refer me to the specific paper, or textbook, or if anyone knows the specific example Prof. Protter is mentioning.

2) There are several examples of local martingales which are not martingales (in continuous-time). Does anyone know an interesting example not involving Brownian motion or specific stochastic processes requiring to use specific properties of those processes? I am looking for a ``vanilla'' example that requires some thinking.

I am aware of one case where, if the probability space is $([0,1], {\cal B}[[0,1]), \lambda), $ and one chooses the filtration $\mathbb{F} = \{{\cal F}_t:t \in \mathbb{R}^+\} $ with ${\cal F}_t = {\cal B}([0, 1]) $ for all $t, $ and $Z $ is a non-integrable r.v. then $\{X_t:= Z: t \in \mathbb{R}^+\} $ is not a martingale as integrability is lost from the start. However, it is a local martingale, because we would have that $X-X_0 \equiv 0. $ But this seems rather uninteresting as there is not even the need to introduce a localizing sequence. I would really appreciate one, where at least a localizing sequence has to be suggested and some simple calculations are involved.

I seem to remember one example I saw some time ago. I think it went along the lines of introducing a filtration $\mathbb{F} = \{{\cal F}_t: t \in \mathbb{R}^+\} $ and a r.v. $Z $ supposed to be ${\cal F}_1$-measurable but having infinite first moment (which takes care of the fact that we will not end up with a martingale). Then one introduces another r.v., say $W, $ which is the classic $\pm 1 $ with probability $1/2, $ and independent of ${\cal F}_{2-} $ and finally we let $X_t = WZ\cdot 1_{\{t\ge 2\}}. $ Then if I am not mistaken one ends up with a local martingale by introducing the localizing sequence given by $$T_n = 1\cdot 1_{\{Z> n\}} + n\cdot 1_{\{Z\le n\}}. $$
Could anyone show me the key details on how to prove that $X_t $ so defined is a local martingale? I think it is even bounded since by our definition of $T_n $ we would end up with an upper limit $n $ for $X_{T_n\wedge t} $ for each choice of $n. $

Thank you and sorry for the long question. I am just trying to clean some examples I collected in the old (sigh!) days of graduate school.


  • $\begingroup$ You can take a look at the book by Jacod and Shiryaev Limit theorem for stochastic processes, in the first chapter, if I remember well, they explain the situation in details. Best regards $\endgroup$
    – TheBridge
    Feb 15, 2016 at 11:13
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    $\begingroup$ A recent paper "Single jump processes and strict local martingales" by M. Herdegen and S. Herrmann might be of interest: arxiv.org/abs/1510.03192 (It has now appeared in Stochastic Processes and Their Applications.) $\endgroup$ Feb 15, 2016 at 15:58

1 Answer 1


To answer your first question, the result was published in 1972 by P.A. Meyer in the "Lecture Notes in Mathematics Series". The title is Martingales and Stochastic Integrals I, and the result is Theorem 42 on page 47.

To answer your second question, let $\Omega = C([0,\infty))$ and let $X(t,\omega) = \omega(t)$, i.e. X is the canonical coordinate process. Define $\mathbb{P}$ to be a probability measure on $\Omega$ such that $X$ is a Wiener process starting from point $1$. Also, let $\tau_0 = \inf \{t: X(t) = 0\}$. Our goal is to define a process $Y = 1/X$, and we will then show that Y is a (integrable) local martingale which is not a martingale.

First, we define another measure $\mathbb{Q}(t)$ using the Radon-Nikodym derivative $$\frac{d\mathbb{Q}(t)}{d\mathbb{P}} = X(t \wedge \tau_0) = X(t)1_{(t < \tau_0)} + X(\tau_0) 1_{(t \geq \tau_0)} = X(t) 1_{(t < \tau_0)}.$$ Because $X^{\tau_0}$ is a martingale under the measure $\mathbb{P}$, $$\mathbb{E}[X(t \wedge \tau_0)|\mathcal{F}_s] = X(s \wedge \tau_0).$$

Furthermore, one needs to check that the measures $(\mathbb{Q}(t))_{t \geq 0}$ are consistent probability measures. I'll leave out the calculation for now, but can add them if you are stuck.

Because we defined $Y = 1/X$, we need to ensure that $X$ is almost surely never zero under $\mathbb{Q}$: $$\mathbb{Q}(\tau_0 \leq t) = \mathbb{Q}(t)1_{\tau_0 \leq t)} = \int_\Omega 1_{(\tau_0 \leq t)} X(\tau_0 \wedge t) d\mathbb{Q} = 0,$$ so $\mathbb{Q}(\tau_0 = \infty) = 1$. Therefore $X > 0$ almost surely, and Y is almost surely well-defined.

Finally we are done with the setup, so we can move onto the interesting parts:

  1. $Y$ is not a martingale under $\mathbb{Q}$
  2. $Y$ is a local martingale under $\mathbb{Q}$

To show the first part, just calculate you the expected value of Y under $\mathbb{Q}$ using the definition of Y, change of measure to $\mathbb{P}$ and you will notice that some parts "cancel out" to give you $$\mathbb{E}[Y(t)|\mathbb{Q}] = \mathbb{P}(t < \tau_0),$$ which is decreasing to $0$.

The second part is the most difficult part. First, let $\varepsilon >0$ and define $$\tau_\varepsilon = \inf\{t: X(t) = \varepsilon\}.$$ Since $X$ is continuous, if $\varepsilon \searrow 0$, then $\tau_\varepsilon (\omega) \nearrow \tau_0(\omega)$ for every outcome $\omega$. Note that $\mathbb{Q}$-almost surely $\tau_\varepsilon \nearrow \infty$.

Moving along, we need to show that the truncated process $Y^{\tau_\varepsilon}$ is a $\mathbb{Q}$-martingale. First, we note that $Y^{\tau_\varepsilon}$ is almost surely bounded by $1 / \varepsilon$, and hence is uniformly integrable. If $s < t$ and $F \in \mathcal{F}_s$, after using the definition of Y, change of measure and some calculations, we obtain $$\int_F Y^{\tau_\varepsilon}(t) d\mathbb{Q} = \frac{1}{\varepsilon} \int_F \varepsilon + (X^{\tau_0}(t) - \varepsilon)1_{t \geq \tau_\varepsilon}d\mathbb{P}.$$

To show that this is a martingale under $\mathbb{Q}$, we can prove that $$M(t) = (X^{\tau_0}(t) - \varepsilon)1_{t \geq \tau_\varepsilon}$$ is a $\mathbb{P}$-martingale. To do so, you will need the normal properties of conditional expectation and the optional sampling theorem.

This example was taken from the textbook "Stochastic Integration Theory" by Medvegeyv, it can be found on page 97ff. I have left out some details in my post here, but they can be found in the book (as well as other useful examples).

  • $\begingroup$ Thank you Olorun for your detailed answer and the reference. $\endgroup$
    – Maurice
    Feb 15, 2016 at 19:24
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    $\begingroup$ @Maurice Any chance you could accept this as the answer? :) $\endgroup$
    – Olorun
    Apr 30, 2020 at 10:52

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