Find an example of a discrete-time local martingale that is not a true martingale.

I was thinking hard for some time about this fun problem. I know that $\mathbb{E}[|M|_t]=\infty \text{ for some } t\geq0$ should hold. Moreover any non-negative local martingale in discrete time is a true martingale, so this restricts my choice even more. I played around with Cauchy distribution, doubling strategy.

  • $\begingroup$ As I remember, in the discrete time the local martingale is always a martingale transform which is usually more easy to construct. Maybe it is also more easy to find an example of a martingale transform which is not a martingale. $\endgroup$
    – SBF
    Commented Mar 2, 2012 at 20:14
  • 1
    $\begingroup$ Bigger hint: you can do it with a local martingale which is constant for $t\geq 1$. $\endgroup$ Commented Mar 4, 2012 at 14:31
  • $\begingroup$ @BenDerrett, if $M_t$=C $\forall t\geq 1$, then $\mathbb{E}[|M_t|]=|C|<\infty$ $\forall t\geq 1$. So $M_0$ is not integrable, and so is $M_0^{T_n}=M_0$ for any stopping time $T_n$. Hence $(M_t^{T_n})_{t\geq0}$ is not a martingale. So $(M_t)_{t\geq0}$ is not a local martingale. What am I doing wrong? $\endgroup$ Commented Mar 4, 2012 at 20:13
  • $\begingroup$ @BenDerrett what is the localising sequence for your example? I do not think any non-trivial example exists. $\endgroup$
    – Lost1
    Commented Nov 27, 2012 at 21:48
  • $\begingroup$ @BenDerrett No, your example is not correct. $M^{T_n}$ is not integrable, because the stopping sequence does not bound the process. Let say at t=1, your process exceed your n, for some n. Then still $M^{T_n}_1=M_1$, which is not integrable. Basically on the event ${t=T_n}$, $M_t^{T_n}=M_t$, this does not make something which was not integrable integrable.... $\endgroup$
    – Lost1
    Commented Nov 28, 2012 at 21:22

1 Answer 1


Let $X$ be a random variable with finite mean and infinite variance. Let $B$ be $1$ with probability half and $−1$ with probability half, independent of $X$. Fix a filtration $\mathcal{F}$ by $\mathcal{F}_0 = \sigma(X)$ and $\mathcal{F}_i = \sigma(X,B)$ for every $i\geq 1$.

Let $M_0=X$ and $M_i=M_0+BM_0^2$ for every $i\geq1$. Then $(M_i)$ is not a true martingale, since $M_i$ is not integrable when $i\geq1$. For every $n$, set $T_n=\inf\{k:|M_k|\ge n\}$.

Fix an $n$. Then $\mathbb{E}[|M^{T_n}_0|]=\mathbb{E}[|X|]<\infty$ and, for every $i\geq1$,

$\begin{align} \mathbb{E}[|M^{T_n}_i|] &= \mathbb{E}[|M^{T_n}_1|\mathbf{1}(T_n=0)]+\mathbb{E}[|M^{T_n}_1|\mathbf{1}(T_n>0)]\\ &= \mathbb{E}[|M_0|\mathbf{1}(T_n=0)]+\mathbb{E}[|M_1|\mathbf{1}(T_n>0)]\\ &\leq \mathbb{E}[|M_0|]+\mathbb{E}[|M_1|\mathbf{1}(M_0\leq n)]\\ &\leq \mathbb{E}[|M_0|] + n+n^2 <\infty \end{align}$

So $M^{T_n}$ is integrable. We may also check that $\mathbb{E}[M^{T_n}_1\mid X]=M^{T_n}_0$, so $(T_n)$ localizes $M$. So $M$ is indeed a local martingale.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .