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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.

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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. –  Ilya Mar 2 '12 at 20:14
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Bigger hint: you can do it with a local martingale which is constant for $t\geq 1$. –  Ben Derrett Mar 4 '12 at 14:31
    
@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? –  Tom Artiom Fiodorov Mar 4 '12 at 20:13
    
@BenDerrett what is the localising sequence for your example? I do not think any non-trivial example exists. –  Lost1 Nov 27 '12 at 21:48
    
@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.... –  Lost1 Nov 28 '12 at 21:22

1 Answer 1

up vote 2 down vote accepted

Let $X$ be a random variable which has a mean, but infinite variance. Let $M_0=X$. Let $B$ be $1$ with probability half and $−1$ with probability half, independent of $X$. Fix a filtration $\mathcal{F}$ by $\mathcal{F}_0 = \sigma(X)$, $\mathcal{F}_i = \sigma(X,B)$, for $i\geq 1$.

Let $M_1=M_0+BM_0^2$. Set $M_i=M_1$ for $i>1$. This isn't a true martingale, since it's not integrable. Set $T_n=\inf\{k:|M_k|>=n\}$.

Fix an $n$.

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

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

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