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I read this question Understanding proof of martingale transform being supermartingale and was wondering if the converse of the statement in the problem is also true. To be precise my question is:

Let $C\bullet X$ (and $Y_n$) be defined as $(C\bullet X)_n:=\sum_{k=1}^{n} C_k(X_k-X_{k-1})=Y_n$. Let $C$ be a bounded non-negative previsible and integrable process so that, for some $K\in[0,\infty)$, $|C_n(\omega)|\le K$ for every $n$ and every $\omega$. Then assume $C\bullet X$ is null at $0$.

Does it follow that $X$ is necessarily a martingale?

Edit:

If the condition is changed to $E[(C\bullet X)_n]=0$, is $X$ necessarily a martingale?

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Under the stated conditions, it is not true as you can easily construct a counterexample since you basically left the process $X$ to be unrestricted. For example, assume for simplicity that $C_{n}(\omega) = 1, \forall n, \omega$, and define $$ X_{n} = \sum_{k=1}^{n}Y_{k}, $$ where $\{Y_{k}\}_{k \in \mathbb{N}}$ is an i.i.d. sequence of Cauchy random variables. Did you perhaps mean to assume that $C \bullet X$ is a (super/sub)martingale that starts at 0?

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    $\begingroup$ thank you! I actually am mostly curious about proper martingales. But would that change if we change the condition to be $E[(C \bullet X)_n]=0$, assuming the transform exists, instead of $C \bullet X=0$? $\endgroup$
    – Jodasilva
    Apr 4 at 10:53
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    $\begingroup$ A variation of what I wrote above would still be a counterexample. Let $C_{n}(\omega)$ be any process that satisfies the conditions you stated, and define $X_{n} = Y$ for $Y$ again a Cauchy random variable, and let $(C \bullet X)_{0} = 0$ (or define $X_{0} = Y$ as well). Then for all positive integers $n$, we have $E[(C\bullet X)_{n}] = 0$ always, but $X$ is again not a martingale. $\endgroup$
    – minginator
    Apr 4 at 16:22
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    $\begingroup$ ok. But does it work because Cauchy is not integrable? Is there an integrable example by any chance? $\endgroup$
    – Jodasilva
    Apr 4 at 17:55
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    $\begingroup$ Indeed, the previous counterexamples work because Cauchy random variables are not integrable, whereas martingales need to be. I can't think of an integrable example currently, will update if I think of something. $\endgroup$
    – minginator
    Apr 4 at 21:25

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