I'm attending a class on stochastic processes and as a “side quest” to understand a discrete martingale transform properly we were given a task:

Provide/construct a not previsible adapted process $C$ such that for a martingale $X,$ the martingale transform $(C\bullet X)_n$ is not a martingale.

The transform in question which has only been defined for discrete processes as: $(C\bullet X)_n:=\sum_{k=1}^{n} C_k(X_k-X_{k-1})$


1 Answer 1


Easy example: $C_k = 1_{X_k > X_{k-1}}$. This is adapted because $X$ is adapted, but $\sum_{k=1}^n C_k (X_k-X_{k-1}) = \sum_{k=1}^n (X_k-X_{k-1})^+$ is non-decreasing and hence cannot be a martingale.

Another example would be $C_k = (X_k-X_{k-1})$, so that $\sum_{k=1}^n C_k (X_k-X_{k-1}) = \sum_{k=1}^n (X_k-X_{k-1})^2$ is the quadratic-variation process of $X$, which is again non-decreasing and hence not a martingale.

  • $\begingroup$ but just to clarify: by $f^{+}$ you mean $\frac{|f| + f}{2}$, right? Also: is the first process not previsible? I still don't have good intuitions on filtrations, so I don't see why $C_{k+1}$ wouldn't be measurable with respect to $\mathcal{F}_k$ $\endgroup$
    – markovian
    Apr 4 at 17:29
  • 1
    $\begingroup$ @markovian I haven't seen it written like that before (usually just $f^+(x) = \max(0,f(x))$), but yes, that is equivalent. And $C_{k+1}$ depends on $X_{k+1}$, so it isn't measurable with respect to $\mathcal F_k$. At time $k$, we don't know whether or not $X_{k+1}$ is greater than $X_k$, so we don't know whether $C_{k+1}$ is $1$ or $0$. $\endgroup$ Apr 4 at 18:19

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