# Example of a process that yields a non-martingale

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: $$(C\bullet X)_n:=\sum_{k=1}^{n} C_k(X_k-X_{k-1})$$

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.
• 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 $\mathcal{F}_k$ Apr 4 at 17:29
• @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 $0$. Apr 4 at 18:19