I am studying discrete time martingale theory and came across the classical "You can't beat the system" theorem: given a martingale $M$ and a previsible process $C=(C_n)_{ n \ge 1}$ such that $C_n$ is limited for every $n$ the martingale transform $C \bullet M $ is a martingale. My question is: can we find a martingale $M$ and $C$ previsible but not limited such that the martingale transform is no longer a martingale?
I have been looking for a similar topic but only found discussions in continuous time which I could not understand (I have no knowledge yet of stochastic continuous time calculus)... Thank you!
You are betting on fair independent coin flips. You must decide your bet $C_{n+1}$ for time $n+1$ at time $n$, i.e. you don't get to view the coin flip before deciding whether you chose heads or tails. Then your profit is given by $X = C \cdot M$.
Consider the strategy of betting a random huge amount each time, e.g. taking $C_n \notin L^1$, then $X_n \notin L^1$ and is thus $X$ is not a martingale because $|X_1|=|C_1| \notin L^1$, i.e. you either won or lost the huge amount you staked.