I want to prove the following statement in discrete time:
Let $(X_k)$ be a local martingale and $(h_k)$ a predictable process, then the discrete stochastic integral is also a local martingale.
First I assume that $(X_k)$ is a true martingale, then
$ M_n:=\int_0^t hdX=\sum_{j=1}^nh_j(X_j-X_{j-1})$
First of all, is this definition of discrete stochastic integral right? I'm not sure since in the integral we have a $t$. Should there be a $n$?
Let's prove that $M_n$ is a martingale
$E[M_n-M_{n-1}|\mathcal{F}_{n-1}]=E[h_n(X_n-X_{n-1})|\mathcal{F}_{n-1}]=h_nE[(X_n-X_{n-1})|\mathcal{F}_{n-1}]=0$
Since $X_k$ was assumed to be a martingale. Now I want to do a localization argument and I'm not sure if my conclusion is right: If $(\tau_n)$ is the sequence of stopping times for which $(X_k)$ is a local martingale then
$E[M_n^{\tau_k}-M^{\tau_k}_{n-1}|\mathcal{F}_{n-1}]=E[h_{n\wedge\tau_k}(X_{n\wedge\tau_k}-X_{\tau_k\wedge n-1})|\mathcal{F}_{n-1}]=h_{n\wedge\tau_k}E[(X_{n_\wedge\tau_k}-X_{\tau_k\wedge n-1})|\mathcal{F}_{n-1}]=0$
The point which confuses me is if $h_n^{\tau_k}$ is still predictable? Thanks for your help
hulik