# discrete stochastic integral is a local martingale

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

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Yes, the stopped process of a predictable process is again a predictable process. It can be shown by a monotone class argument and the fact that the predictable sigma-field is generated by sets of the form $$A\times \{0\}, \;A\in\mathcal{F}_0\quad\text{and}\quad A\times (s,t],\;A\in\mathcal{F}_s,\;0\leq s<t.$$

Also I'm pretty sure that your definition of $M_n$ should be $\int_0^n h_s\,\mathrm{d} X_s$.

To show the above result we let $$\mathcal{H}=\{(X_t)_{t\geq 0}\mid \text{if } (X_t)_{t\geq 0} \text{ is predictable, then }(X_t^{\tau})_{\geq 0} \text{ is also predictable for every s.t. }\tau \},$$ and $$\mathcal{K}=\{(X_t)_{t\geq 0}\mid X_t(\omega)=1_{A\times \{0\}}(\omega,t),\; A\in\mathcal{F}_0\text{ or } X_t(\omega)=1_{A\times (s,r]}(\omega,t),\;0\leq s<r,\; A\in\mathcal{F}_s\},$$ i.e. $\mathcal{K}$ is the elementary predictable processes. Then $\mathcal{H}$ is vector space containing all constant functions and it is closed under monotone bounded convergence (i.e. if $(f_n)\subseteq \mathcal{H}$ is a bounded increasing sequence, then $f=\sup_n f_n\in\mathcal{H}$) and $\mathcal{K}$ is stable under multiplication. Then we are in the scope of the monotone class theorem.

Now if we can show that $\mathcal{K}\subseteq \mathcal{H}$, then every bounded predictable process is in $\mathcal{H}$. For a general predictable process $X=(X_t)_{t\geq 0}$, we let $X^n=X\wedge n\vee -n$ and hence $X^n$ are bounded predictable processes for each $n\in\mathbb{N}$. In particular $X^n\in\mathcal{H}$ for each $n$ and therefore $(X^n)^{\tau}$ is predictable for each $n$, where $\tau$ is any stopping time. Now we get that $X^{\tau}$ is predictable because $$X^{\tau}_t(\omega)=\lim_{n\to\infty} (X^n)^{\tau}_t(\omega),\quad (\omega,t)\in \Omega\times [0,\infty).$$

What is left is to show that $\mathcal{K}\subseteq\mathcal{H}$.

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Thank you for your answer. Can you prove that predictable processes are stable under stopping or could you give me a reference? – user20869 Sep 30 '12 at 12:16
@hulik: See the edit. Try to show the inclusion yourself. I'm not entirely sure how the proof of that goes. – Stefan Hansen Sep 30 '12 at 13:32
Thank you Stefan. I will think about it. – user20869 Sep 30 '12 at 13:46

For $n\geq 1$, $$X_{\tau\wedge n}=\sum_{j=0}^{n-1}1_{(\tau=j)}\,X_j+1_{(\tau>n-1)}\, X_n.$$ If $\tau$ is a stopping time and $(X_n)$ predictable, then all the summands are ${\cal F}_{n-1}$ measurable.

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