For the definition of Itô integral, one uses simple stochastic processes. I have found two definitions for simple stochastic process, given a filtration $(\mathcal{F}_t)_{t\geq0}$, an interval $[0,T]$ and a sample space $\Omega$:
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$u_t=\sum_{i=1}^p \phi_i 1_{(t_{i-1},t_i]}(t)$,
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$u_t=\sum_{i=1}^p \phi_i 1_{[t_{i-1},t_i)}(t)$,
where $0=t_0<t_1<\ldots<t_p=T$ is a partition, and $\phi_i$ is an $\mathcal{F}_{t_{i-1}}$-measurable random variable such that $E[\phi_i^2]<\infty$, for $i=0,\ldots,p$. The first definition corresponds to the book Introduction to Stochastic Calculus Applied to Finance, by Lamberton and Lapeyre. The second definition corresponds to my lecture notes. In both cases, given a general adapted stochastic process $u$ in $L^2(\Omega\times[0,T])$, it is approximated by simple processes $(u^n)_{n=1}^\infty$ in $L^2(\Omega\times[0,T])$ and the Itô integral is defined as a limit in $L^2(\Omega)$ of $\int_0^T u_t^n \,dB_t$.
My question is whether the two definitions are equivalent to define the Itô integral.
Best Answer
There is a small difference between the two processes. Note that process (1) is left continuous with right limits (caglad process), while process (2) is right continuous with left limits (cadlag process).
For integration with respect to continuous martingales (such as Brownian motion), which one you take as a simple process will not matter, as mentioned by Will in his answer. Indeed, Oksendal (2010) defines the stochastic integral by constructing the integral first for the processes of type (2), while Cont and Tankov (2004) or Karatzas and Shreve (1998) consider first processes of type (1).
In a more general theory of stochastic integration based on semimartingales (which can be discontinuous), one typically considers integration of caglad processes. Therefore, one starts to build integrals with simple processes of type (1). I found two justifications for it in the literature.
(1) As shown in Protter (2004), if we consider a stochastic integral of a cadlag process with respect to semimartingale then the resulting integral may not be a martingale even if the process with respect to which we define the integral is. On the other hand, an integral of a caglad process with respect to a martingale will be a martingale.
As an example (page 65), he takes $M_t=N_t-\lambda t$, a compensated Poisson process with intensity $\lambda$, with jump times denoted by $(T_i)_{i\geq 1}$ and an adapted process $H_t=1_{[0,T_1)}$. In that case, we have
$$ \int_0^t H_s dM_s = -\lambda(t\land T_1),$$
which is not a martingale. But if we consider $H_t=1_{(0,T_1]}$ then the above integral is equal to $$ \int_0^t H_s dM_s = N_{t\land T_1}-\lambda(t\land T_1),$$ which is a martingale (since a stopped martingale is a martingale).
(2) Cont and Tankov (2004) motivate restriction to caglad functions by showing that if we allow cadlag functions to be trading strategies then in the presence of jumps in the price of a stock there would be arbitrage opportunity and one would be able to earn an infinite rate of return (see Example 8.1, page 250). This justifies restriction to caglad processes when considering a suitable class of functions which we want to integrate.
PS. I am self-learning stochastic integrations so my apologies for any inaccuracies in my answer. And I am curious to hear answers' of other users!