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Ito integral

An ordinary Riemann (Riemann-Stieltjes) integral is defined as a limit of sum approximations. Let {$t^{(n)}_k~,k=0,1,\ldots,n$} for each $n\gt 0$ be an increasing (in k) sequence of time-points with $t^{(n)}_0=T_0$ and $t^{(n)}_n=T_f$, such that $$\max_{k}~t^{(n)}_{k+1}-t^{(n)}_k\rightarrow 0, n\rightarrow \infty.$$ Then $$\int_{T_0}^{T_f}f(t)dt=\lim_{n \rightarrow \infty}\sum^{n-1}_{k=0}f(t^{(n)*}_k)(t^{(n)}_{k+1}-t^{(n)}_k),$$ where $t^{(n)*}_k$ is some arbitrary point in the interval [$t^{(n)}_k,t^{(n)}_{k+1}$]. Now assume that we want to define a stochastic integral in which we replace the infinitesimal $dt$ by $dW(t)$, where the stochastic process{$W(t,\omega)$} is a Brownian motion defined on a filtered probability space ($\Omega,\{{\mathcal F}_t\},{\mathcal F},P$). The above definition is not possible to employ for defining the stochastic integral, $$I[f](\omega)="\int_{T_0}^{T_f}f(t,\omega) dW(t,\omega)",$$ since the Brownian motion {W(t)} does not have finite variation. As a consequence of this, we cannot in general define stochastic integrals $\omega$ by $\omega$.
Assume that:
  1. $f(t,\omega)$ is jointly measurable in $(t,\omega)$ for $T_0 \leq t \leq T_f$,
  2. $f(t),~T_0 \leq t \leq T_f$, is adapted to the filtration ${\mathcal F}_t=\sigma(W(s), T_0\leq s \leq t),$
  3. The Riemann integral $\int_{T_0}^{T_f} f(t)^2dt$ has finite expectation.
We can then define the stochastic integral as a limit in mean square sense, i.e. $$I[f]=\text{$L^2$-lim}_{n\rightarrow\infty}~\sum^{n-1}_{k=0} f(t^{(n)*}_k)(W(t^{(n)}_{k+1})-W(t^{(n)}_{k})).$$ However, this limit depends on what point $t^{(n)*}_k$ in each time-interval we choose to evaluate the function f(.) at. A classical example of this following Øksendahl (1995) is if we take $f(\cdot)=W(\cdot)$ and compare the limits $I_1[W]$ and $I_2[W]$ resulting from choosing $t^{(n)*}_k=t^{(n)}_k$ and $t^{(n)*}_k=t^{(n)}_{k+1}$. Thus define \begin{eqnarray*} I_1[f] & = & =\text{$L^2$-lim}_{n\rightarrow\infty}\sum_{k=0}^{n-1}W(t^{(n)}_k)(W(t^{(n)}_{k+1})-W(t^{(n)}_k)),\\ I_2[f] & = & \text{$L^2$-lim}_{n\rightarrow\infty}\sum_{k=0}^{n-1}W(t^{(n)}_{k+1})(W(t^{(n)}_{k+1})-W(t^{(n)}_k)). \end{eqnarray*} Then, by the independent increments of Brownian motion, \[EI_1[f]=E\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}W(t^{(n)}_{k})(W(t^{(n)}_{k+1})-W(t^{(n)}_k))=0,\] but \begin{eqnarray*} EI_2[f]&=&E\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}W(t^{(n)}_{k+1})(W(t^{(n)}_{k+1})-W(t^{(n)}_k))\\ &=&\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}E[(W(t^{(n)}_{k+1})-W(t^{(n)}_k))^2]\\ &=&\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}(t^{(n)}_{k+1}-t^{(n)}_k)\\ &=&\lim_{n\rightarrow\infty} (t^{(n)}_n-t^{(n)}_0)\\ &=&T_f-T_0. \end{eqnarray*} The two most common evaluation points are (a) $t^{(n)\ast}_k=t^{(n)}_k$, which leads to the Ito integral, and (b) $t^{(n)\ast}_k=(t^{(n)}_k+t^{(n)}_{k+1})/2$, which leads to the Stratonovich integral. These integrals are denoted by $\int_{T_0}^{T_f} f(t)dW(t)$ and $\int_{T_0}^{T_f} f(t) \circ\!dW(t)$ respectively. The Stratonovich calculus obeys the usual chain rule while the Ito calculus does not (see Ito's formula). The Ito integral as opposed to the Stratonovich integral is a martingale.

 

Questions: Magnus Wiktorsson
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