Stochastic Integration: A Quick Summary

What follows is a heuristic derivation of the Stochastic Integral, Stochastic Differential Equations and Itô’s Formula.

First note that for $(B_t: t \geq 0)$ a standard Brownian motion argue that, for all $T$ and for $\delta$ sufficiently small and positive,

$\label{SI:BMprop} \sum_{t\in \{0,\delta,..,T\}} \left(B_{t+\delta} - B_t\right)= B_T\quad \text{and}\quad \sum_{t\in \{0,\delta,..,T\}} \left(B_{t+\delta} - B_t\right)^2 \approx T$

The 1st sum is an interpolating sum. By independent increments property of Brownian motion, the 2nd sum adds IIDRVs with each with mean $\delta$ . Thus the strong law of large numbers gives the approximation. From this it is reasonable to expect that

$\sum_{t\in \{0,\delta,..,T\}} \sigma(X_t)\left(B_{t+\delta} - B_t\right) \approx \int_0^T \sigma(X_t)dB_t$

and

$\sum_{t\in \{0,\delta,..,T\}} \mu(X_t)\left(B_{t+\delta} - B_t\right)^2 \approx \int_0^T \mu(X_t)dt.$

The first sum, above, is approximation from a Riemann-Stieltjes integral, i.e.

$\int_{0}^{T} f(t)dg(t) \approx \sum_{t\in \{0,\delta,..,T\}} f(t) (g(t+\delta)-g(t)).$

So one might expect a integral limit. (This is unrigorous because Riemann-Stieltjes Integration only applies to functions with finite variation – while Brownian motion does not have finite variation.) The second sum is a Riemann integral upon using the approximation $\left(B_{t+\delta} - B_t\right)^2\approx \delta$ . This is, very roughly, how a stochastic integral is defined.

We can also define stochastic differential equations. If we inductively define $X_t$ by the recursion

$\label{SDE:Recurs} X_{t+\delta} - X_t = \sigma(X_t) (B_{t+\delta} - B_t) + \mu(X_t) \delta, \qquad t=0,\delta, 2\delta,....$

then, by summing over values of $t\in \{ 0, \delta,....,T-\delta\}$ , we expect $X_t$ to approximately obey an equation of the form

$X_T = X_0 + \int_0^T \sigma(X_t) dB_t + \int_0^T \mu(X_t) dt.$

This gives a Stochastic Differential Equation.

Often in differential and integration, we apply chain rule, $\frac{d f(x_t)}{dt} = f'(x_t) \frac{dx_t}{dt}$ . Ito’s the analogous result for Stochastic Integrals. Let $X_t$ be as above. For a twice continuously differentiable function $f$ and $\delta>0$ small, we can apply a Taylor approximation

$\begin{aligned} &f(X_{t+\delta})-f(X_t)\\ =& f(X_t + \sigma(X_t) (B_{t+\delta} - B_t) + \mu(X_t) \delta ) - f(X_t)\\ =& f'(X_t) \left\{\mu \delta + \sigma \cdot (B_{t+\delta}-B_t) \right\} + \frac{f''(X_t)}{2} \left\{ \mu \delta + \sigma\cdot (B_{t+\delta} - B_t) \right\}^2 + o(\delta)\\ =& f'(X_t) \left\{\mu \delta + \sigma \cdot (B_{t+\delta}-B_t) \right\} + \frac{f''(X_t)}{2} \sigma^2\cdot (B_{t+\delta} - B_t)^2 + o(\delta) \end{aligned}$

In the last equality we use that $(B_{t+\delta}-B_t)=o(\delta^{1/2})$ (which follows from ). Thus we see that

$f(X_{t+\delta}) - f(X_t) \approx \left[f'(X_t)\mu(X_t) + \frac{\sigma(X_t)^2}{2}f''(X_t)\right] \delta + f'(X_t) \sigma(X_t) \left(B_{t+\delta}-B_t\right) .$