April 25, 2024
A large class of stochastic processes can be defined using standard Brownian motion (B_t)_{t\ge 0}.
The stochastic differential equation dX = \mu\,dt + \sigma\,dB, X_{t_0} = x_0 is shorthand for X_t = x_0 + \int_{t_0}^t \mu\,ds + \int_{t_0}^t \sigma\,dB_s, which is shorthand for X_t(\omega) = x_0 + \int_{t_0}^t \mu(s,\omega)\,ds + \int_{t_0}^t \sigma(s,\omega)\,dB_s(\omega), where \mu,\sigma\colon [0,\infty)\times\Omega\to\boldsymbol{R} are the drift and diffusion coefficients, respectively. The first integral is (pointwise in \omega) Riemann and the second is the Ito integral.
Just as an integral is a continuous linear transformation from a vector space of functions to the real numbers, a stochastic integral is a continuous linear transformation from a vector space of functions to a vector space of random variables.
Recall that Brownian motion is a stochastic process B_t\colon\Omega\to\boldsymbol{R}, t\ge 0, where B_t(\omega) = \omega(t) for \omega\in\Omega = C[0,\infty), the space of continuous function from [0,\infty) to the real numbers \boldsymbol{R}. The information available at time t is specified by \mathcal{F}_t, the smallest \sigma-algebra for which (B_s)_{0\le s\le t} are measurable.
If \omega\mapsto\sigma(t, \omega) is \mathcal{F}_t measurable (only depends on the information available at time t) then so is the Ito integral \int_0^t \sigma(s, \omega)\,dB_s(\omega) defined as a limit \int_0^t \sigma(s, \omega)\,dB_s(\omega) = \lim_{\Delta T\to 0} \sum_{0\le j < n} \sigma(t_j, \omega)\,\Delta B_j(\omega) where 0 = t_0 < \cdots < t_n = t is a partition of 0,t and \Delta B_j = B_{t_{j+1}} - B_{t_j}. We let T = \{t_j\} denote the partition and define \Delta T = \max_{0\le j < n}\{\Delta t_j\}. The set of all partitions of 0,t is a net and the limit is defined as described therein.
Exercise. Show E\\sum\_{0\\le j \< n} B\_{t_j}\\Delta B_j = 0 for any partition 0 = t_0 < \cdots < t_n = t of 0,t.
This shows E\\int_0\^t B_s\\,dB_s = 0. Later we will see \int_0^t B_s dB_s = (B_t^2 - t)/2.
Exercise. Show E\\sum\_{0\\le j \< n} B\_{t\_{j+1}}\\Delta B_j = t \not= 0.
Hint: EB_t B_u = \min\{t, u\}. Note the integrand \omega\mapsto B_{t_{j+1}}(\omega) is not \mathcal{F}_{t_j} measurable.
The Riemann integral \int_0^t f(s)\,ds = \lim_{\Delta T\to 0} \sum_j f(t_j^*)\Delta t_j converges to the same value if f is continuous and t_j^* is any point in t_j, t\_{j+1}. Unlike the Riemann integral, the Ito integral requires the left endpoint be used.
Exercise. Show E\\sum\_{0\\le j \< n} (\\Delta B_j)\^2 = t for any partition.
Hint: E(\\Delta B)\^2 = \Delta t.
Exercise. Show \lim_{\Delta T\to 0}\sum_{0\le j < n} (\Delta t_j)^2 = 0.
Hint: \sum_j (\Delta t_j)^2 \le (\max_j \Delta t_j) \sum_j \Delta t_j = (\max_j \Delta t_j)t.
Shorthand notation for this is (dt)^2 = 0.
Exercise Show \lim_{\Delta T\to 0}\operatorname{Var}\\sum\_{0\\le j \< n} (\\Delta B_j)\^2 = 0.
Hint: E(\\Delta B)\^4 = 3(\Delta t)^2.
Although \sum_{0\le j < n} (\Delta B_j)^2 is random for any given partition, it always has expected value t and its variance tends to 0 as the partition gets finer. Shorthand notation for this is (dB)^2 = dt.
Exercise Show \sum_{0\le j < n} \Delta B_j \Delta t_j has mean 0 and its variance tends to 0 as \Delta T\to 0.
Shorthand notation for this is dB\,dt = 0.
The Ito integral can be extended to m-dimensional Brownian motion, B_t = (B_t^1,\ldots,B_t^m), where the B_t^j are independent standard Brownian motions. In this case \mu\colon [0,\infty)\times\Omega\to\boldsymbol{R}^n is vector-valued and \sigma\colon [0,\infty)\times\Omega\to\boldsymbol{R}^{n,m} is matrix-valued. The \sigma dB term is the matrix-vector product.
The Ito integral can be generalized to stochastic process other than Brownian motion. If X_t is any stochastic process we can define dY_t = \sigma\,dX_t, Y_0 = y, by Y_t(\omega) = y + \lim_{\Delta T\to 0}\sum_j \sigma(t_j, \omega) \Delta X_j(\omega) where \Delta X_j = X_{t_{j+1}} - X_{t_j}. For Y_t to be \mathcal{F}_t-measurable we need \sigma to be adapted. Of course there must be restrictions on X_t too in order to ensure convergence and continuity. If X_t is a martingale that is (almost surely) continuous from the right and has left limits then a well-behaved stochastic integral can be define. Note Brownian motion satisfies this since it is almost everywhere continuous. This can be generalized by adding a stochastic process to the martingale that has bounded variation. As shown in @Pro04, this is the most general process for which a well-behaved stochastic integral can be defined.
An Ito process X_t:\Omega\to\boldsymbol{R}, t\ge0, satisfies the SDE dX_t(\omega) = \mu(t,\omega)\,dt + \sigma(t,\omega)\,dB_t(\omega), X_{t_0} = x_0, t\ge t_0. where \mu,\sigma:[0,\infty)\times\Omega\to\boldsymbol{R} are functions of time and the Brownian sample path \omega\in\Omega. If \omega\mapsto \mu(t, \omega) is \mathcal{F}_t measurable for all t then so is the Riemann integral \omega\mapsto \int_0^t \mu(s, \omega)\,ds. If \omega\mapsto \sigma(t, \omega) is \mathcal{F}_t measurable for all t then so is the Ito integral \omega\mapsto \int_0^t \sigma(s, \omega)\,dB_s(\omega).
Exercise. In this case the Ito integral \int_0^t \sigma dB is a martingale.
Hint: EB_u - B_t\\mid\\mathcal{F}\_s = 0 if s\le t\le u.
Exercise. If dX = \mu\,dt + \sigma\,dB is a martingale then \mu = 0.
Hint: Show \int_t^u E_t\\mu(s,\\omega)\,ds = 0 for t\le u, \omega\in\Omega.
Exercise. The sum of Ito processes is an Ito process.
Hint: If dX^j = \mu_j\,dt + \sigma_j\,dB are Ito processes what are \mu and \sigma for X = \sum_j X_j? Write out the integrals in full.
If X_t and Y_t are Ito processes then so is their product X_tY_t.
Exercise. Show d(XY) = Y\,dX + X\,dY + dX\,dY.
Hint: Show X_t Y_t = X_0 Y_0 + \lim_{\Delta T\to 0}\sum_j Y_j\Delta X_j + Y_j\Delta Y_j + \Delta X_j\Delta Y_j where where \Delta X_j = X_{t_{j+1}} - X_{t_j}, etc., using (X_j + \Delta X_j)(Y_j + \Delta Y_j) = X_jY_j + Y_j\Delta X_j + X_j\Delta Y_j + \Delta X_j \Delta Y_j.
Exercise. If dX = \mu\,dt + \sigma\,dB and dY = \nu\,dt + \tau\,dB show d(XY) = (\mu Y + \nu X + \sigma\tau)\,dt + (\sigma + \tau)\,dB.
Hint: Show \begin{align*} X_t(\omega)Y_t(\omega) = &X_0 Y_0 \\ &+ \int_0^t (\mu(s,\omega) Y_s(\omega) + \nu(s,\omega) X_s(\omega) + \sigma(s, \omega)\tau(s,\omega))\,ds \\ &+ \int_0^t (\sigma(s,\omega) + \tau(s, \omega))\,dB_s(\omega) \\ \end{align*}
The Ito calculus uses (dt)^2 = 0, dt\,dB = 0 = dB\,dt, and (dB)^2 = 0 to simplify such calculations. \begin{align*} d(XY) &= Y\,dX + X\,dY + dX\,dY \\ &= Y(\mu\,dt + \sigma\,dB) + X(\nu\,dt + \tau\,dB) + (\mu\,dt + \sigma\,dB)(\nu\,dt + \tau\,dB) \\ &= (\mu Y + \nu X + \sigma\tau)\,dt + (\sigma + \tau)\,dB. \end{align*}
An Ito diffusion \bar{X}_t(\omega) satisfies d\bar{X}_t(\omega) = \bar{\mu}(t,\bar{X}_t(\omega))\,dt + \bar{\sigma}(t,\bar{X}_t(\omega))\,dB_t(\omega), \bar{X}_{t_0} = x, t \ge t_0 where \bar{\mu},\bar{\sigma}\colon [0,\infty)\times\boldsymbol{R}\to\boldsymbol{R}.
Ito diffusions satisfy the Markov property. Loosely speaking, Ef(t, \\bar{X}\_t)\\mid\\mathcal{F}\_t = Ef(t, \\bar{X}\_t)\\mid \\bar{X}\_t. The conditional expectation does not depend on the trajectory s\mapsto X_s(\omega), 0\le s\le t, it only depends on the final value X_t(\omega).
If X_t is an Ito diffusion and f\colon[0,\infty)\times\boldsymbol{R}\to\boldsymbol{R} then Y_t = f(t, X_t) is also an Ito diffusion satisfying the SDE dY_t = f_t(t, X_t)\,dt + f_x(t, X_t)\,dX_t + \frac{1}{2} f_{xx}(t, X_t) (dX_t)^2 If X_t = B_t is standard Brownian motion then dY_t = (f_t + \frac{1}{2}f_{xx})\,dt + f_x\,dB.
Exercise. If f_t + \frac{1}{2}f_{xx} = 0 then f(t,B_t) is a martingale.
Exercise. Show f(t, x) = x^2 - t, f(t, x) = se^{\sigma x - \sigma^2t/2}, and f(t, x) = e^{-x^2/2t}/\sqrt{2\pi t} satisfy f_t + \frac{1}{2}f_{xx} = 0.