title: Ito Processes author: Keith A. Lewis institution: KALX, LLC email: [email protected] classoption: fleqn abstract: Brownian motion and the heat equation. ...
A large class of stochastic processes can be defined using standard Brownian motion [(B_t)_{t\ge 0}]{.math .inline}.
The stochastic differential equation [ dX = \mu\,dt + \sigma\,dB, X_{t_0} = x_0 ]{.math .display} is shorthand for [ X_t = x_0 + \int_{t_0}^t \mu\,ds + \int_{t_0}^t \sigma\,dB_s, ]{.math .display} 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), ]{.math .display} where [\mu,\sigma\colon [0,\infty)\times\Omega\to\boldsymbol{R}]{.math .inline} are the drift and diffusion coefficients, respectively. The first integral is (pointwise in [\omega]{.math .inline}) Riemann and the second is the Ito integral.
::: {#ito-integral .section .level2}
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}]{.math .inline}, [t\ge 0]{.math .inline}, where [B_t(\omega) = \omega(t)]{.math .inline} for [\omega\in\Omega = C[0,\infty)]{.math .inline}, the space of continuous function from [[0,\infty)]{.math .inline} to the real numbers [\boldsymbol{R}]{.math .inline}. The information available at time [t]{.math .inline} is specified by [\mathcal{F}_t]{.math .inline}, the smallest [\sigma]{.math .inline}-algebra for which [(B_s)_{0\le s\le t}]{.math .inline} are measurable.
If [\omega\mapsto\sigma(t, \omega)]{.math .inline} is [\mathcal{F}_t]{.math .inline} measurable (only depends on the information available at time [t]{.math .inline}) then so is the Ito integral [\int_0^t \sigma(s, \omega)\,dB_s(\omega)]{.math .inline} 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) ]{.math .display} where [0 = t_0 < \cdots < t_n = t]{.math .inline} is a partition of [[0,t]]{.math .inline} and [\Delta B_j = B_{t_{j+1}} - B_{t_j}]{.math .inline}. We let [T = \{t_j\}]{.math .inline} denote the partition and define [\Delta T = \max_{0\le j < n}\{\Delta t_j\}]{.math .inline}. The set of all partitions of [[0,t]]{.math .inline} 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]{.math .inline} for any partition [0 = t_0 < \cdots < t_n = t]{.math .inline} of [[0,t]]{.math .inline}.
This shows [E[\int_0^t B_s\,dB_s] = 0]{.math .inline}. Later we will see [\int_0^t B_s dB_s = (B_t^2 - t)/2]{.math .inline}.
Exercise. Show [E[\sum_{0\le j < n} B_{t_{j+1}}\Delta B_j] = t \not= 0]{.math .inline}.
Hint: [E[B_t B_u] = \min\{t, u\}]{.math .inline}. Note the integrand [\omega\mapsto B_{t_{j+1}}(\omega)]{.math .inline} is not [\mathcal{F}_{t_j}]{.math .inline} measurable.
The Riemann integral [\int_0^t f(s)\,ds = \lim_{\Delta T\to 0} \sum_j f(t_j^*)\Delta t_j]{.math .inline} converges to the same value if [f]{.math .inline} is continuous and [t_j^*]{.math .inline} is any point in [[t_j, t_{j+1}]]{.math .inline}. 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]{.math .inline} for any partition.
Hint: [E[(\Delta B)^2] = \Delta t]{.math .inline}.
Exercise. Show [\lim_{\Delta T\to 0}\sum_{0\le j < n} (\Delta t_j)^2 = 0]{.math .inline}.
Hint: [\sum_j (\Delta t_j)^2 \le (\max_j \Delta t_j) \sum_j \Delta t_j = (\max_j \Delta t_j)t]{.math .inline}.
Shorthand notation for this is [(dt)^2 = 0]{.math .inline}.
Exercise Show [\lim_{\Delta T\to 0}\operatorname{Var}[\sum_{0\le j < n} (\Delta B_j)^2] = 0]{.math .inline}.
Hint: [E[(\Delta B)^4] = 3(\Delta t)^2]{.math .inline}.
Although [\sum_{0\le j < n} (\Delta B_j)^2]{.math .inline} is random for any given partition, it always has expected value [t]{.math .inline} and its variance tends to 0 as the partition gets finer. Shorthand notation for this is [(dB)^2 = dt]{.math .inline}.
Exercise Show [\sum_{0\le j < n} \Delta B_j \Delta t_j]{.math .inline} has mean 0 and its variance tends to 0 as [\Delta T\to 0]{.math .inline}.
Shorthand notation for this is [dB\,dt = 0]{.math .inline}. :::
::: {#higher-dimensions .section .level2}
The Ito integral can be extended to [m]{.math .inline}-dimensional Brownian motion, [B_t = (B_t^1,\ldots,B_t^m)]{.math .inline}, where the [B_t^j]{.math .inline} are independent standard Brownian motions. In this case [\mu\colon [0,\infty)\times\Omega\to\boldsymbol{R}^n]{.math .inline} is vector-valued and [\sigma\colon [0,\infty)\times\Omega\to\boldsymbol{R}^{n,m}]{.math .inline} is matrix-valued. The [\sigma dB]{.math .inline} term is the matrix-vector product.
::: {#stochastic-integral .section .level3}
The Ito integral can be generalized to stochastic process other than Brownian motion. If [X_t]{.math .inline} is any stochastic process we can define [dY_t = \sigma\,dX_t]{.math .inline}, [Y_0 = y]{.math .inline}, by [ Y_t(\omega) = y + \lim_{\Delta T\to 0}\sum_j \sigma(t_j, \omega) \Delta X_j(\omega) ]{.math .display} where [\Delta X_j = X_{t_{j+1}} - X_{t_j}]{.math .inline}. For [Y_t]{.math .inline} to be [\mathcal{F}_t]{.math .inline}-measurable we need [\sigma]{.math .inline} to be adapted. Of course there must be restrictions on [X_t]{.math .inline} too in order to ensure convergence and continuity. If [X_t]{.math .inline} 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]]{.citation cites="Pro04"}, this is the most general process for which a well-behaved stochastic integral can be defined. ::: :::
::: {#ito-process .section .level2}
An Ito process [X_t:\Omega\to\boldsymbol{R}]{.math .inline}, [t\ge0]{.math .inline}, 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. ]{.math .display} where [\mu,\sigma:[0,\infty)\times\Omega\to\boldsymbol{R}]{.math .inline} are functions of time and the Brownian sample path [\omega\in\Omega]{.math .inline}. If [\omega\mapsto \mu(t, \omega)]{.math .inline} is [\mathcal{F}_t]{.math .inline} measurable for all [t]{.math .inline} then so is the Riemann integral [\omega\mapsto \int_0^t \mu(s, \omega)\,ds]{.math .inline}. If [\omega\mapsto \sigma(t, \omega)]{.math .inline} is [\mathcal{F}_t]{.math .inline} measurable for all [t]{.math .inline} then so is the Ito integral [\omega\mapsto \int_0^t \sigma(s, \omega)\,dB_s(\omega)]{.math .inline}.
Exercise. In this case the Ito integral [\int_0^t \sigma dB]{.math .inline} is a martingale.
Hint: [E[B_u - B_t\mid\mathcal{F}_s] = 0]{.math .inline} if [s\le t\le u]{.math .inline}.
Exercise. If [dX = \mu\,dt + \sigma\,dB]{.math .inline} is a martingale then [\mu = 0]{.math .inline}.
Hint: Show [\int_t^u E_t[\mu(s,\omega)]\,ds = 0]{.math .inline} for [t\le u]{.math .inline}, [\omega\in\Omega]{.math .inline}.
Exercise. The sum of Ito processes is an Ito process.
Hint: If [dX^j = \mu_j\,dt + \sigma_j\,dB]{.math .inline} are Ito processes what are [\mu]{.math .inline} and [\sigma]{.math .inline} for [X = \sum_j X_j]{.math .inline}? Write out the integrals in full.
If [X_t]{.math .inline} and [Y_t]{.math .inline} are Ito processes then so is their product [X_tY_t]{.math .inline}.
Exercise. Show [d(XY) = Y\,dX + X\,dY + dX\,dY]{.math .inline}.
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]{.math .inline} where where [\Delta X_j = X_{t_{j+1}} - X_{t_j}]{.math .inline}, 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]{.math .inline}.
Exercise. If [dX = \mu\,dt + \sigma\,dB]{.math .inline} and [dY = \nu\,dt + \tau\,dB]{.math .inline} show [d(XY) = (\mu Y + \nu X + \sigma\tau)\,dt + (\sigma + \tau)\,dB]{.math .inline}.
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*} ]{.math .display}
The Ito calculus uses [(dt)^2 = 0]{.math .inline}, [dt\,dB = 0 = dB\,dt]{.math .inline}, and [(dB)^2 = 0]{.math .inline} 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*} ]{.math .display} :::
::: {#ito-diffusion .section .level2}
An Ito diffusion [\bar{X}_t(\omega)]{.math .inline} 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 ]{.math .display} where [\bar{\mu},\bar{\sigma}\colon [0,\infty)\times\boldsymbol{R}\to\boldsymbol{R}]{.math .inline}.
Ito diffusions satisfy the Markov property. Loosely speaking, [E[f(t, \bar{X}_t)\mid\mathcal{F}_t] = E[f(t, \bar{X}_t)\mid \bar{X}_t]]{.math .inline}. The conditional expectation does not depend on the trajectory [s\mapsto X_s(\omega)]{.math .inline}, [0\le s\le t]{.math .inline}, it only depends on the final value [X_t(\omega)]{.math .inline}. :::
::: {#ito-formula .section .level2}
If [X_t]{.math .inline} is an Ito diffusion and [f\colon[0,\infty)\times\boldsymbol{R}\to\boldsymbol{R}]{.math .inline} then [Y_t = f(t, X_t)]{.math .inline} 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 ]{.math .display} If [X_t = B_t]{.math .inline} is standard Brownian motion then [dY_t = (f_t + \frac{1}{2}f_{xx})\,dt + f_x\,dB]{.math .inline}.
Exercise. If [f_t + \frac{1}{2}f_{xx} = 0]{.math .inline} then [f(t,B_t)]{.math .inline} is a martingale.
Exercise. Show [f(t, x) = x^2 - t]{.math .inline}, [f(t, x) = se^{\sigma x - \sigma^2t/2}]{.math .inline}, and [f(t, x) = e^{-x^2/2t}/\sqrt{2\pi t}]{.math .inline} satisfy [f_t + \frac{1}{2}f_{xx} = 0]{.math .inline}. :::