Stochastic Calculus

def. Ito Integral. Let the following:

• Brownian Motion $\mathfrak{B}=\{{\mathfrak{B}}_s\mid s\ge 0\}$, an Ito process $X=\{X_s\mid s\ge 0\}$
• $0=t_1<t_2<\cdots <t_n=t$ and thus $t_i=\frac{t}{n}i$
• Then the Ito integral $Y_t$ of $X(t)$ with respect to $\mathfrak{B}(t)$ is defined as such:

$$Y(t)={\int }_0^{t}X(s)d\mathfrak{B}(s):=\underset{n\to \infty }{\lim }\sum _{i=0}^{n}X\left(t_{i-1}\right)[{\mathfrak{B}}_{t_i}-{\mathfrak{B}}_{t_{i-1}}]$$

• Visualization. Think of the 3D visualization of the Riemann–Stieltjes integral, but both $f(x)$ and $g(x)$ are zigzags.
• The solution to the Ito integral is also a stochastic process $Y=\{Y_t\mid t\ge 0\}$. It also has the following properties because the integrator is ${\mathfrak{B}}_t$
  • $Y_t$ is a martingale
  • $\mathbb{E}(Y_t)=0$
  • let $Z_t={\int }_0^{t}X(s{)}^{2}d\mathfrak{B}(s)$
    • Ito Isometry: $\text{Var}(Y_t)=\mathbb{E}(Z_t)$
    • $Y_t\sim \mathcal{N}(0,\mathbb{E}(Z_t))$

Remark. We often use shorthand notation (abuse of notation) to write an integral term like this:

• $d{\mathfrak{B}}_t:={\int }_0^{t}1d{\mathfrak{B}}_s:={\sum }_{i=1}^{\infty }1\cdot ({\mathfrak{B}}_{t_i}-{\mathfrak{B}}_{t_{i-1}})$
• $X_{t}d{\mathfrak{B}}_t:={\int }_0^t{X}_{s}d{\mathfrak{B}}_s:={\sum }_{i=0}^{\infty }{X}_t({\mathfrak{B}}_{t_i}-{\mathfrak{B}}_{t_{i-1}})$
• $dt:={\int }_0^{t}1dt:={\sum }_{i=0}^{\infty }1(t_{i+1}-t_i)=t$ by definition of the Reimann Integral
• $X_{t}dt:={\int }_0^t{X}_{s}ds:=$ Now, these also exist but they are trivial (Ito Isometry):
• $d\mathfrak{B}dt=dtd\mathfrak{B}=0$
• $(dt{)}^2=0$
• $(d\mathfrak{B}{)}^2=dt$

Reference Table of Common Ito Integrals

Stochastic Integral	Result	Variance
${\int }_0^{t}d{B}_s$	$B_t$	$t$
${\int }_0^{t}sd{B}_s$	$t{B}_t-{\int }_0^t{B}_{s}ds$	$\frac{1}{3}{t}^3$
${\int }_0^t{B}_{s}d{B}_s$	$\frac{1}{2}{B}_t^2-\frac{1}{2}t$	$\frac{1}{2}{t}^2$
${\int }_0^t{B}_s^{2}d{B}_s$	$\frac{1}{3}{B}_t^3-{\int }_0^t{B}_{s}ds$	$3t^2$
${\int }_0^t{e}^{B_s-\frac{1}{2}s}d{B}_s$	$e^{B_t-\frac{1}{2}t}-1$	$e^t-1$

Motivation. Unfortunately we cannot calculate ito intergrals directly. For example, as we saw in the above example finding ${\int }_0^t\mathfrak{B}(s)d\mathfrak{B}(s)$ using just the definition was a laborious process. In order to make this easier, we use the following lemma.

thm. Ito’s Lemma. (Ito’s Chain Rule) The stochastic calculus chain rule. Stated in two ways.

• $X(t)$ is an Ito process $dX(t)=a(X,t)dt+b(X,t)d\mathfrak{B}$
• $f(X,t)$ is a normal function
• Then:

$$df(X,t)=\left(f_t+a{f}_x+\frac{b^2}{2}{f}_{xx}\right)dt+b{f}_{x}d\mathfrak{B}$$

• $f_x=\frac{\partial f}{\partial x}{|}_{x=X}$, $f_t=\frac{\partial f}{\partial t}$

Example. Suppose we want to find the integral ${\int }_0^t{\mathfrak{B}}_u^2-ud{\mathfrak{B}}_u$. Since it’s nearly impossible to find it from the definition of the Ito integral, let us use Ito’s lemma. Arbitrarily set an ordinary function $f(x,t)=\frac{x^3}{3}-tx$.

• ! Now, it is important to note that we normally don’t know what function to use, but in this case assume we are given a nice $f(x,t)$ that will lead to the answer. Given this function, we then have:
• $f_x=x^2-t$, $f_{xx}=2x$, and $f_t=-x$
• Because $\mathfrak{B}$ is a standard brownian motion it conforms to the form $d\mathfrak{B}=0\cdot dt+1\cdot d\mathfrak{B}$ which means for Ito’s lemma $a=0,b=1$. Then, we use Ito’s lemma:

$$\begin{array}{rl}df(\mathfrak{B},t)& =\left(f_t+a{f}_x+\frac{b^2}{2}{f}_{xx}\right)dt+b{f}_{x}d\mathfrak{B}\\ f(B_t,t)-\text{cancelto}0f(B_0,0)& =\text{cancelto}0\left(-x+0+\frac{1}{2}\cdot 2x\right)dt+(x^2-t)d\mathfrak{B}\\ \frac{{\mathfrak{B}}_t^3}{3}-t{\mathfrak{B}}_t& =({\mathfrak{B}}^2-t)d\mathfrak{B}\\ & ={\int }_0^t{\mathfrak{B}}_u-ud{\mathfrak{B}}_u\end{array}$$

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Reference.

Motion: $X_t$	Differential: $d{X}_t=udt+ud{B}_t$
$B_t$	$d{B}_t$
$B_t^2$	$2B_{t}d{B}_t+dt$
$B_t^2-t$	$2B_{t}d{B}_t$
$B_t^3$	$3B_t^{2}d{B}_t+3B_{t}dt$
$e^{B_t}$	$e^{B_t}d{B}_t+\frac{1}{2}{e}^{B_t}dt$
$e^{B_t-\frac{1}{2}t}$	$e^{B_t-\frac{1}{2}t}d{B}_t$
$e^{\frac{1}{2}t}\sin B_t$	$e^{\frac{1}{2}t}\cos B_{t}d{B}_t$
$e^{\frac{1}{2}t}\cos B_t$	$-e^{\frac{1}{2}t}\sin B_{t}d{B}_t$
$(B_t+t)e^{-B_t-\frac{1}{2}t}$	$(1-B_t-t)e^{-B_t-\frac{1}{2}t}d{B}_t$

thm. Ito’s Product Rule. for two stochastic processes $X_t,Y_t$

$$d(X_t{Y}_t)=X_{t}d{Y}_t+Y_{t}d{X}_t+d{X}_t{d}_t$$

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Ito Processes §

def. Ito Process. An Ito process $X(t)=\{X_t\mid t\ge 0\}$ is a type of Stochastic Process that is defined by a certain $a(X_t,t),b(X_t,t)$:

$$X(t)=X(0)+{\int }_0^{t}a(X_s,s)ds+{\int }_0^{t}b(X_s,s)d\mathfrak{B}$$

• $a(X_t,t)$ must be integrable in the ordinary sense
• $b(X_t,t)$ must be integrable in the stochastic sense
  • It actually needs to be integrable twice As an abuse of notation we can write:

$$dX=a(X_t,t)\cdot ds+b(X_t,t)\cdot d\mathfrak{B}$$

• This is called a Stochastic Differential Equation (even if it’s not actually a differential equation). Essentially, a function of an Ito process is also an Ito process.

Properties.

• $\mathbb{E}(dY\mid {\mathcal{F}}_t)=(f_t+a{f}_x+\frac{b^2}{2}{f}_{xx})\cdot dt$
• $\text{Var}(dY\mid {\mathcal{F}}_t)=(b{f}_x{)}^2$