📂Stochastic Differential Equations

Ito Multiplication Table

Build-up

$s< t < t+u$ Suppose, meet the following conditions for a stochastic process $\left\{ W_{t} \right\}$ to be called a Wiener process.

• (i): $W_{0} = 0$
• (ii): $\left( W_{t+u} - W_{t} \right) \perp W_{s}$
• (iii): $\left( W_{t+u} - W_{t} \right) \sim N ( 0, u )$
• (iv): Sample paths of $W_{t}$ are almost surely continuous.

The Wiener process has the following properties:

• [1]: $\displaystyle W_{t} \sim N ( 0 , t )$
• [2]: $\displaystyle E ( W_{t} ) = 0$
• [3]: $\displaystyle \operatorname{Var} ( W_{t} ) = t$
• [4]: $\displaystyle \text{cov} ( W_{t} , W_{s} ) = {{1} \over {2}} (|t| + |s| - |t-s|) = \min \left\{ t , s \right\}$

Consider a very short infinitesimal interval $\left[ t , t + d t \right]$ of the Wiener process $\left\{ W_{t} \right\}_{t \ge 0}$. Although it is not an analytically rigorous assumption, let $dt > 0$ be $\left( dt \right)^{1/2} > 0$, and small enough to be considered as $\left( dt \right)^{k} = 0$ for any $k = 2 , 3, \cdots$. In algebraic terms, under this assumption, we treat $\alpha + \beta dt$ as an infinitesimal.

Now, suppose $dW_{t} := W_{t + dt} - W_{t}$, and we want to consider the multiplications between $dt$ and $d W_{t}$.

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Part 1. $\left( dt \right)^{2} = 0$

Of course, $dt > 0$ holds, but let’s assume $dt$ is small enough such that $\left( dt \right)^{2} = 0$.

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Part 2. $dt d W_{t} = 0$

Since $W_{t}$ is assumed to follow a Wiener process, it adheres to a normal distribution as $d W_{t} \sim N \left( 0, \sqrt{dt}^{2} \right)$.

Properties of mean and variance:

• [2]: $E(aX + b) = a E(X) + b$
• [5]: $\operatorname{Var} (aX + b) = a^2 \operatorname{Var} (X)$

The expectation of $dt d W_{t}$ brings the constant $dt$ outside, $$ E \left( dt d W_{t} \right) = dt E \left( d W_{t} \right) = dt \cdot 0 = 0 $$ Similarly, the variance of $dt d W_{t}$ also removes the squared $dt$, $$ \operatorname{Var} \left( dt d W_{t} \right) = (dt)^{2} \operatorname{Var} \left( d W_{t} \right) = 0 \cdot \operatorname{Var} \left( d W_{t} \right) = 0 $$ Thus, since $dt d W_{t}$ has a variance of $0$ and an expectation of $0$, it must be $$ dt d W_{t} = d W_{t} dt = 0 $$

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Part 3. $\left( d W_{t} \right)^{2} = dt$

From $\operatorname{Var} \left( d W_{t} \right) = dt$, computing the expectation of $d W_{t} \cdot d W_{t}$, $$ \begin{align*} dt =& \operatorname{Var} \left( d W_{t} \right) \\ =& E \left( \left( d W_{t} \right)^{2} \right) - \left[ E \left( d W_{t} \right) \right]^{2} \\ =& E \left( \left( d W_{t} \right)^{2} \right) - 0^{2} \end{align*} $$ Thus, it is $E \left( \left( d W_{t} \right)^{2} \right) = dt$.

Expectation of even powers of normally distributed random variables with zero mean: Given a random variable $X$ following a normal distribution as $N \left( 0 , \sigma^{2} \right)$, the expectation of its even power $X^{n}$ can be recursively expressed as $$ E \left( X^{n} \right) = (n - 1) \sigma^{2} E \left( X^{n-2} \right) $$ The equation $E \left( X^{n} \right)$ holds $n$ when it is odd, but for even, it is given by $$ E \left( X^{2n} \right) = \left( 2n - 1 \right)!! \sigma^{2n} $$ Here, the double exclamation mark $k!! = k \cdot \left( k - 2 \right) \cdots$ denotes the double factorial.

Assuming $d W_{t} \sim N \left( 0, \sqrt{dt}^{2} \right)$, the variable $d W_{t}$ follows a normal distribution with zero mean, and the variance of $\left( d W_{t} \right)^{2}$ follows from the expectation of the squared random variable $E \left( X^{2n} \right) = \left( 2n - 1 \right)!! \sigma^{2n}$, $$ \begin{align*} \operatorname{Var} \left( \left( d W_{t} \right)^{2} \right) =& E \left( \left[ \left( d W_{t} \right)^{2} \right]^{2} \right) - \left[ E \left( \left( d W_{t} \right)^{2} \right) \right]^{2} \\ =& E \left( \left( d W_{t} \right)^{2 \cdot 2} \right) - \left[ dt \right]^{2} \\ =& \left( 2 \cdot 2 - 1 \right) \sqrt{dt}^{2 \cdot 2} - dt^{2} \\ =& 3 dt^{2} - dt^{2} \\ =& 2 dt^{2} \\ =& 0 \end{align*} $$ Therefore, $\left( d W_{t} \right)^{2}$ should carry an expected value of $dt$ followed by $$ \left( d W_{t} \right)^{2} = dt $$

Summary ¹

Assuming $\alpha + \beta dt$ to be an infinitesimal, the product of $dt$ and $d W_{t}$ is given by, $$ \begin{align*} \left( dt \right)^{2} =& 0 \\ dt d W_{t} =& 0 \\ d W_{t} dt =& 0 \\ \left( d W_{t} \right)^{2} =& dt \end{align*} $$