1. Derive Ito’s Lemma

Recall that the Chain Rule for non-stochastic functions:

\[ \begin{aligned} \frac{d}{dx}f(g(x)) &= f'(g(x))g'(x) \\ &= f'(g(x))dg(x) \end{aligned} \]

where \(f\) and \(g\) are differentiable. It can also be expressed as

\[\frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx}\] where \(y\) is a function of \(x\) and \(z\) is a function of \(y\).

Extended to integral:

\[\int_a^bf(g(x))g'(x)dx = \int_{g(a)}^{g(b)} f(u)du = \int_a^bf(g(x))dg(x)\] where \(u = g(x)\)

However, all of the above only works when function \(g\) is differentiable. In a case where \(g(t) = W(t)\) which is a Brownian motion, a non-differentiable function, we can’t use the Chain Rule as described but we need a new rule.

Luckily Taylor’s Expansion come to save us. The Taylor’s Expansion is:

\[ \begin{aligned} d(f(g(t))) &= f(g(t)+dg(t)) - f(g(t)) \\ &= f'(g(t))dg(t) + \frac{1}{2}f^{(2)}(g(t)) \cdot [dg(t)]^2 + ..\frac{1}{k!}f^{(k)}(g(t)) \cdot [dg(t)]^k + ... \end{aligned} \]

Plug in \(g(t) = W(t)\) and ignoring the higher order terms (\(k>=3\)), then we can get the Ito’s Lemma:

\[ \begin{aligned}df(W(t)) &= f(W(t)+dW(t)) - f(W(t)) \\ &= f'(W(t))dW(t) + \frac{1}{2} f^{(2)}(W(t))[dW(t)]^2\\ &= f'(W(t))dW(t) + \frac{1}{2} f^{(2)}(W(t))dt \\ \end{aligned} \]

The 3rd step of the above equation is an important property of the Wiener process and will be shown in the following paragraph.

2. Wiener Process

Properties of Wiener Process:

  • \(W_0 = 0\)
  • \(W_n = W_0 + \sum_{t=1}^n \Delta W_t\)
  • Increments \(\Delta W_t\) are i.i.d. and \(W_t-W_s \sim \mathcal{N}(0,t-s)\) where \(t-s\) is the variance

3. \((dW_t)^2 = dt\)

Given that \(t_0 = 0 < t_1 < t_2 < ...t_n = t\) and \(t_{i}-t_{i-1} = \frac{t}{n}\) and \(n \to \infty\) then \[W_{t_i}-W_{t_{i-1}} \sim \mathcal{N}(0,\frac{t}{n})\]

\[\int_0^tdW_{t_i} = \sum_{i=0}^{n}(W_{t_i}-W_{t_{i-1}})\]

\[ \begin{aligned} \int_0^t(dW_s)^2 &= \sum_{i=0}^{n}(W_{t_i}-W_{t_{i-1}})^2 \\ when \ \ {n \to \infty} \ \ & =n \cdot E[(W_{t_i}-W_{t_{i-1}})^2]\\ &= n \cdot Var(W_{t_i}-W_{t_{i-1}}) \\ &= n \cdot \frac{t}{n} = t\\ \end{aligned} \]

By taking the derivative, we can derive that \((dW_t)^2 = dt\)