Ito integral

確定性函數的鏈鎖律(Chain rule for deterministic function)

• $f$ is a deterministic function, then
• $F(x) = \int_s^t f(x)dx = \int_s^t dF(x) = F(t) - F(s)$.
• Assuming $f$, $g$ are differentiable, by the chain rule

• $f(g(x))^{'} = f^{'}(g(x))\cdot g^{'}(x)$ or $df(g) = f^{'}dg$.

  • E.g. $f(x) = x^3$, $g(x) = e^{2x}$.
  • $df(g) = (e^{2x})^3 = (e^{6x})^{'} = 6e^{6x}$.
• $f(g(t)) - f(g(0)) = \int_0^t f^{'}(g(s))g^{'}(s)ds = \int_0^t f^{'}(g(r))dg(r)$.

Weiner integral

• $\{ X(t), \ t \geq 0 \}$ is Brownian process, $X(t) \sim N(0, \sigma^2 t)$.
• $f$ is a deterministic function, continuous and differentiable on $[a,b]$.
• Let $\Delta: a=t_0 < t_1 < t_2 < \cdots < t_n = b$, and $\max(t_k - t_{k-1}) \rightarrow 0$.
• $\therefore X(t_k) - X(t_{k-1}) \sim N(0, \sigma^2 (t_k - t_{k-1})$.
• 可得$\int_a^b f(t) dX(t) = f(b)X(b) - f(a)X(a) - \int_a^b X(t)df(t)$.

$\begin{array}{rcl} \sum_{k=1}^n f(t_{k-1})(X(t_k) - X(t_{k-1}) & = & f(t_0)(X(t_1) - X(t_0)) + f(t_1)(X(t_2) - X(t_1))+ \cdots + f(t_{n-1})(X(t_n) - X(t_{n-1})) \\ & = & f(t_n)X(t_n) - f(t_0)X(t_0) - X(t_1)(f(t_1) - f(t_0)) - \cdots X(t_n)(f(t_n) - f(t_{n-1})) \\ & = & f(b)X(b) - f(a)X(a) - \sum_{k=1}^n X(t_k)(f(t_k)- f(t_{k-1})). \end{array}$

• $\because \lim_{n \rightarrow \infty} \sum_{k=1}^n X(t_k)(f(t_k) - f(t_{k-1})) = \int_a^b X(t)df(t)$ and $\max(t_k - t_{k-1}) \rightarrow 0$.
• $\therefore \int_a^b f(t)dX(t) = f(b) X(b) - f(a) X(a) - \int_a^b X(t) df(t)$.

Weiner integral example

• $f(t) = 1$

  • $\int_a^b f(t) dX(t) = \int_a^b dX(t) = X(b) - X(a) \sim N(0, \sigma^2(b-a))$.

• $f(t) = t$, $a=0$, $b=1$.

  • $\int_0^1 tdX(t) = X(1) - \int_0^1 X(t)dt$.

Weiner integral properties

• $\int_a^b X(t)dt = \lim_{(t_k - t_{k-1}) \rightarrow 0} \sum_{k=1}^n X(t_k)(t_k - t_{k-1})$ is normal distribution.

• $\int_a^b f(t) dX(t) = f(b)X(b) - f(a)X(a) - \int_a^b X(t)df(t)$ is also normal distribution.

• $E\left( \int_a^b f(t) dX(t) \right) = 0$.

  • $\int_a^b f(t)dX(t) = \sum_{k=1}^n f(t_{k-1})(X(t_k) - X(t_{k-1}))$ and $E(X(t_k) - X(t_{k-1})) =0$.
  • $\therefore E \left( \int_a^b f(t)dX(t) \right) = 0$.

• If $f$, $g$ are differentiable and continuous on $[a,b]$.

• $E\left( \int_a^b f(t)dX(t) \int_a^b g(t)dX(t) = \sigma^2 \int_a^b f(t)g(t)dt \right)$
• If $a < b$ and $g=f$ then $Var\left( \int_a^b f(t) dX(t) \right) = \sigma^2 \int_a^b (f(t))^2 dt$.

Taylor級數 (Taylor series)

• $\begin{array}{rcl} f(x) & = & f(h) + f^{(1)}(x-h)+ \frac{f^{(2)}(h)}{2!}(x-h)^2+\cdots \\ & = & \sum_{k=0}^n \frac{f^{(k)}(h)}{k!}(x-h)^k. \end{array}$

  • E.g. $f(x) = \sin x$, $f^{(1)}(x) = \cos x$, $f^{(2)}(x) = -\sin x$
  • Let $h=0$, $\sin x = 0 + x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$.
  • 使用不同的$h$時，$f(x)$有不同的多項式逼近式。

• $f(x+h) = f(x) + f^{(1)}(x)h + \frac{f^{(2)}(x)}{2!}h^2 + \frac{f^{(3)}(x)}{3!}h^3+\cdots$

• $\therefore f(x+h) - f(x) = f^{(1)}(x)h + \frac{f^{(2)}(x)}{2!}h^2 + \frac{f^{(3)}(x)}{3!}h^3+\cdots$.
• Let $x = g(t)$, $h= dg(t) = g(t+dt) - g(t)$, $h$為$g$的微小變化量。
• $df(g) = f(g(t)+dg(t)) - f(g(t))$ and $df(g) = f^{'}dg$.
• $\Rightarrow \begin{array}{rcl} df(g) & = &f(g(t)+dg(t)) - f(g(t)) \\ & = &f^{(1)}(g(t))dg(t) + \frac{f^{(2)}(g(t))}{2!}\left( dg(t) \right)^2 + \cdots \end{array}$

• 比較上式後，可發現在確定性函數$g(t)$中，當$dt$很小，即時間變化量很小時， $g(t)$在$[t,t+dt]$的變量化$dg(t)$也很小，所以$\left( dg(t) \right)^2$以上的高次方均可省略不計。
• 但是若$g(t)=B(t)$為Brownian process時，$\left( dg(t) \right)^2$不可省略，因為$\left( dg(t) \right)^2 = (dB(t))^2 = dt$.

Ito lemma 1

• Let $g(t) = B(t) \sim N(0, t)$ is standard Brownian process.
• $(dg(t))^2 = (dB(t))^2$ and $\lim_{n \rightarrow \infty} E \left(\sum_{k=1}^n (\Delta B_k)^2 - t)^2 \right) = 0$.
• $\lim_{n \rightarrow \infty} \sum_{k=1}^n (\Delta B_k)^2 = \int_0^t (dB(s))^2 = t = \int_0^t ds$.

• Derivative form: $(dB(t))^2 = (B(t+dt) - B(t))^2 = dt$. a.s.

  • $\because B(t)$ continuous but not differentiable on $[a,b]$.
  • $\therefore dB(t) \neq B^{'}(t)dt$.
  • $E(dB(t)) = E(B(t+dt) - B(t)) = E(B(t+dt)) - E(B(t)) = 0$.
  • $E((dB(t))^2) = Var(dB(t)) + E^2(dB(t)) = Var(dB(t)) = dt$.
  • $\Rightarrow \therefore B(t+dt) - B(t) \sim N(0, dt)$.

  • $Var((dB(t))^2) = 0$.

    $\begin{array}{rcl} Var((dB(t))^2) & = & E((dB(t))^4) - E^2((dB(t))^2) \\ & = & 3(Var(dB(t))^2 - (dt)^2 \\ & = & 3(dt)^2 - (dt)^2 \\ & = & 2(dt)^2 \\ (\because \ dt \rightarrow 0) & \approx & 0. \end{array}$

  • $\because E((dB(t))^2)=dt$ and $Var((dB(t))^2) = 0$.

  • $\Rightarrow (dB(t))^2 = dt$ a.s.

• 一般若$X$為隨機變數，則連續函數$g(X)$也是隨機變數。

• 然而隨機變數$dB(t)$的函數$(dB(t))^2$的變異數為$2(dt)^2$，在很短的時間變化量$dt$下，變異數趨近於0，因此$(dB(t))^2$可視為與常數$dt$之值相等。

Ito lemma 2