A bit about Stochastic Calculus

Itô's Integrals

$(\mathrm{\Omega },\mathcal{F},\mathbb{P},{\mathcal{F}}_{t\in [0,T]})$$(\Omega, \mathcal{F}, \mathbb{P},{\mathcal{F}_{t\in [0,T]}})$. Each represents outcomes, information, probability, and a set of information until T.

The Itô's Integral is,

$\underset{It{ô}^{\prime }sIntegral}{\underbrace{I_t^{\mathrm{\Delta }}}}={\int }_0^T{\mathrm{\Delta }}_{t}d{W}_t$$\underbrace{I_t^{\Delta}}_{Itô's Integral}=\int_0^T \Delta_t\ dW_t$

${\pi }_n=\{0=t_0,t_1,...,t_n=T\}$$\pi_n=\{0=t_0,t_1,...,t_n=T\}$. We evenly partition the time duration. (There are "n" periods, from 0 to n-1)

$\begin{array}{c}{\mathrm{\Delta }}_{t_i}^{(n)}=\{\begin{array}{l}{\mathrm{\Delta }}_0,[t_0,t_1)\\ {\mathrm{\Delta }}_1,[t_1,t_2)\\ {\mathrm{\Delta }}_2,[t_2,t_3)\\ ...\\ {\mathrm{\Delta }}_{n-1},[t_{n-1},t_n)\end{array}\end{array}$$\Delta_{t_i}^{(n)}= \begin{cases} \Delta_0 \quad ,[t_0,t_1) \\ \Delta_1 \quad ,[t_1,t_2)\\ \Delta_2 \quad ,[t_2,t_3) \\... \\ \Delta_{n-1} \quad ,[t_{n-1},t_n) \end{cases}$

Thus the integral could be understood in a discrete way,

$\begin{array}{c}{I}_t^{\mathrm{\Delta }}={\int }_0^T{\mathrm{\Delta }}_{t}d{W}_t\\ =\underset{integral}{\underbrace{\underset{n\to \mathrm{\infty }}{lim}\sum _{i=0}^{n-1}}}\underset{partition\mathrm{\Delta }}{\underbrace{{\mathrm{\Delta }}_{t_i}^{(n)}}}\underset{d{W}_t}{\underbrace{(W_{t_{i+1}}-W_{t_i})}}\end{array}$$I^{\Delta}_t =\int_0^T \Delta_t\ dW_t \\ =\underbrace{\lim_{n\to \infty} \sum_{i=0}^{n-1}}_{integral}\underbrace{ \Delta^{(n)}_{t_i}}_{\ partition \Delta }\underbrace{(W_{t_{i+1}}-W_{t_i}) }_{dW_t}$

The limit exists if ${\mathrm{\Delta }}_t$$\Delta_t$:

1. Adapted to $\{{\mathcal{F}}_t{\}}_{t\in [0,T]}$$\{\mathcal{F}_t\}_{t\in[0,T]}$.
2. Squared - Integrable in $L^2$$L^2$ Space. $\mathbb{E}[{\int }_0^T{\mathrm{\Delta }}_t^{2}dt{]}^{1/2}<\mathrm{\infty }$$\mathbb{E}[\int^T_0 \Delta^2_t dt]^{1/2}<\infty$, so ${\mathrm{\Delta }}_t\in L_T^2,L_T^2\equiv L^2([0,T],\mathrm{\Omega })$$\Delta_t \in L^2_T,\quad L^2_T \equiv L^2([0,T],\Omega)$

General Itô's Process

A stochastic process, $I:[0,T]\times \mathrm{\Omega }\to \mathbb{R}$$I:[0,T]\times \Omega \to \mathbb{R}$

$I_t=I_0+\underset{\text{ Riemann Integral}}{\underbrace{{\int }_0^t{\mu }_{u}du}}+\underset{\text{ Itô’ Integral}}{\underbrace{{\int }_0^t{\sigma }_{u}d{W}_u}}$$I_t=I_0 + \underbrace{\int_0^t \mu_u\ du}_{\text{ Riemann Integral}} +\underbrace{\int_0^t \sigma_u \ dW_u}_{\text{ Itô' Integral}}$

The Riemann Integral is the ordinal integral we normally used, and that term is the Drift Term. The last term is the stochastic/ diffusion term. P.S. $\sigma \in L^2$$\sigma \in L^2$

One Example of an Itô's Process

• Geometric Brownian Motion

$d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t$$dS_t=\mu S_t dt +\sigma S_t dW_t$

, where the first term in RHS is a drift and the second is a stochastic term.

For that GBM, we could take an integration.

${\int }_0^{t}d{S}_u={\int }_0^t\mu S_{u}du+{\int }_0^t\sigma S_{u}d{W}_u$$\int_0^t dS_u=\int_0^t \mu S_u du + \int_0^t \sigma S_u dW_u$

We consider a financial contract $C_t=f(S_t,t)$$C_t=f(S_t,t)$. For example, that contract could be a call option, then $S_t$$S_t$ is the underlying assets' price of that contract.

Since $S_t$$S_t$ follows a stochastic process, $C_t$$C_t$ must also follow a stochastic process but is a bit different.

$\begin{array}{rl}d{C}_t& =df(S_t)\\ & =f(S_{t+dt})-f(S_t)\\ & =f(S_t+dt{S}_t)-f(S_t)\end{array}$$\begin{aligned}dC_t &= df(S_t)\\ &= f(S_{t+dt})-f(S_t) \\ &= f(S_{t}+dtS_t)-f(S_t) \end{aligned}$

Taking a Taylor Expansion as dt is very small,

$\begin{array}{}\text{(1)}& df(S_t)=\frac{\partial f}{\partial S_t}d{S}_t+\frac{1}{2}\frac{{\partial }^{2}f}{\partial S_t^2}(d{S}_t{)}^2+\underset{dropit}{\underbrace{O(d{S}_t{)}^3}}\end{array}$$\begin{equation} df(S_t)=\frac{\partial f}{\partial S_t}dS_t + \frac{1}{2}\frac{\partial^2 f}{\partial S_t^2} (dS_t)^2 +\underbrace{O(dS_t)^3}_{drop\ it} \end{equation}$

Or in Partial Differential Equation (PDE),

$\begin{array}{}\text{(2)}& df(S_t,t)=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial S_t}d{S}_t+\frac{1}{2}\frac{{\partial }^{2}f}{\partial S_t^2}(d{S}_t{)}^2+\underset{dropit}{\underbrace{O(d{S}_t{)}^3}}\end{array}$$\begin{equation} df(S_t,t)=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial S_t}dS_t + \frac{1}{2}\frac{\partial^2 f}{\partial S_t^2} (dS_t)^2 +\underbrace{O(dS_t)^3}_{drop\ it} \end{equation}$

As $dt\to 0$$dt \to 0$, the followings terms would go to zero even faster. So we drop them.

1. $(dt{)}^k\to 0$$(dt)^k\to 0$ for k>1.
2. $dtdW\to 0$$dt\ dW \to 0$.
3. $(dW{)}^k\to 0$$(dW)^k \to 0$ for k>3.

However, the following three terms would be kept.

4. dt
5. dW
6. $dt\sim (dW{)}^2$$dt \sim (dW)^2$

There are some intuitions by Paul Willmot

We shouldn’t really think of dX^2 as being the square of a single Normally distributed random variable, mean zero, variance dt. No, we should think of it as the sum of squares of lots and lots (an infinite number) of independent and identically distributed Normal variables, each one having mean zero and a very, very small (infinite small) variance. What happens when you add together lots of i.i.d. variables? In this case we get a quantity with a mean of dt and a variance which goes rapidly to zero as the ‘lots’ approach ‘infinity.’

Why we emphasis the 6th term? Because it's useful in calculating $df(S_t)$$df(S_t)$.

We could square $d{S}_t$$dS_t$ first,

$\begin{array}{}\text{(3)}& d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t\end{array}$$\begin{equation} dS_t=\mu S_t dt +\sigma S_t dW_t \end{equation}$

$(d{S}_t{)}^2=(\mu S_{t}dt{)}^2+2\mu S_{t}dt\sigma S_{t}d{W}_t+(\sigma S_{t}d{W}_t{)}^2$$(dS_t)^2=(\mu S_t dt)^2 +2 \mu S_t dt\sigma S_t dW_t+ (\sigma S_t dW_t)^2$

In the RHS, the first term and the second term vanish, as 1&2, there are squared dt and cross term dt dW. Only the last term is left. Thus we get,

$\begin{array}{}\text{(4)}& \begin{array}{rl}(d{S}_t{)}^2& =(\sigma S_{t}d{W}_t{)}^2\\ & ={\sigma }^2{S}_t^2(d{W}_t{)}^2\\ & ={\sigma }^2{S}_t^{2}dt\end{array}\end{array}$$\begin{equation}\begin{aligned}(dS_t)^2 &=(\sigma S_t dW_t)^2 \\ &=\sigma^2 S_t^2 (dW_t)^2 \\ &=\sigma^2 S_t^2\ dt\end{aligned}\end{equation}$

Plug (3) & (4) into (2), we get,

$\begin{array}{rl}df(S_t,t)& =\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial S_t}d{S}_t+\frac{1}{2}\frac{{\partial }^{2}f}{\partial S_t^2}(d{S}_t{)}^2\\ & =\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial S_t}(\mu S_{t}dt+\sigma S_{t}d{W}_t)+\frac{1}{2}\frac{{\partial }^{2}f}{\partial S_t^2}({\sigma }^2{S}_t^{2}dt)\\ & =(\frac{\partial f}{\partial t}+\frac{\partial f}{\partial S_t}\mu S_t+\frac{1}{2}\frac{{\partial }^{2}f}{\partial S_t^2}{\sigma }^2{S}_t^2)dt+\frac{\partial f}{\partial S_t}\sigma S_{t}d{W}_t\end{array}$$\begin{aligned} df(S_t, t)&=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial S_t}dS_t + \frac{1}{2}\frac{\partial^2 f}{\partial S_t^2} (dS_t)^2 \\ &=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial S_t}\bigg(\mu S_t dt +\sigma S_t dW_t\bigg) + \frac{1}{2}\frac{\partial^2 f}{\partial S_t^2} \bigg(\sigma^2 S_t^2\ dt \bigg) \\ &=\bigg( \frac{\partial f}{\partial t}+ \frac{\partial f}{\partial S_t}\mu S_t + \frac{1}{2}\frac{\partial^2 f}{\partial S_t^2} \sigma^2 S_t^2 \bigg) \ dt + \frac{\partial f}{\partial S_t}\sigma S_t\ dW_t\end{aligned}$

So we get,

$\begin{array}{}\text{(5)}& df(S_t)=(\frac{\partial f}{\partial t}+\frac{\partial f}{\partial S_t}\mu S_t+\frac{1}{2}\frac{{\partial }^{2}f}{\partial S_t^2}{\sigma }^2{S}_t^2)dt+\frac{\partial f}{\partial S_t}\sigma S_{t}d{W}_t\end{array}$$\begin{equation} df(S_t)=\bigg( \frac{\partial f}{\partial t}+\frac{\partial f}{\partial S_t}\mu S_t + \frac{1}{2}\frac{\partial^2 f}{\partial S_t^2} \sigma^2 S_t^2 \bigg) \ dt + \frac{\partial f}{\partial S_t}\sigma S_t\ dW_t\end{equation}$

• A Specific Form $f(S)=ln(S)$$f(S)=ln(S)$

$f(S)=ln(S),f^{\prime }(S)=\frac{1}{S},f^{″}(S)=-\frac{1}{S^2}$$f(S)=ln(S),\quad f'(S)=\frac{1}{S},\quad f''(S)=-\frac{1}{S^2}$

Plus them into (5) we would get,

$\begin{array}{}\text{(6)}& dln(S_t)=(\mu -\frac{1}{2}{\sigma }^2)dt+\sigma d{W}_t\end{array}$$\begin{equation} d ln(S_t)=\bigg( \mu -\frac{1}{2}\sigma^2 \bigg)dt+\sigma \ dW_t \end{equation}$

We can also take integration.

$\begin{array}{rl}{\int }_0^{t}d(ln(S_u))& ={\int }_0^t(\mu -\frac{1}{2}{\sigma }^2)du+{\int }_0^t\sigma d{W}_u\end{array}$$\begin{aligned}\int_0^t d\bigg(ln(S_u)\bigg)&=\int_0^t \bigg( \mu -\frac{1}{2}\sigma^2 \bigg)du+ \int_0^t\sigma \ dW_u \end{aligned}$

$\begin{array}{r}ln(S_t)=ln(S_0)+(\mu -\frac{1}{2}{\sigma }^2)t+\sigma W_t\end{array}$$\begin{aligned}ln(S_t)= ln(S_0)+ \bigg( \mu -\frac{1}{2}\sigma^2 \bigg)t+ \sigma \ W_t \end{aligned}$

• We would know then, $ln(S_t)\sim N(ln(S_0)+(\mu -\frac{1}{2}{\sigma }^2)t,{\sigma }^{2}t)$$ln(S_t)\sim N\Bigg(ln(S_0)+\bigg(\mu -\frac{1}{2}\sigma^2\bigg)t,\sigma^2t\Bigg)$
• By taking exponential, we could then get $S_t=S_0{e}^{(\mu -\frac{1}{2}{\sigma }^2)t+\sigma W_t}$$S_t=S_0 \ e^{ ( \mu -\frac{1}{2}\sigma^2 )t+ \sigma \ W_t }$

By Fanyu Zhao