See the HTML file for full detals.

#### Key Takeaways

- Property of \{W_t\}:

- $ W_t – W_s \sim N(0, t-s) $
- $(W_t – W_{t-1})$ and $W_{t-i} – W_{t-i+1}$ are uncorrelated. So, $\int_0^t dW_u = \sum^t dW_u = W_t$

- For,

$$ dS_t = \mu S_t \ dt +\sigma S_t \ dW_t $$

Why the Geometric Brownian Motion of \{S_t\} is designed in that form?

The answer might be,

$$ dS_t /S_t = \mu \ dt +\sigma \ dW_t $$

$$\int_0^t dS_t /S_t = \int_0^t \mu \ dt + \int_0^t \sigma \ dW_t $$

$$ log(S_t) = \mu \ t +\sigma \ W_t $$

Taking the first difference is similar to differentiation. (d(log(S_T)) = log(S_t/S_t-1) = log(1+r_t) \ approx r_t).

$$ r_t = \mu + \sigma \ \Delta W_t $$

The return, \{r_t\}, is equal to a mean, \mu, plus a stochastic term. That is a random walk.

- We apply a transformation, if S_t follows a Geometric Brownian Motion, then f(S_t) follows another,

In calculating d f(S_t), we would get, (by Taylor Expansion)

$$ df(S_t) = \bigg( \frac{\partial f}{\partial t} + \frac{\partial f}{\partial S_t}\mu S_t +\frac{1}{2}\frac{\partial^2 f}{\partial S_t^2}\sigma^2 S_t^2 \bigg)dt + \frac{\partial f}{\partial S_t}\sigma S_t \ dW_t $$

- A Special Form of f(\cdot) is f(S) = log(S),

$$ d\ log(S_t) = \bigg( \mu – \frac{1}{2}\sigma^2 \bigg)dt + \sigma \ dW_t $$

Integrate the above equation from time 0 to time t, then we would get,

$$ log(S_t) = log(S_0) + (\mu – \frac{1}{2}\sigma^2)t + \sigma W_t $$

## 2 thoughts on “A bit Stochastic Calculus”

Comments are closed.