Ito Differential Equation example

Question

Could someone explain Ito through an example as following?

How to use Ito differential equation to find $dy$ , where $y = e^{w(t)}$

Answer 1

Sure

assume $dy = \mu dt + \sigma dZ$ is an Ito process.

$f(y) = e^y$

Normally by the chain rule,

$df = \frac {\partial f}{\partial t}dt + \frac {\partial f}{\partial y}dy$

However, because $(dy)^2$ ~ $O(dt)$ we cannot neglect $(dy)^2$ terms.

Taylor expand $df$ to order $2$ in $dy$

$\Rightarrow df = \frac {\partial f}{\partial t}dt + \frac {\partial f}{\partial y}dy + \frac 12 \frac {\partial^2 f}{\partial y^2}(dy)^2$

Now, sub in $dy = \mu dt + \sigma dZ$

$\Rightarrow df = \frac {\partial f}{\partial t}dt + \frac {\partial f}{\partial y}(\mu dt + \sigma dZ) + \frac 12 \frac {\partial^2 f}{\partial y^2}(\mu dt + \sigma dZ)^2$

$(\mu dt + \sigma dZ)^2 = \mu^2 dt^2 + 2\mu \sigma dt dZ + \sigma^2 dZ^2$

Now, we eliminate all terms that are of higher order than $dt$ and use the convention $dZ = \sqrt {dt}$

$\Rightarrow dZ^2 = dt, dZdt = 0, dt^2=0$ (Ito Box rule)

$\Rightarrow (\mu dt + \sigma dZ)^2 = \sigma^2 dt$

So,

$\Rightarrow df = \frac {\partial f}{\partial t}dt + \frac {\partial f}{\partial y}(\mu dt + \sigma dZ) + \frac 12 \frac {\partial^2 f}{\partial y^2}\sigma^2 dt$

Now,

$\frac {\partial f}{\partial t} = 0$

$\frac {\partial f}{\partial y} = e^y$

$\frac {\partial^2 f}{\partial y^2} = e^y$

$\Rightarrow df = e^y(\mu + \frac 12 \sigma^2) dt + e^ydZ$

$f=e^y$

so,

$$df = f(\mu + \frac 12 \sigma^2) dt + fdZ$$ $$\frac {df}{f} = (\mu + \frac 12 \sigma^2) dt + dZ$$

Answer 2

Recall Ito's formula, written in differential form, $$ df(X) = f'(X)dX + \frac12f''(X)d\langle X\rangle. $$

Assuming $w(t)$ is a semimartingale, since $y(x):=e^x$ is a $C^2$ function, we can compute $$ dy(w(t)) = e^{w(t)}dw(t) + \frac12e^{w(t)}d\langle w \rangle(t). $$

If by $w(t)$ you meant a brownian motion, then the quadratic variation term is just $dt$ so $$ dy(w(t)) = e^{w(t)}(dw(t) + \frac12dt). $$