# Applying Itos Lemma to show one SDE is related to another.

1. Feb 10, 2013

### jend23

Hello,

1. The problem statement, all variables and given/known data
Given the process
$$d\sqrt{z} = (\alpha - \beta\sqrt{z})dt + \delta dW$$

$\alpha$, $\beta$ and $\delta$ are constants.

Use Ito's Lemma to show that:
$$dz = (\delta^2 + 2\alpha\sqrt{z} - 2\beta z)dt + 2\delta\sqrt{z}dW$$

2. Relevant equations
Itos Lemma:
$$df = \left(\frac{∂f}{∂t} + \frac{1}{2}\frac{∂^2f}{∂W^2}\right)dt + \frac{∂f}{∂W}dW$$
where $t$ is time and $W$ a Wiener process.

3. The attempt at a solution

Essentially, I have a bit of a mental block about how to start going about solving this. I'm sure it's relatively simple though if I had a hint on the right direction to take.

In the first equation, the diffusion coefficient is:

$$\frac{∂f}{∂W} = \delta$$

The drift is:
$$\frac{∂f}{∂t} + \frac{1}{2}\frac{∂^2f}{∂W^2} = \alpha - \beta\sqrt{z}$$

I also know that $dW^2$ can be replaced with $dt$.

To simplify, let $y=\sqrt{z}$ so the SDE becomes:

$$dy = (\alpha - \beta y)dt + \delta dW$$

Can we use the simpler form of Ito's Lemma since drift does not seem to be a function of t i.e. $\frac{∂f}{∂t}$ is 0 and the partial derivatives can become ordinary derivatives?

I don't really know where to go from there. I'm sure I'm missing something simple and obvious. Ideally, if possible I'd like to be given a hint about how to go about solving this problem.

Any help appreciated. Thanks.

2. Feb 11, 2013

### Ray Vickson

You are mis-stating Ito's Lemma. You should have: if $$dX = a(X,t) dt + b(X,t) dW$$ and if $Y = f(X,t)$, then
$$dY = f_{x}(X,t)\, dX + f_t(X,t)\, dt + \frac{1}{2} f_{xx}(X,t) b^2(X,t) dt$$
$$X = \sqrt{z}, \: Y = X^2.$$

3. Feb 11, 2013

### jend23

Thanks for the pointer and after reviewing my notes a few more times, it has now become embarrassingly obvious. Much appreciated.

So, we have
$$dX = (\alpha - \beta X)dt + \delta dW$$

$$\frac{∂Y}{∂t} = 0, \frac{∂Y}{∂X} = 2x, \frac{∂^2Y}{∂X^2} = 2$$

therefore, from (slightly different notation and arrangement to your one):

$$dY = \left(\frac{∂Y}{∂t} + a(X,t)\frac{∂Y}{∂X} + \frac{1}{2}b(X,t)^2\frac{∂^2Y}{∂X^2}\right)dt + b(X,t)\frac{∂Y}{∂X}dW$$

we have:

$$dY = \left( 0 + (\alpha - \beta X)(2X) + \frac{1}{2}\delta^2(2) \right)dt + \delta(2X)dW$$

$$= \left(2\alpha X - 2\beta X^2 + \delta^2 \right)dt + 2\delta X dW$$

substituting back $\sqrt{z}$ for $X$ yields the required result.

Thanks again!