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

[jend23]

Hello,

Homework Statement

Given the process
<br /> d\sqrt{z} = (\alpha - \beta\sqrt{z})dt + \delta dW<br />

\alpha, \beta and \delta are constants.

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

Homework Equations

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

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:

<br /> \frac{∂f}{∂W} = \delta<br />

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

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

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

<br /> dy = (\alpha - \beta y)dt + \delta dW<br />

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.

 

[Ray Vickson]

Hello,

Homework Statement

Given the process
<br /> d\sqrt{z} = (\alpha - \beta\sqrt{z})dt + \delta dW<br />

\alpha, \beta and \delta are constants.

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

Homework Equations

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

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:

<br /> \frac{∂f}{∂W} = \delta<br />

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

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

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

<br /> dy = (\alpha - \beta y)dt + \delta dW<br />

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.

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
In your case,
X = \sqrt{z}, \: Y = X^2.

 

[jend23]

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
In your case,
X = \sqrt{z}, \: Y = X^2.

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

So, we have
<br /> dX = (\alpha - \beta X)dt + \delta dW<br />

<br /> \frac{∂Y}{∂t} = 0, \frac{∂Y}{∂X} = 2x, \frac{∂^2Y}{∂X^2} = 2<br />

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

<br /> 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<br />

we have:

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

<br /> = \left(2\alpha X - 2\beta X^2 + \delta^2 \right)dt + 2\delta X dW<br />

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

Thanks again!