• Support PF! Buy your school textbooks, materials and every day products Here!

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

  • Thread starter jend23
  • Start date
  • #1
12
0
Hello,

Homework Statement


Given the process
[tex]
d\sqrt{z} = (\alpha - \beta\sqrt{z})dt + \delta dW
[/tex]

[itex]\alpha[/itex], [itex]\beta[/itex] and [itex]\delta[/itex] are constants.

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

Homework Equations


Itos Lemma:
[tex]
df = \left(\frac{∂f}{∂t} + \frac{1}{2}\frac{∂^2f}{∂W^2}\right)dt + \frac{∂f}{∂W}dW
[/tex]
where [itex]t[/itex] is time and [itex]W[/itex] 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:

[tex]
\frac{∂f}{∂W} = \delta
[/tex]

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

I also know that [itex]dW^2[/itex] can be replaced with [itex]dt[/itex].

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

[tex]
dy = (\alpha - \beta y)dt + \delta dW
[/tex]

Can we use the simpler form of Ito's Lemma since drift does not seem to be a function of t i.e. [itex]\frac{∂f}{∂t}[/itex] 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.
 

Answers and Replies

  • #2
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
Hello,

Homework Statement


Given the process
[tex]
d\sqrt{z} = (\alpha - \beta\sqrt{z})dt + \delta dW
[/tex]

[itex]\alpha[/itex], [itex]\beta[/itex] and [itex]\delta[/itex] are constants.

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

Homework Equations


Itos Lemma:
[tex]
df = \left(\frac{∂f}{∂t} + \frac{1}{2}\frac{∂^2f}{∂W^2}\right)dt + \frac{∂f}{∂W}dW
[/tex]
where [itex]t[/itex] is time and [itex]W[/itex] 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:

[tex]
\frac{∂f}{∂W} = \delta
[/tex]

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

I also know that [itex]dW^2[/itex] can be replaced with [itex]dt[/itex].

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

[tex]
dy = (\alpha - \beta y)dt + \delta dW
[/tex]

Can we use the simpler form of Ito's Lemma since drift does not seem to be a function of t i.e. [itex]\frac{∂f}{∂t}[/itex] 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 [tex]dX = a(X,t) dt + b(X,t) dW [/tex] and if ##Y = f(X,t)##, then
[tex] dY = f_{x}(X,t)\, dX + f_t(X,t)\, dt + \frac{1}{2} f_{xx}(X,t) b^2(X,t) dt [/tex]
In your case,
[tex] X = \sqrt{z}, \: Y = X^2.[/tex]
 
  • #3
12
0
You are mis-stating Ito's Lemma. You should have: if [tex]dX = a(X,t) dt + b(X,t) dW [/tex] and if ##Y = f(X,t)##, then
[tex] dY = f_{x}(X,t)\, dX + f_t(X,t)\, dt + \frac{1}{2} f_{xx}(X,t) b^2(X,t) dt [/tex]
In your case,
[tex] X = \sqrt{z}, \: Y = X^2.[/tex]
Thanks for the pointer and after reviewing my notes a few more times, it has now become embarrassingly obvious. Much appreciated.

So, we have
[tex]
dX = (\alpha - \beta X)dt + \delta dW
[/tex]

[tex]
\frac{∂Y}{∂t} = 0, \frac{∂Y}{∂X} = 2x, \frac{∂^2Y}{∂X^2} = 2
[/tex]

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

[tex]
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
[/tex]

we have:

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

[tex]
= \left(2\alpha X - 2\beta X^2 + \delta^2 \right)dt + 2\delta X dW
[/tex]

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

Thanks again!
 

Related Threads on Applying Itos Lemma to show one SDE is related to another.

  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
0
Views
3K
Replies
2
Views
897
Replies
8
Views
546
Replies
6
Views
2K
  • Last Post
Replies
2
Views
328
Replies
5
Views
4K
Replies
7
Views
4K
Replies
3
Views
1K
Top