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

  1. Feb 10, 2013 #1
    Hello,

    1. The problem statement, all variables and given/known data
    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]

    2. Relevant 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.


    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:

    [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.
     
  2. jcsd
  3. Feb 11, 2013 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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]
     
  4. Feb 11, 2013 #3
    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!
     
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