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Math Help - Ornstein-Uhlenbeck equation

  1. #1
    Junior Member elenaas's Avatar
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    Ornstein-Uhlenbeck equation

    I find out the solution of Ornstein-Uhlenbeck equation is


    My problem is how to find and

    It seems like it is easy to find the mean and variance.But I was trying and got no solution
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  2. #2
    MHF Contributor

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    Quote Originally Posted by elenaas View Post
    I find out the solution of Ornstein-Uhlenbeck equation is


    My problem is how to find and

    It seems like it is easy to find the mean and variance.But I was trying and got no solution
    All you need to know is E[\int_0^t f(s) dB_s]=0, because this integral is a martingale (as a function of t), and E[\left(\int_0^t f(s) dB_s\right)^2]=\int_0^t f(s)^2 ds, as follows from the definition of the stochastic integral.
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  3. #3
    Junior Member elenaas's Avatar
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    I suppose that I got the correct solution, but I would like you to confirm me.
    I started with an explicit solution of the SDE




    because this integral is an Ito integral and it is martingale
    So the mean is:

    and for the variance I got:


    I used that


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  4. #4
    MHF Contributor

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    Quote Originally Posted by elenaas View Post
    I suppose that I got the correct solution, but I would like you to confirm me.
    I started with an explicit solution of the SDE




    because this integral is an Ito integral and it is martingale
    So the mean is:

    and for the variance I got:


    I used that


    It depends on what is X_0. If it is constant, then you are correct (and E[X_t]=X_0e^{\mu t}). If it is a random variable, then it must appear in the variance: X_t-E[X_t]=(X_0-E[X_0])e^{\mu t} +\int_0^t (\cdots) and both terms are independent and centered hence E[(X_t-E[X_t])^2]={\rm Var}(X_0)e^{2\mu t}+E[(\int_0^t(\cdots))^2] = {\rm Var}(X_0)e^{2\mu t}+\frac{\sigma^2}{2\mu}(e^{2\mu t}-1). I think.
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