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I am given the stochastic process $X_t$ to be the unique process starting at $X_0$ and solution of the following SDE:

$$dX_t = (a-bX_t)\,dt + \sigma \sqrt{X_t} \, dW_t,$$

where $W_t$ is a real standard Brownian Motion. I am asked to decomposed $X_t^2$ as an Itô process, but I don't really know how to do it. I think i did something completely wrong by expanding $dX_t^2$, I don't even know if that makes sense mathematically speaking. I also tried to use Itô formula that says if $f: [0,T] \times D \to \Bbb{R}$ (D an open connected set) is a twice continuously differentiable function then

$$f(t,X_t) = f(0,X_0) + \int_0^t \frac{\partial f}{\partial s}(s,X_s)\,ds + \int_0^t \frac{\partial f}{\partial x}(s,X_s) \, dX_s + \frac12 \int_0^t\frac{\partial f}{\partial x^2}(s,X_s) \, d\langle X,X\rangle,$$

by taking $f(t,x) = x^2$, but this does not work.

I would be really grateful if someone could help me figuring this out. I am very confused.

The solution to the exercise is

$$X_t^2 = X_0^2 + 2\int_0^t X_s(a-bX_s)\,ds + 2\sigma \int_0^t X_s\sqrt{X_s} \, dW_s + \sigma^2\int_0^t X_s \, ds$$

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  • $\begingroup$ By Itô, $d(X^2)=2XdX+d\langle X\rangle$. In your case, $d\langle X\rangle=$ $____$ $dt$ hence $dX=$ $____$ $dW$ $+$ $____$ $dt$. $\endgroup$
    – Did
    Feb 8, 2015 at 16:09
  • $\begingroup$ @Ale Why does it not work out to take $f(t,x)= x^2$? $\endgroup$
    – saz
    Feb 8, 2015 at 16:12
  • $\begingroup$ @saz I think i did bad my calculation. Is it correct now? (I think so, but I'm not sure since I'm not that much familiar with stochastic processes :)) $\endgroup$
    – Bman72
    Feb 8, 2015 at 16:36
  • $\begingroup$ @Ale Except for some typos (e.g. $\int_0^t X_t \, ds$): Yes. $\endgroup$
    – saz
    Feb 8, 2015 at 16:38
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    $\begingroup$ @Ale Instead of writing the solution as an edit, you could write it as an answer to your question. $\endgroup$
    – saz
    Feb 8, 2015 at 21:23

1 Answer 1

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after reading the comments i think i did some mistakes in my calculations. Taking $f(t,x) = x^2$ i get $\frac{\partial f}{\partial t}= 0, \frac{\partial f}{\partial t}=2x, \frac{\partial f}{\partial t}=2$. By the Itô formula i get

\begin{align*} X_t^2 &= X_0^2 +0 + 2\int_0^t X_s \, dX_s + \frac12 \int_0^t 2 d\langle X,X\rangle \\ &= X_0^2 + \int_0^t X_s \left[ (a-b X_s \, ds + \sigma \sqrt{X_t} \, dW_t\right] + \int_0^t d\langle X,X\rangle \\ &= X_0^2 + 2 \int_0^t X_s(a-bX_s) \, ds+2 \sigma \int_0^t X_s\sqrt{X_s} \, dW_s+\int_0^td\langle X,X\rangle \end{align*}

This seems very similar to the solution. Now I tried to evaluate $\int_0^td\langle X,X\rangle$:

\begin{align*} \int_0^t d\langle X,X\rangle &= \int_0^t \left|\sigma \sqrt{X_s} \right|^2 \, ds \\ &= \sigma^2 \int_0^t X_s \, ds \end{align*} Thank you for the help.

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