0
$\begingroup$

I can't seem to find the solution of the following SDE:

$dX_t = X_t^4dt + 2X_tdW_t \\ X_0 = \beta$

where $W_t$ is a Wiener process. The question gives me a hint that I should consider an integrating factor of the form $I_t = e^{\int_{0}^{t}c(s)dW_s-\frac{1}{2}\int_{0}^{t} [c(s)]^2 ds}$. I know that I should proceed by applying Itô's Formula to $d(X_tI_t)$, but I can't seem to know how to deal with the exponential. Shall I treat it as a different process?

Thanks in advance for any help given.

$\endgroup$
  • 1
    $\begingroup$ If you want to follow the hint, you need Itô's formula for Itô processes; more precisely you need to compute the stochastic differential $d(X_t Y_t)$ for two Itô processes $X$, $Y$. Then put $Y=I$. $\endgroup$ – saz Jun 29 at 20:25
  • $\begingroup$ Sorry, I meant formula there. Thanks a lot for the explanations. $\endgroup$ – Leonardo Lima Jun 29 at 20:57
2
$\begingroup$

Assume that $(X_t)_{t \geq 0}$ is a solution to the SDE, and set $$Z_t := X_t^{\delta}$$ for some $\delta$ which we will choose in a minute. By Itô's formula,

\begin{align*} Z_t-Z_0 &= \delta \int_0^t X_s^{\delta-1} \, dX_s + \frac{\delta(\delta-1)}2 \int_0^t X_s^{\delta-2} \, d\langle X \rangle_s \\ &= 2\delta \int_0^t X_s^{\delta} \, dW_s + \int_0^t \left(\delta X_s^{\delta+3} + 2 \delta (\delta-1) X_s^{\delta} \right) \, ds. \end{align*}

For $\delta=-3$ this becomes

$$Z_t-Z_0 = -6 \int_0^t Z_s \, dW_s + \int_0^t (24 Z_s -3) \, ds. \tag{1}$$

Since $(1)$ is a linear SDE, there is a general formula for its solution:

$$Z_t = \exp(6t - 6W_t) \left( Z_0 -3 \int_0^t \exp \left[-6s+6W_s \right] \, ds \right)$$

and so

$$X_t = \frac{1}{Z_t^{1/3}}$$

is a candidate for the solution of the original SDE. It remains to check using Itô's formula that it is indeed a solution.

Remark: See this question to get an idea how to come up with the transform $Z_t = X_t^{\delta}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.