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.

  • 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 '19 at 20:25
  • $\begingroup$ Sorry, I meant formula there. Thanks a lot for the explanations. $\endgroup$
    – 6c656c
    Jun 29 '19 at 20:57

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 $Z_t = X_t^{\delta}$.


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