7
$\begingroup$

I am trying to solve the following SDE

$$dX_t=(c+\sigma_\zeta W'_tX_t)dt + \sigma_\epsilon dW_t$$ $c\in \mathbb{R}$ is a constant, $X_t$ is a stochastic process, $\sigma_\zeta,\sigma_\epsilon \in \mathbb{R}^+$ are constants and $W'_t,W_t$ are independent Wiener processes. Let $x_0\in \mathbb{R}$ be the starting value of $X_t$.

I tried using mathematica (see https://mathematica.stackexchange.com/questions/90088/solving-sde-fracdytdt-c-sigma-wwtyt-epsilont-in-mathematica). This failed. I guess because $X_t$ is not an Ito process. I think it is not an Ito process because there is a random process in the $dt$ term. Sadly, I have no knowledge about solving non Ito process SDEs.

If a solution is to hard to obtain, it would also be sufficient to show that $X_t$ is not a Gaussian Process.

$\endgroup$
2
$\begingroup$

Using a (random) integrating factor, the solution is explicitly: $$ X_t =\exp\left( \sigma_{\zeta} \int_0^t W'_s ds \right) \times \left[ x_0+\int_0^t \exp\left(- \sigma_{\zeta} \int_0^s W'_r dr \right) c ds + \sigma_{\epsilon} \int_0^t \exp\left(-\sigma_{\zeta} \int_0^s W'_r dr \right) dW_s \right] $$ For example, using the fact that $\int_s^t W'_r dr \sim \mathcal{N}(0,(s-t)^2 (2 s +t)/3)$, its expected value is: $$ \mathbb{E} \{ X_t \} = \exp\left( \frac{\sigma_{\zeta}^2}{6} t^3 \right) x_0 + c \int_0^t \exp\left( \frac{\sigma_{\zeta}^2}{6} (t-s)^2 (2 s + t) \right) ds $$

$\endgroup$
  • $\begingroup$ Great. You provided a formula for the first the expectation of every $X_t$. Is it also possible to state the distribution of every $X_t$? Such as $X_t$ is a Gaussian Process with this mean function and that covariance function? $\endgroup$ – Julian Karls Sep 1 '16 at 13:11
  • $\begingroup$ No. Even in the special case $\sigma_{\epsilon}=0$, the solution process is not Gaussian in general. However, you can use the formula to calculate higher moments of the solution; it's a bit tedious though. $\endgroup$ – Nawaf Bou-Rabee Sep 1 '16 at 13:16
  • $\begingroup$ To be clear, this answer gives an explicit formula for the pathwise unique solution to the given SDE with random coefficients. $\endgroup$ – Nawaf Bou-Rabee Sep 1 '16 at 13:58
  • $\begingroup$ Damn, I made a mistakes with the brackets. I have fixed it. It would be great if you could help me again and even better point me to the methods you have used such that i can unterstand how you obtained the solution. $\endgroup$ – Julian Karls Dec 15 '16 at 21:06
  • $\begingroup$ I'm not sure what else to add to the pathwise unique solution that I wrote down. It is a fairly explicit formula for the solution to the given SDE. $\endgroup$ – Nawaf Bou-Rabee Dec 15 '16 at 23:34

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.