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.


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: $$ \mathbb{E} \{ X_t \} = \exp\left( \frac{\sigma_{\zeta}^2}{6} t^3 \right) x_0 + c \int_0^t \exp\left( \frac{\sigma_{\zeta}^2}{6} (t-s)^2 (2 s + t) \right) ds $$

  • $\begingroup$ 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

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