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I want so solve the following SDE. Specifically, I want to know if $y(t)$ is a Gaussian Process and if so the corresponding mean and covariance function.

$$\frac{dy(t)}{dt}=(c+\sigma_wW(t))y(t)+\epsilon(t) $$ 11

where $W(t)$ is the Wiener process, $\epsilon(t) \sim \mathcal{N}(0,\sigma^2_e)$ for all $t$ and all the $e(t)$ random variables are mutually indepedent, $\sigma_w$ is a non-negative scalar and $c \in \mathbb{R}$.

I don't have any background in solving stochastic differential equations. Thus, my approach so far has been trying mathematica, which was not too successful yet.

Julian Karch
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  • Is any information about $\epsilon (t)$ ? – Khosrotash Aug 05 '15 at 17:13
  • Can you write it as :$dy(t)=c dt +\epsilon (t) dt +\sigma_w y(t)dw(t)$ ? – Khosrotash Aug 05 '15 at 17:16
  • @Khosrotash I added more information about $\epsilon(t)$. If one wants to write it in your form I think it should be $$dy(t) = cdt + e(t)dt + \sigma_w y(t)W(t)dt$$ The wiener process is used a bit differently as it is used usually. – Julian Karch Aug 06 '15 at 12:07
  • Is $\epsilon(t)$ a continuous process? – Calculon Aug 06 '15 at 12:37
  • yes, $\epsilon(t)$ is actually a gaussian process with mean function $m(t)=0$ and kernel function $m(t,t')=\delta_{tt'}\sigma_\epsilon^2$ – Julian Karch Aug 06 '15 at 12:39
  • Well such a process cannot exist. See this question: http://math.stackexchange.com/questions/1207453/impossible-stochastic-process – Calculon Aug 06 '15 at 12:41
  • Can you elaborate on this a bit? In my understanding $y(t)$ does not fullfill the first two assumptions in your question. I do not understand the third assumption. – Julian Karch Aug 06 '15 at 13:13
  • I am talking about $\epsilon(t)$ itself. You cannot force it to be continuous. – Calculon Aug 06 '15 at 14:09
  • Hey, i guess my comment about $\epsilon(t)$ being continuous was wrong. $\epsilon(t)$ is a stochastic process. More specifically it is a Gaussian white noise process, i.e., it is a Gaussian process with mean function $m(t,t')=0$ and covariance function $k(t,t')=\sigma_\epsilon^2$ if $t=t'$ and $0 $ otherwise. – Julian Karch Aug 06 '15 at 16:01

1 Answers1

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By conditioning on $W(t)$, this is a regular ODE

$$y'=a(t)y(t)+b(t)$$

and it has the solution First-order equation with variable coefficients

$$y(t)=ce^{\int^{t} a(s)ds}+e^{\int^{t} a(s)ds}\int b(s)e^{-\int^{t} a(s)ds}.$$

Thomas Kojar
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