1

I want so solve the following SDE

\begin{align}\dot{y}(t)=r(t)y(t)+\epsilon_1(t) \\ \dot{r}(t)=r(t)+\epsilon_2(t)\end{align}

with $r,y$ both being stochastic processes and $e_1,e_2$ both being gaussian white noise processes with variance $\sigma^2_1$ and $\sigma^2_2$ respectively. A gaussian white noise process is a gaussian process with mean function $m(t)=0$ and kernel function $k(t,t')=\delta_{tt'}\sigma^2$, where $\delta_{ij}$ is the kronecker delta. $\sigma^2$ is called variance of the gaussian white noise process.

I think an alternative problem description using wiener processes (Brownian motion) would be

\begin{align} dy(t)=r(t)y(t)dt+\sigma_1dW_1(t) \\ dr(t)=r(t)dt+\sigma_2dW_2(t) \end{align}

where $W_1$ and $W_2$ are wiener processes.

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
  • 151
  • 4
  • What is your approach? – baharampuri Jul 27 '15 at 16:58
  • I don't have any background in stochastic differential equations, so I don't know if one can use classic methods (i.e. for deterministic DEs) to solve them. If one can, then they are both of the form solvable using the integrating factor method, which result in: $$r(t) = e^{t}\int e^{-t}\epsilon_2(t),\mathrm{d}t $$ and $$y(t) = e^{R(t)}\int e^{-R(t)}\epsilon_1(t),\mathrm{d}t$$ where $R(t)$ is any anti-derivative of $r(t)$. – Eff Jul 27 '15 at 17:00
  • What's your definition of "white noise process"? Do you mean Brownian motion? – saz Jul 27 '15 at 18:26

1 Answers1

3

The solution in terms of Brownian motions is: $$Y_t=e^{\int_0^t\, r_s\,ds} Y_0 + \int_0^t\,\sigma_1\,e^{\int_s^t\,r_u\,du}\,dW_s^1$$ and $$r_t=e^t r_0 + \int_0^t\,\sigma_2\,e^{t-s}\,dW_s^2.$$ The solution works even if the Brownian Motions are correlated $(dW_t^1dW_t^2=\rho\,dt)$. You can prove this by applying the product rule to $e^{-\int_0^t\, r_s\,ds}Y_t$ and $e^{-t}r_t$.

Esteban Chávez
  • 31