Consider the following SDE:
$$dx = (-ax)dt + \sigma db\\ dy = (-by+e^{-x})dt$$
where $a,b,\sigma>0$ and $b$ is a Brownian motion.
PROBLEM: What is the solution of this system? How can I estimate the variance and expected value of $y$?
I not sure if I should solve the first equation for $x$ and then solve the second one;
$$x(t) = e^{-at}x(0)+\int\limits_{0}^{t}{e^{-a(t-s)}db_s}$$
Yes it's good start! Once we solved the Ornstein Ulenbeck process $x_t$. We then solve the ODE on $y$. Let suppose that we have no 2nd member first. We solve the homogeneous ode: $dy_t = -by_tdt$ where an obvious solution is $y_t = C_t\exp(-bt)$.
Now we will use the variation parameter method. To do so, we apply the integration by part: \begin{align*} dy_t &= \exp(-bt)dC_t -b\exp(-bt)C_tdt \\ -by_tdt + \exp(-x_t)dt &= \exp(-bt)dC_t -b\exp(-bt)C_tdt \\ dC_t = \exp(b_t-x_t)dt \end{align*} Knowning that $C_0 = y_0$, we have: $C_t = y_0 + \int_0^t \exp(bs - x_s)ds$. Thus, \begin{equation*} y_t = y_0\exp(-bt) + \int_0^t \exp(-b(t-s) - x_s)ds \end{equation*} The second part of your question is to compute the expectation and the variance of $y$ for any $t$. Bear in mind that the stochastic integral in $x$, i.e. $\int_0^te^{-a(t-s)}db_s$, is a Wiener process. It follows a centered Gaussian variable with variance $\int_0^te^{-2a(t-s)}ds$. Then $x_t \sim \mathcal{N}(e^{-at}x_0, \int_0^te^{-2a(t-s)}ds)$. Now that we know the distribution of $x$ computing the expectation and the variance of $y$ is very simple. \begin{align*} \forall t\in R_+, m_t = E[y_t] &= e^{-bt}y_0 + E[\int_0^t \exp(-b(t-s) - x_s)ds] \\ &= e^{-bt}y_0 + \int_0^t\exp(-b(t-s))E[\exp(- x_s)]ds \quad \text{Fubini-Tonelli} \\ &= e^{-bt}y_0 + \int_0^t\exp\left(-b(t-s) - e^{-as}x_0 + \frac12\int_0^s e^{-2a(s-u)}du\right)ds \end{align*} For the variance, we calculate: \begin{equation*} Var(y_t) = E\left[\left(\int_0^t \exp(-b(t-s) - x_s)ds\right)^2\right] - (m_t)^2 \end{equation*} I will let you finish the computation. (Hint :We still need to use Fubini in order to bring the expectation inside of the integral and differentate all the cases).