Let $W_{t}$ be a Wiener process. And let $\mu$, $\sigma$, $\lambda$ $\psi$ be real valued positive constants. Consider the following system of differential equations:
\begin{equation}\tag1 \frac{\dot{S}(t)}{S(t)} = \psi\mu dt + \psi\sigma dW_{t} -\psi B(t) \end{equation}
\begin{equation} \dot{B}(t) = \frac{1}{\lambda}(S(t) - B(t)) \end{equation}
First question: how can we argue that a solution exists? Second question:does $S(t)$ have an analytical solution in terms of $W_{t}$?
Here is my (failed!!) attempt using guess and verify. I first solve for $B(t)$, which is straight-forward:
\begin{equation}\tag 2 B(t) = \frac{1}{\lambda}\int_{s=0}^{t}S(s)e^{\frac{s-t}{\lambda}}ds \end{equation}
Then I plug in (2) into (1) and solve out for $S(t)$ to get:
\begin{equation}\tag 3 S(t) = S(0)\exp\big\{\psi\mu t +\psi \sigma W_{t} - \frac{\psi}{\lambda}\int_{s=0}^{t}\int_{v=0}^{s}S(v)e^{\frac{v-s}{\lambda}}dv\big\} \end{equation}
I guessed that:
\begin{equation}\tag 4 S(t) = S(0)\exp\big\{\psi\sigma W_{t} - \frac{\psi}{\lambda}\int_{0}^{t}W_{s}e^{\frac{s-t}{\lambda}}ds\big\} \end{equation}
For the sake of demonstration, I also assume $\mu =0$ for now. Now equating expressions in the exponent of (3) and (4), we get:
\begin{equation} \int_{0}^{t}W_{s}e^{\frac{s-t}{\lambda}}ds = \int_{s=0}^{t}\int_{v=0}^{s}S(v)e^{\frac{v-t}{\lambda}}dv \end{equation}
Taking the derivative of both sides twice and after some algebra, we end up with $S(t) = dW_{t}$, so clearly my guess is incorrect.
I am starting to think there is no analytical solution? I would appreciate any guidance.
OK I will answer my questions as I seem to have figured it out.
I will modify the question slightly (assuming $\mu =0$ and the modified $\dot{B}$ as below) so that the algebra is neater, but the same idea should work for the original question. Assume $\frac{\psi}{\lambda}<1$. We seek a solution to the system:
\begin{equation}\tag1 \frac{\dot{S}(t)}{S(t)} = \psi\sigma dW_{t} -\psi B(t) \end{equation}
\begin{equation} \dot{B}(t) = \frac{1}{\lambda}(\log(S(t)) - B(t)) \end{equation}
I first solve for $B(t)$, which is straight-forward:
\begin{equation}\tag 2 B(t) = \frac{1}{\lambda}\int_{s=0}^{t}\log(S(s))e^{\frac{s-t}{\lambda}}ds \end{equation}
Now plugging in the solution for $B(t)$ into the expression for $\frac{\dot{S}(t)}{S(t)}$, we get:
\begin{equation}\tag 3 S(t) = S(0)\exp\big(\psi \sigma W_{t} - \frac{\psi}{\lambda}\int_{s=0}^{t}\int_{v=0}^{s}\log(S(v))e^{\frac{v-s}{\lambda}}dv ds\big) \end{equation}
Define $\textbf{L}$ to be an operator that integrates the smooth value of its argument as follows. For (suitable) stochastic process X:
\begin{equation} \textbf{L}X(t) = \int_{0}^{t}\int_{0}^{s}X(v)e^{\frac{s-t}{\lambda}}dvds \end{equation}
We can apply $\textbf{L}$ recursively, for $L^{i}X$, $L^{i+1}X$ is defined as:
\begin{equation} L^{i+1}X(t) = \int_{0}^{t}\int_{0}^{s} L^{i}X(v)e^{\frac{s-t}{\lambda}}dvds \end{equation}
I claim the following solution for random function $S$ satisfies (3):
\begin{equation}\tag 4 S(t) = S(0)\exp\big(\sigma\sum_{i=0}^{\infty}\frac{\psi^{i+1}}{\lambda^{i}(-1)^{i+2}}L^{i}W(t)\big), S(0) = 1 \end{equation}
(I slightly abuse notation, writing W_{t} as W(s)) Plug in (4) into (3) to get
\begin{align} S(t)& = \exp\big(\psi \sigma W_{t} - \frac{\psi}{\lambda}\int_{s=0}^{t}\int_{v=0}^{s}\log(S(v))e^{\frac{v-s}{\lambda}}dv ds\big) \\\\ & = \exp\big(\psi \sigma W_{t} - \frac{\psi}{\lambda}\int_{s=0}^{t}\int_{v=0}^{s}\sigma\sum_{i=0}^{\infty}\frac{\psi^{i+1}}{\lambda^{i}(-1)^{i+2}}L^{i}W(v)e^{\frac{v-s}{\lambda}}dv ds\big) \\\\ & = \exp\big(\psi \sigma W_{t} + \sum_{i=0}^{\infty}\sigma\frac{\psi^{i+2}}{\lambda^{i+1}(-1)^{i+3}}\int_{s=0}^{t}\int_{v=0}^{s}L^{i}W(v)e^{\frac{v-s}{\lambda}}dv ds\big)\\\\ & = \exp\big(\psi \sigma W_{t} + \sum_{i=0}^{\infty}\sigma\frac{\psi^{i+2}}{\lambda^{i+1}(-1)^{i+3}}L^{i+1}W(t)\big)\\\\ & = \exp\big(\psi \sigma L^{0}W(t) + \sum_{i=1}^{\infty}\sigma\frac{\psi^{i+2}}{\lambda^{i}(-1)^{i+2}}L^{i}W(t)\big)\\\\ & = \exp\big(\sigma\sum_{i=0}^{\infty}\frac{\psi^{i+1}}{\lambda^{i}(-1)^{i+2}}L^{i}W(t)\big) \end{align}
As was to be shown.
(The assertions I have not proved are 1) that the infinite sum at (4) is well-defined and 2) that the exchange of summation and integrals at the third step above is valid. I do no think these should be a problem)