Hello, I was trying to prove this proposition for CIR model. I am able to follow the proof but then couldn't solve that last ODE. Any help would be great.
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2$\begingroup$ The first of the ODEs is just the Riccati equation, so please look up the solution of the Riccati equation. In general terms, to solve the system, note the first equation is autonomous in $\psi$ so you can solve it for $\psi$, and then plug this solution into the second ODE and simple integration (which usually comes down to application of partial fraction) then solves the second. $\endgroup$– Magic is in the chainFeb 12, 2020 at 18:27
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Looking at the final ODEs we see that all we need to do is solve the ODE for $\psi$ and then the rest is easy. To solve this ODE we notice that with a little re-arranging we can complete the square on the right hand side, and then effectively separate this into a known integral. Completing the square I get: \begin{equation} \psi' = \mu + \frac{b^2}{2\sigma^2} - \frac{\sigma^2}{2}\left(\psi + \frac{b}{\sigma^2}\right)^2 \end{equation} Dividing by the right hand side and integrating with respect to $t$ gives \begin{equation} \int_{\lambda}^{\psi(t)} \frac{1}{\mu + \frac{b^2}{2\sigma^2} - \frac{\sigma^2}{2}\left(\psi + \frac{b}{\sigma^2}\right)^2} \mathrm{d}\psi = t \end{equation} A simple change of variables and you can re-arrange the left hand side to an integral of the form $\int \tfrac{1}{1 - x^2} \mathrm{d}x$ which has a known solution involving logarithms. The rest you should be able to do.
