Solving a Riccati ODE in the Cox–Ingersoll–Ross Model

Summary

The document outlines a way to solve the final ordinary differential equation arising in a Cox–Ingersoll–Ross model proof. The answer rearranges the equation by completing the square, then separates variables and integrates from the initial value to the value at time t. A change of variables reduces the integral to the standard form involving one over one minus the square of the variable, which can be integrated using logarithms.

This is a sketch of the method rather than a full worked solution: it does not carry out the change of variables, derive the resulting expression, or discuss parameter restrictions and special cases. It may help identify the integration technique, but a reader seeking an explicit CIR solution must complete those algebraic steps and check the applicable initial conditions.

Key ideas

• Completing the square rewrites the CIR model’s Riccati equation in a form suitable for separation of variables.
• Separating variables turns the problem into an integral between the initial and current values of the unknown function.
• A change of variables reduces the integral to a logarithmic antiderivative involving one over one minus a square.
• The post sketches the technique but does not provide the complete explicit solution or cover parameter edge cases.

Tags

# Cox Ingersoll Ross Model


# Cox Ingersoll Ross Model












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.

## Answer by oliversm (score 1)

https://quant.stackexchange.com/a/51120

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.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.