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Let $$ \begin{align*} dY_{t} &= \left(r - \frac{1}{2} V_{t}\right) dt + \sqrt{V_{t}}dW_{t}\\ dV_{t} &= \kappa(\theta - V_{t}) dt + \rho \sigma \sqrt{V_{t}}dW_{t} + \sigma\sqrt{1-\rho^{2}}\sqrt{V_{t}}dB_{t} \end{align*} $$ where $W$ and $B$ are two Brownian motions correlated by $\rho$. Due to the affinity of the log-transformed Heston model Duffie, Pan, and Singleton (2000) show a way to find the characteristic function of the Heston model. The characteristic function is then $$ f(u) = \mathrm{e}^{a(t) + b_{1}(t)X_{0} + b_{2}(t)V_{0}}. $$ They also specify the Riccati ODEs which in this case will be $$ \begin{align*} \frac{db_{1}(t)}{dt} &= 0,\\ \frac{db_{2}(t)}{dt} &= \frac{1}{2}b_{1}(t) + \frac{1}{2}\kappa b_{2}(t) - \frac{1}{2}b_{1}(t)^{2} - \rho\sigma b_{1}(t)b_{2}(t) - \frac{1}{2}\sigma^{2}b_{2}(t)^{2},\\ \frac{da(t)}{dt} &= r - rb_{1}(t) - \sigma\theta b_{2}(t).\\ \end{align*} $$ with initial conditions $a(0) = 0$, $b_{1}(0) = u$, and $b_{2}(0) = u$. So, the solution for $b_{1}$ would be $b_{1}(t) = u$, then inserting this into the Riccati for $b_{2}$ we get $$ \frac{db_{2}(t)}{dt} = \frac{1}{2}(u - u^{2}) + \left(\frac{1}{2}\kappa - \rho\sigma u\right)b_{2}(t) - \frac{1}{2}\sigma^{2}b_{2}(t)^{2}. $$ My issue is how do I solve this? I would appreciate if someone would help me derive the solution for $b_{2}(t)$ and maybe tell me if I also did it the right way this far. I would also like to note that I'm new to Riccati equations so any help would be appreciated.

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2 Answers 2

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For the proof, it suffices to follow this procedure with $$\begin{align} q_0(t) &= \frac{1}{2}u(1-u)\\ q_1(t) &= \frac{1}{2}\kappa-\rho\sigma u\\ q_2(t) &= -\frac{1}{2}\sigma^2\\ \end{align}$$

For this specific equation, all the $q_i(t)$ are constant functions, then you can even use this answer to solve the problem.

We note that for Riccacti equation, we need to know a particular solution first. You can use $f(t) = \text{const} :=f$ as the solution, then $f$ satisfies $$ 0 = \frac{1}{2}(u - u^{2}) + \left(\frac{1}{2}\kappa - \rho\sigma u\right)f - \frac{1}{2}\sigma^{2}f^2 \tag{1} $$

and $(1)$ is just a quadratic equation and can be solved easily.

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  • $\begingroup$ Thank you so much! I following your reference made it easier to understand. $\endgroup$
    – Marc Allan
    May 1, 2023 at 21:55
  • $\begingroup$ @MarcAllan You're welcome! $\endgroup$
    – NN2
    May 1, 2023 at 23:37
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Partial Answer

In [1] on p. 290-291 you find a discussion of the Cox-Ingersoll-Ross model in which the the Riccati equation $$ \textstyle n_t(t,T)-\frac{1}{2}\sigma^2\, n^2(t,T)-b\,n(t,T)+1=0\,,\quad n(T,T)=0 $$ is said to have the solution $$ n(t,T)=\frac{\sinh(\gamma t)}{\gamma\cosh(\gamma t)+\frac{1}{2}b\sinh(\gamma t)}\,,\quad\gamma=\sqrt{b^2+2\sigma^2}\,,\quad \tau=T-t\,. $$

[1] M. Musiela, M. Rutkowski, Martingale Methods in Financial Modelling.

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  • $\begingroup$ Thank you Kurt! I will check out the book reference. I also found NN2's answer very helpful. $\endgroup$
    – Marc Allan
    May 1, 2023 at 21:54

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