Before we start today, let’s recap a little bit. In the previous blog I showed you an interesting and important interest rate model called Vasicek model. We have also seen the merits & demerits of this model. So to improve we go to the next model. It’s the CIR model.
The CIR model or Cox- Ingersoll- Ross model describes the evolution of interest rate driven by market risk. It was introduced in 1985 by John C. Cox, Jonathan E. Ingersoll and Stephen A. Ross as an extension of the Vasicek model.
Definition: Let \(\mathbb{W(t),\ t\geq 0}\) be a Brownian motion. The CIR model for interest rate \(\mathbb{R(t)}\) is given by-
\[\mathbb{dR(t)}\ =\ \mathbb{(\alpha\ -\ \beta\ R(t))\ dt\ +\ \sigma\ \sqrt{R(t)}\ dW(t) }\]
where \(\mathbb{\alpha}\), \(\mathbb{\beta}\), \(\mathbb{\sigma}\) are positive constants.
Just for the time being forget about everything you have seen here. Forget Ito process, Ito- Doeblin formula, Stochastic differential equation.
Just focus on the above written equation and think what is the domain of the random variable \(\mathbb{R(t)}\)?
\[\mathbb{dR(t)}\ =\ \mathbb{(\alpha\ -\ \beta\ R(t))\ dt\ +\ \sigma\ \sqrt{R(t)}\ dW(t) }\]
In the equation, we have \(\mathbb{\sqrt{R(t)}}\). Therefore \(\mathbb{R(t)}\) can not take any negative values. \(\mathbb{R(t)}\) has domain \(\mathbb{[0, \infty)}\).
So it seems like the first problem or demerit of the Vasicek model, which is- \(\mathbb{\Pr[R(t) < 0]\ > 0}\), is removed in the CIR model.
Now we have to see if CIR model have the merits of the Vasicek model.
The drift term in CIR model is similar to the Vasicek model, \(\mathbb{(\alpha - \beta\ R(t))}\). It’s also mean-reverting.
First, there is no closed form solution of CIR model (the reason you will see later, when I show you how to solve SDEs). So we are going (or trying) to derive the expectation and variance of \(\mathbb{R(t)}\). For this we are gonna use- \(\mathbb{f(t,x)}\ =\ \mathbb{e^{\beta\ t}\ x}\) and the Ito- Doeblin formula. To apply Ito- Doeblin formula we need three things- \(\mathbb{f_{t}(t,x)}\), \(\mathbb{f_{x}(t,x)}\), \(\mathbb{f_{xx}(t,x)}\).
\[\begin{aligned} & \mathbb{f_{t}(t,x)}\ =\ \mathbb{\beta\ e^{\beta\ t}\ x}\ =\ \mathbb{\beta\ f(t,x)} \\ & \mathbb{f_{x}(t,x)}\ =\ \mathbb{e^{\beta\ t}} \\ & \mathbb{f_{xx}(t,x)}\ =\ \mathbb{0} \\ \end{aligned}\]
\[\begin{aligned} \mathbb{d\ e^{\beta\ t} R(t)}\ &=\ \mathbb{d\ f(t,R(t))} \\ &=\ \mathbb{f_{t}(t,R(t))\ dt}\ +\ \mathbb{f_{x}(t,R(t))\ dR(t)}\ +\ \frac{1}{2}\ \mathbb{f_{xx}(t,R(t))\ dR(t)\ dR(t)}\ \\ &=\ \mathbb{\beta\ e^{\beta\ t}\ R(t)\ dt}\ +\ \mathbb{e^{\beta\ t}\ dR(t)}\ \\ &=\ \mathbb{\beta\ e^{\beta\ t}\ R(t)\ dt}\ +\ \mathbb{e^{\beta\ t}\ [(\alpha\ -\ \beta\ R(t))\ dt\ +\ \sigma\ \sqrt{R(t)}\ dW(t) ]}\ \\ &=\ \mathbb{\alpha\ e^{\beta\ t}\ dt}\ +\ \mathbb{\sigma\ e^{\beta\ t}\ \sqrt{R(t)}\ dW(t)}\ \\ \end{aligned}\]
Now, integrating both sides of the above equation we get-
\[\begin{aligned} &\mathbb{\int_{0}^{t}\ d\ (e^{\beta\ s} R(s))}\ =\ \mathbb{\int_{0}^{t}\ \alpha\ e^{\beta\ s}\ ds}\ +\ \mathbb{\int_{0}^{t}\ \sigma\ e^{\beta\ s}\ \sqrt{R(s)}\ dW(s)} \\ \implies & \mathbb{e^{\beta\ t} R(t)}\ =\ \mathbb{R(0)}\ +\ \mathbb{\frac{\alpha}{\beta}\ (e^{\beta\ t}\ -\ 1 )}\ +\ \mathbb{\int_{0}^{t}\ \sigma\ e^{\beta\ s}\ \sqrt{R(s)}\ dW(s)} \\ \end{aligned}\]
So R.H.S. is some constant (given a value of \(\mathbb{t, \alpha, \beta}\)) and an Ito integral. I have already showed you- Ito integral has mean zero. So we get-
\[\begin{aligned} \mathbb{e^{\beta\ t}\ E\ R(t)}\ & =\ \mathbb{E\ [\mathbb{R(0)}\ +\ \mathbb{\frac{\alpha}{\beta}\ (e^{\beta\ t}\ -\ 1 )}\ +\ \mathbb{\int_{0}^{t}\ \sigma\ e^{\beta\ s}\ \sqrt{R(s)}\ dW(s)}]} \\ & =\ \mathbb{\mathbb{R(0)}\ +\ \mathbb{\frac{\alpha}{\beta}\ (e^{\beta\ t}\ -\ 1 )}\ } \\ \end{aligned}\]
Therefore,
\[\begin{aligned} \mathbb{E\ R(t)}\ & =\ \mathbb{e^{- \beta\ t}\ \mathbb{R(0)}\ +\ \mathbb{\frac{\alpha}{\beta}\ (1 -\ e^{- \beta\ t} )}\ } \\ \end{aligned}\]
This is the expectation of \(\mathbb{R(t)}\) in the Vasicek model.
To calculate variance, we use \(\mathbb{X(t)\ =\ e^{\beta\ t}\ R(t)\ }\), for which we have already calculated the diffferential equation-
\[\mathbb{d\ X(t)}\ =\ \mathbb{\alpha\ e^{\beta\ t}\ dt}\ +\ \mathbb{\sigma\ e^{\beta\ t}\ \sqrt{R(t)}\ dW(t)\ } =\ \mathbb{\alpha\ e^{\beta\ t}\ dt}\ +\ \mathbb{\sigma\ e^{\frac{\beta\ t}{2}}\ \sqrt{X(t)}\ dW(t)\ }\]
Here \(\mathbb{E(X(t))} =\ \mathbb{\mathbb{R(0)}\ +\ \mathbb{\frac{\alpha}{\beta}\ (e^{\beta\ t}\ -\ 1 )}\ }\)
Now we will apply Ito- Doeblin formula with \(\mathbb{f(t,x)}\ =\ \mathbb{x^2}\).
\[\begin{aligned} & \mathbb{f_{t}(t,x)}\ =\ \mathbb{0} \\ & \mathbb{f_{x}(t,x)}\ =\ \mathbb{2x} \\ & \mathbb{f_{xx}(t,x)}\ =\ \mathbb{2} \\ \end{aligned}\]
Therefore-
\[\begin{aligned} \mathbb{d\ X^2(t)}\ &=\ \mathbb{2\ X(t)\ dX(t)}\ +\ \mathbb{dX(t)\ dX(t)} \\ &=\ \mathbb{2 \alpha\ e^{\beta t}\ X(t)\ dt}\ +\ \mathbb{2\sigma\ e^{\frac{\beta t}{2}}\ X^{3/2}(t)\ dW(t)}\ +\ \mathbb{\sigma^2\ e^{\beta t}\ X(t)\ dt} \end{aligned}\]
Integrating both sides of the above equation we get-
\[\begin{aligned} & \mathbb{\int_{0}^{t} d\ X^2(u)}\ =\ \mathbb{\int_{0}^{t} 2 \alpha\ e^{\beta u}\ X(u)\ du}\ +\ \mathbb{\int_{0}^{t} 2\sigma\ e^{\frac{\beta u}{2}}\ X^{3/2}(u)\ dW(u)}\ +\ \mathbb{\int_{0}^{t} \sigma^2\ e^{\beta u}\ X(u)\ du} \\ \implies & \mathbb{X^2(t)}\ =\ \mathbb{X^2(0)}\ +\ \mathbb{(2\alpha + \sigma^2)\ \mathbb{\int_{0}^{t} e^{\beta u}\ X(u)\ du}\ } +\ \mathbb{\int_{0}^{t} 2\sigma\ e^{\frac{\beta u}{2}}\ X^{3/2}(u)\ dW(u)}\ \end{aligned}\]
So, \(\mathbb{X^2(t)}\) also contains an Ito integral. Therefore if we take expectation of \(\mathbb{X^2(t)}\), we get-
\[\begin{aligned} \mathbb{E\ X^2(t)}\ &=\ \mathbb{X^2(0)}\ +\ \mathbb{(2\alpha + \sigma^2)\ \mathbb{\int_{0}^{t} e^{\beta u}\ E\ X(u)\ du}\ } \\ &=\ \mathbb{X^2(0)}\ +\ \mathbb{(2\alpha + \sigma^2)\ \mathbb{\int_{0}^{t} e^{\beta u}\ (R(0)\ +\ \frac{\alpha}{\beta}\ (e^{\beta\ u}\ -\ 1 ))\ du}\ } \\ &=\ \mathbb{X^2(0)}\ +\ \mathbb{\frac{(2\alpha + \sigma^2)}{\beta}}\ \mathbb{(R(0)\ -\ \frac{\alpha}{\beta})\ (e^{\beta\ t}\ -\ 1)}\ +\ \mathbb{\frac{(2\alpha + \sigma^2)}{2\beta}}\ \frac{\alpha}{\beta}\ (e^{2 \beta\ t}\ -\ 1) \\ \end{aligned}\]
Therefore,
\[\begin{aligned} \mathbb{E\ R^2(t)}\ &=\ \mathbb{e^{-2 \beta\ t} \ E\ X^2(t)}\ \\ &=\ \mathbb{ \ R^2(0)}\ +\ \mathbb{\frac{(2\alpha + \sigma^2)}{\beta}}\ \mathbb{(R(0)\ -\ \frac{\alpha}{\beta}).\ (e^{-\beta\ t}\ -\ e^{-2 \beta\ t})}\ +\ \mathbb{\frac{(2\alpha + \sigma^2)}{2\beta}}.\ \frac{\alpha}{\beta}\ (1\ -\ e^{-2 \beta\ t}) \\ \end{aligned}\]
Therefore, we can write-
\[\begin{aligned} \mathbb{Var(R(t))}\ &=\ \mathbb{E\ R^2(t)\ -\ E^2\ R(t)} \\ &=\ \mathbb{ \ R^2(0)}\ +\ \mathbb{\frac{(2\alpha + \sigma^2)}{\beta}}\ \mathbb{(R(0)\ -\ \frac{\alpha}{\beta}).\ (e^{-\beta\ t}\ -\ e^{-2 \beta\ t})}\ \\ &+\ \mathbb{\frac{(2\alpha + \sigma^2)}{2\beta}}.\ \frac{\alpha}{\beta}\ (1\ -\ e^{-2 \beta\ t}) - (\mathbb{e^{- \beta\ t}\ \mathbb{R(0)}\ +\ \mathbb{\frac{\alpha}{\beta}\ (1 -\ e^{- \beta\ t} )}\ })^2 \\ &=\ \mathbb{\frac{\sigma^2}{\beta}}\ \mathbb{R(0).\ (e^{-\beta\ t}\ -\ e^{-2 \beta\ t})}\ +\ \mathbb{\frac{\alpha \sigma^2}{2 \beta^2}\ (1\ -\ 2e^{- \beta t}\ +\ e^{- 2\beta t})} \end{aligned}\]
In particular, if \(\mathbb{t \rightarrow \infty}\) we get-
\[\mathbb{\underset{t \rightarrow \infty}{\lim}\ Var(R(t))}\ =\ \mathbb{\frac{\alpha \sigma^2}{2 \beta^2}}\]
Hence we get mean and variance of \(\mathbb{R(t)}\) following CIR model. These were all interest rate models. Now we will move to another well known model called Black- Scholes- Merton’s Asset Pricing Model. Happy reading.