stochastic processes - CIR Process from Ornstein–Uhlenbeck process

The wikipedia entry on the CIR Model states that "this process can be defined as a sum of squared Ornstein–Uhlenbeck process" but provides no derivation or reference. Can any one do that? I could only derive equilibrium level for special numbers proportional to natural numbers and not for arbitrary real numbers.

Answer

I don't think that the statement you reference is correct for general $n \in \mathbb{R}$ but only for $n \in \mathbb{N}$.

The intuition behind this is that each Ornstein-Uhlenbeck (OU) process is normally distributed. Thus the sum of $n$ squared OU processes is chi-squared distributed with $n$ degrees of freedom. Define $X$ to be a $n$-dimensional vector valued OU process with

$$\begin{equation} \mathrm{d}X_t^i = \alpha X_t^i \mathrm{d}t + \beta \mathrm{d}W_t^i, \end{equation}$$

where $W$ is a $n$-dimensional vector of independent Brownian motions. Let

$$\begin{equation} Y_t = \sum_{i = 1}^n \left( X_t^i \right)^2. \end{equation}$$

Note that

$$\begin{eqnarray} \mathrm{d} \left( X_t^i \right)^2 & = & 2 X_t^i \mathrm{d}X_t^i + 2 \mathrm{d} \langle X^i \rangle_t\\ & = & \left( 2 \alpha \left( X_t^i \right)^2 + \beta^2 \right) \mathrm{d}t + 2 \beta X_t^i \mathrm{d}W_t^i \end{eqnarray}$$

Thus

$$\begin{eqnarray} \mathrm{d}Y_t & = & \mathrm{d} \left( \sum_{i = 1}^n \left( X_t^i \right)^2 \right)\\ & = & \sum_{i = 1}^n \mathrm{d} \left( X_t^i \right)^2\\ & = & \left( 2 \alpha Y_t + n \beta^2 \right) \mathrm{d}t + 2 \beta \sum_{i = 1}^n X_t^i \mathrm{d}W_t^i, \end{eqnarray}$$

where the second step follows from the independence of the Brownian motions. Next note that the process

$$\begin{equation} Z_t = \int_0^t \sum_{i = 1}^n X_u^i \mathrm{d}W_u^i \end{equation}$$

is a martingale with quadratic variation

$$\begin{eqnarray} \langle Z \rangle_t & = & \int_0^t \sum_{i = 1}^n \left( X_u^i \right)^2 \mathrm{d}u\\ & = & \int_0^t Y_u \mathrm{d}u. \end{eqnarray}$$

Consequently, by Levy's characterization theorem, the process

$$\begin{equation} \tilde{W}_t = \int_0^t \frac{1}{\sqrt{Y_u}} \sum_{i = 1}^n X_u^i \mathrm{d}W_u^i \end{equation}$$

is a Brownian motion. Thus

$$\begin{eqnarray} \mathrm{d}Y_t & = & \left( 2 \alpha Y_t + n \beta^2 \right) \mathrm{d}t + 2 \beta \sqrt{Y_t} \mathrm{d}\tilde{W}_t\\ & = & \kappa \left( \theta - Y_t \right) \mathrm{d}t + \xi \sqrt{Y_t} \mathrm{d}W_t, \end{eqnarray}$$

where $\kappa = -2 \alpha$, $\theta = -n \beta^2 / 2 \alpha$ and $\xi = 2 \beta$.

This can be generalized to $n \in \mathbb{R}$ by considering a time-change of a squared Bessel process. A comprehensive reference is Chapter 6 in Jeanblanc, Yor and Chesney (2009) "Mathematical Methods for Financial Markets", Springer.