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.


1 Answer 1


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}


\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.

  • $\begingroup$ Actually, I should have put my question is more specific form, because I did the same exact thing but could not derive $\theta$ for arbitrary value given $(\alpha,\beta,n)$. So $n\in R$ is what I was after. You stated, this general case comes from a squared Bessel process rather than the OU procss. Does it mean in general, the Wikipedia statement is not true? $\endgroup$
    – Hans
    Jan 10, 2017 at 16:25
  • $\begingroup$ In the special case of integer $n$, a squared Bessel process can be defined as the sum of $n$ squared Brownian motions. In the general case its SDE still has a unique strong solution but you can't use the above construction any longer. See the reference that I pointed to. I presents all details that you are looking for I think. $\endgroup$ Jan 10, 2017 at 17:32
  • $\begingroup$ Thank you very much, LocalVolatility, for the detailed reference. Before I go find the book, would you please confirm that for arbitrary $n\in R$ the Wikipedia statement is wrong? If you do confirm, would you think it would make the matter absolutely clear by stating that at the top of your answer so as to leave no doubt? $\endgroup$
    – Hans
    Jan 10, 2017 at 18:11
  • $\begingroup$ My understanding is that yes - for arbitrary $n \in \mathbb{R}$ you can't construct it as a sum. $\endgroup$ Jan 10, 2017 at 18:15
  • $\begingroup$ Would you mind making that absolutely clear that the wikipedia statement is not generally true at the top of your answer so that I can accept this as a completely correct answer? Thank you. $\endgroup$
    – Hans
    Jan 10, 2017 at 18:33

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