CIR model: is the short rate really non-central $\chi^2$ distributed?

Question

Probably simple question. Consider the CIR (1985) model for interest rates $$ dr = k(\theta - r)dt + \sigma \sqrt{r}dz $$ Then it is known in closed form the conditional pdf $f(r(s),s|r(t),t)$ ($s \geq t$) $$ f(r(s),s|r(t),t) = ce^{-u-v}\left(\frac{v}{u}\right)^{q/2}I_{q}(2\sqrt{uv}) $$ where \begin{equation} \begin{aligned} c &=\frac{2k}{\sigma^{2}\left(1-e^{-k(s-t)}\right)}\\ u &=cr(t)e^{-k(s-t)}\\ v &=cr(s)\\ q &=\frac{2k\theta}{\sigma^2}-1 \end{aligned} \end{equation} and $I_{q}(\cdot)$ is a modified Bessel function of the first kind of order $q$.

Then authors state:

<< The distribution function is the non central chi-square $\chi^2[2 c r(s); 2q + 2, 2u]$, with $2q+2$ degrees of freedom and parameter of non centrality $2u$ proportional to the current spot rate. >>

Then my questions:

1) Is it correct to say that what is (conditionally on $r(t)$) non-central $\chi^2$ distributed is the variable $2cr(s)$?

I can answer by my own to this question: Since the conditional expectation $E(r(s)|r(t))$ and variance $Var(r(s)|r(t))$ are provided in the paper (Eq. 19), it'easy to check the validity of 1) verifying that: \begin{equation} \begin{aligned} (2q+2) + (2u) &= E(2cr(s)|r(t)) = 2c E(r(s)|r(t))\\ 2[(2q+2) + 2(2u)] &= Var(2cr(s)|r(t)) = 4c^2Var(r(s)|r(t)) \end{aligned} \end{equation} where l.h.s. of both equations are expressions for the first two moments of a non-central $\chi^2$ variable with $2q+2$ and parameter of non-centrality $2u$ (you may want to check Wikipedia).

2) If 1), which is the conditional distribution of $r(s)$ alone? Is it still non-central $\chi^2$?

I want to be crystal clear: we know that $2cr(s) \stackrel{|r(t)}{\sim} \chi^2(2q+2,2u)$. Moreover, we know in closed form the (conditional on $r(t)$) pdf of $r(s)$ (the $f(r(s),s|r(t),t)$ above)... but then, is $r(s)$ a KNOWN random variable ($|r(t)$)? In particular, is it still non-cenral $\chi^2$ distributed? (*)

Thanks for your attention

(*) I'm afraid $r(s)$ cannot still be non-central $\chi^2$ since this would imply that the non-central $\chi^2$ would be close w.r.t. scaling of the variable, and - I'm not sure - this should not be the case.

Answer 1

To answer this I sum up a paragraph of "Interest rate models - An Introduction" by A.Cairns: For $i=1,\ldots,d$ consider the OU-processes $$ dX^i_t = -\frac 12 \alpha X^i_t dt + \sqrt{\alpha} dW^i_t. $$ Looking at the squared radius $R_t = \sum_{i=1}^d (X^i_t)^2 $ (in $\mathbb{R}^d$) of this process we get by Ito: $$ dR_t = \sum_{i=1}^d (2 X^i_t dX^i_t) + d \alpha dt. $$ Using the definition of $R_t$ introducing a new Brownian motion $B_t$ we get in distribution that that $$ dR_t = \alpha (d - R_t) dt + \sqrt{4 \alpha R_t} dB_t. $$ Defining $r_t = R_t/\theta$ with $\theta = 4\alpha/\sigma^2$ this is the CIR model. This gives a nice geometric interpretation. I am aware that not all details are covered here.

Recall the definition of the non central chi-squared distribution. Let $$ R = \sum_{i=1}^d (W_i + \delta_i)^2 $$ and $\lambda = \sum_{i=1}^d \delta_i^2$, then $R$ has a non-central chi-squared distribution with $d$ degrees of freedom and non-centrality parameter $\lambda$.

Since the $X_i^t$ above are all normally distributed with variance $1- e^{-\alpha t}$ we see that $R_t/(1- e^{-\alpha t})$ has non-central chi-squared distribution. Finally we have that for $d = 4 \alpha \mu/\sigma^2$ that $4 \alpha r_t/(\sigma^2 (1- e^{-\alpha t}))$ has a non-central chi-squared distribution with $d$ degrees of freedom and non-centrality parameter $\lambda = 4 \alpha r_0/(\sigma^2 (1- e^{-\alpha t}))$.

Conditionally on $r_t$ replace $r_0$ by $r_t$.

The answers then are: i) Yes, the variable that has non-central chi-squared distribution is the complicated expression that you mention.

ii) Only this complicated expression is non-central chi-squared distributed - $r_s$ itself is not. As you see in the link the non-central chi-squared distribution relates to standardized Gaussians (variance equals 1). Maybe the Generalized chi-squared distribution could be of help. But I don't know this.