# demonstrate that a Square-root process is Non-central Chi-squared distributed

how can i prove that the value at some future time $t'$, $x_{t'}$, of the Square-root process at current time $t$, $x_t$, is Chi-squared distributed?

$dx_t = k(\theta - x_t)dt + \beta \sqrt{x_t}dz_t$

explicitely:

$x_{t'} = x_t e^{-k(t'-t)} + \theta (1 - e^{-k(t'-t)}) + \beta \int_t^{t'} e^{-k(t'-u)}\sqrt{x_u}dz_u$

I just got to mean and variance by Ito's isometry:

$E_t[x_{t'}] = \theta + (x_t - \theta) e^{-k(t'-t)}$

and

$Var_t[x_{t'}] = \frac{\beta^2 x_t}{k} (e^{-k(t'-t)} - e^{-2k(t'-t)}) + \frac{\beta^2 \theta}{2k} (1 - e^{-k(t'-t})^2$

in the Ornstein-Uhlenbeck case there's no $\sqrt{x_t}$ in the volatility and therefore the stochastic integral is Normally distributed and everything is fine

unfortunately, I've never met the Non-central Chi-squared distribution before, so I'm not able to understand how to get to it (also because it seems itself pretty a mess to me)

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.

• This is the answer from the question which I marked as a dublicate ... Feb 10, 2016 at 7:03

Hint

let $s\le t$ and $$P(s,y\,;\,t,x)=P(r_t\le x|r_s=y)$$ The transition probability $P(s,y\,;\,t,x)$ satisfy Kolmogorov bachward equation $$\frac{\partial P}{\partial s}+\kappa (\theta -{{r}_{s}})\frac{\partial P}{\partial r}+\frac{1}{2}{{\sigma }^{2}}{{r}_{s}}\frac{{{\partial }^{2}}P}{\partial {{r}^{2}}}=0\quad \,\,(1)$$ $$P(s,y\,;\,t,x)=\delta_x\,\,,\,\,s\to t$$ we define $$f(\tau,u,{{r}_{s}})=\mathbb{E}[e^{\large{iur_t}}|\,r(s)]$$ where $\tau=t-s$. We know CIR process is a affine process thus $$f(\tau ,u,{{r}_{s}})=\exp [A(\tau\,,\,u)+B(\tau\,,\,u){{r}_{s}}]$$ where \begin{align} & A(0\,,\,u)=0 \\ & B(0\,,\,u)=iu \\ \end{align} Substitute for $f$ in $(1)$
$$\frac{\partial f}{\partial s}+\kappa (\theta -{{r}_{s}})\frac{\partial f}{\partial r}+\frac{1}{2}{{\sigma }^{2}}{{r}_{s}}\frac{{{\partial }^{2}}f}{\partial {{r}^{2}}}=0\,.$$ Note \begin{align} & \frac{\partial f}{\partial s}=-\left( \frac{\partial A}{\partial r}+{{r}_{s}}\frac{\partial B}{\partial r} \right)f \\ & \frac{\partial f}{\partial r}=B\,f \\ & \frac{{{\partial }^{2}}f}{\partial {{r}^{2}}}={{B}^{2}}f \\ \end{align} as a result $$\frac{1}{2}{{\sigma }^{2}}{{r}_{s}}{{B}^{2}}+\kappa (\theta -{{r}_{s}})B-\frac{\partial B}{\partial s}{{r}_{s}}-\frac{\partial A}{\partial s}=0$$ set $r_s=0$, then $$\frac{\partial A}{\partial s}=\kappa \theta B$$ setting $r_s=1$,then $$\frac{\partial B}{\partial s}+\kappa B=\frac{1}{2}{{\sigma }^{2}}{{B}^{2}}$$ The first equation is an ordinary differential equation the second is Riccati equation. If You solve these equations then $$f(t-s,u,{{r}_{s}})={{\left( 1-\frac{iu}{{{c}_{t}}} \right)}^{-\frac{2\kappa \theta }{{{\beta }^{2}}}}}\exp \left( \frac{i\,u{{e}^{-\kappa (t-s)}}}{1-\frac{iu}{c_t}}{{r}_{s}} \right)$$ where \begin{align} & {{c}_{t}}=\frac{2\kappa }{{{\beta }^{2}}\,[1-{{e}^{-\kappa (t-s)}}]} \\ & k=\frac{4\kappa \theta }{{{\beta }^{2}}} \\ & {{\lambda }_{\,t}}=2{{c}_{t}}\,{{r}_{s}}{{e}^{-\kappa (t-s)}} \\ \end{align} By application Inverse Fourier Transform we have $$p(s,y;\,t,x)=\frac{1}{2}{{e}^{-\frac{1}{2}(2{{c}_{t}}{{r}_{t}}+{{\lambda }_{t}})}}{{\left( \frac{2{{c}_{t}}{{r}_{t}}}{{{\lambda }_{t}}} \right)}^{\frac{k}{4}-\frac{1}{2}}}{{I}_{\frac{k}{2}-1}}(\sqrt{2{{c}_{t}}{{r}_{t}}{{\lambda }_{t}}})$$

Proof in the paper by Feller: http://www.jstor.org/stable/1969318

• If you can summarize the major steps, it will be more helpful. Feb 8, 2016 at 20:49
• Gordon's right.. I'd prefer a user's explanation rather than an academic paper which I have to pay.. Feb 9, 2016 at 22:45