CIR process characteristic function

Question

what is the characteristic function of the CIR process given by $dv_t = \kappa (\theta - v_t)dt + \sigma \sqrt{v_t}dW_t$ Unfortunately, I could not find the answer in the literature. I know it is in the class of affine diffusion processes, but how can we find the characteristic function?

Answer 1

The distribution of the variance $v_t$ is known, see here. We have $$v_t=\frac{1}{2c_t}Y,$$ where $$c_t=\frac{2\kappa}{(1-e^{-\kappa t})\sigma^2}$$ and $Y$ follows a non-central $\chi^2$ distribution with $k=\frac{4\kappa\theta}{\sigma^2}$ degrees of freedom and non-centrality parameter $\lambda_t=2c_tv_0e^{-\kappa t}$. The characteristic function of $Y$ is known to be $$\varphi_Y(u)=\frac{1}{(1-2iu)^{k/2}}\exp\left(\frac{iu\lambda_t}{1-2iu}\right).$$ The characteristic function of $v_t$ is thus $$\varphi_{v_t}(u)=\varphi_{Y/2c_t}(u)=\varphi_Y\left(\frac{u}{2c_t}\right)=\frac{1}{(1-iu/c_t)^{k/2}}\exp\left(\frac{iu\lambda_t/2}{c_t-iu}\right).$$

This is the characteristic function of $v_t$, i.e. the variance in the Heston (1993) model or the short rate $r_t$ in the Cox et al. (1985) model. This is very different to the characteristic function of the (log-)stock price in the Heston (1993) model that you need for usual option pricing.

Answer 2

In addition to @Kevin ‘s splendid answer, you can make use of the affine properties you have just described. The relevant machinery is found in DPS2000, equations (2.4) thru (2.6).

We are looking for

$$ \phi(u,t) \equiv E\left(e^{iuv_t}\right)$$

where $E(\cdot)\equiv E(\cdot|\mathcal{F_0})$. For affine processes, this expectation can be solved thru a cleverly chosen set of differential equations, i.e. Feynman-Kac (see paper):

$$ \phi(u,s) = E\left(e^{\alpha_s + \beta_s v_s}\right) \mathrm{s.t.} \alpha_t=0, \beta_t=ui$$

Where the functions $\alpha(t), \beta(t)$ satisfy ordinary differential equations in case of a CIR process:

$$\frac{\partial \beta}{\partial \tau} = -\kappa\beta_\tau + \frac{1}{2}\sigma^2\beta_\tau^2$$ subject to $\beta_0 = ui$ and $$ \frac{\partial \alpha}{\partial \tau}=\kappa\theta\beta_\tau$$ subject to $\alpha_0=0$

NB: I have flipped the time dimension from $s:0\to t$ to $\tau:t\to 0$, hence you find different signs when comparing to the paper

You can now solve the ODEs, starting with $\beta$ and plugging the result back into the ODE for $\alpha$.