Continuous dependence of Cox-Ingersoll-Ross (CIR) process on its initial value and model parameters

Question

Suppose we are considering a Cox-Ingersoll-Ross (CIR) process $$\mathrm{d}\nu_t=\kappa\left(\theta-\nu_t\right)\mathrm{d}t+\sigma\sqrt{\nu_t}\mathrm{d}W_t,$$ with deterministic initial value $\nu_0>0$, where $W$ is a standard Brownian motion. You can assume that Feller's condition holds to guarantee the positivity of its unique solution.

My question is, for some fixed $T>0$, is there any result about continuous dependence (with an order) of the resulting CIR process on model parameters $\nu_0$, $\kappa$, $\theta$ and $\sigma$? To be more concrete, let us take $\nu_0$ for example. Suppose there are another CIR process $$\mathrm{d}\mu_t=\kappa\left(\theta-\mu_t\right)\mathrm{d}t+\sigma\sqrt{\mu_t}\mathrm{d}W_t$$ with initial value $\mu_0=\nu_0+\epsilon>0$, where $\epsilon$ is a deterministic shift on the initial value which can be sufficiently small. I wonder whether there are results claiming that $$\sup_{0\leq t\leq T}\mathbb{E}\left[\left|\nu_t-\mu_t\right|^2\right]$$ is $O(\sqrt{\left|\epsilon\right|})$ (or $O(\left|\epsilon\right|)$ or $O(\epsilon^2)$ or another power of $\epsilon$ if possible)?

Answer 1

Let's define $X_t=\sqrt{\nu_t}$. By using the Ito's lemma, it's easy to prove that $$\cases{dX_t=\left(\left(\frac{\kappa\theta }{2}-\frac{\sigma^2 }{2} \right)\frac{1}{X_t}-\frac{\kappa}{8}X_t \right)dt + \frac{\sigma}{2}dW_t \tag{1}\\ X_0 = \sqrt{\nu_0} }$$

Define $X_{\epsilon,t}=\sqrt{\mu_t}$, we have also $$\cases{dX_{\epsilon,t}=\left(\left(\frac{\kappa\theta }{2}-\frac{\sigma^2 }{8} \right)\frac{1}{X_{\epsilon,t}}-\frac{\kappa}{2}X_{\epsilon,t} \right)dt + \frac{\sigma}{2}dW_t\tag{2}\\ X_{\epsilon,0} = \sqrt{\nu_0+\epsilon} }$$

Define $Z_{t} = X_{\epsilon,t}-X_t$, take $(1)-(2)$, we deduce $$dZ_t=d(X_{\epsilon,t}-X_t)=-\left(\left(\frac{\kappa\theta }{2}-\frac{\sigma^2 }{8} \right)\frac{1}{X_t\cdot X_{\epsilon,t}}+\frac{\kappa}{2} \right)Z_tdt $$

We remind that the CIR process has unique solution. So, for $\epsilon \ne 0$, the sign of $Z_t$ must remain unchanged over the domain $[0,T]$ (if not, suppose there exists a time $t_0 \in (0,T]$ such that $X_{\epsilon,t_0}=X_{t_0} \iff \nu_{t_0}=\mu_{t_0}$ then $\nu_t = \mu_t$ for all $t \in [0,T]$, in particular, $\nu_0 = \mu_0$: contradiction).

Then, WOLG, let's suppose that $Z_t>0$, we compute $d(\ln(Z_t))$ by using Ito's lemma, we have: $$\begin{align} d(\ln(Z_t))&=\frac{1}{Z_t}dZ_t +\frac{1}{2} \left( -\frac{1}{Z_t^2} \right)d[Z]_t\\ &=-\left(\left(\frac{\kappa\theta }{2}-\frac{\sigma^2 }{8} \right)\frac{1}{X_t\cdot X_{\epsilon,t}}+\frac{\kappa}{2} \right)dt \le -\frac{\kappa}{2} dt \tag{3} \end{align}$$ because $X_t$ and $X_{\epsilon,t}$ are positive and by the Feller condition we have $\left(\frac{\kappa\theta }{2}-\frac{\sigma^2 }{8} \right) >0$

From $(3)$, we deduce that $$\ln Z_t - \ln Z_0 \le -\frac{1}{2}\kappa t \iff \color{red}{Z_t \le (\sqrt{\nu_0+\epsilon} - \nu)\cdot e^{-\frac{1}{2}\kappa t}}\tag{4}$$

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Return now to the main question, we have: $$\begin{align} \mathbb{E} \left(|\nu_t - \mu_t|^2 \right) &= \mathbb{E} \left(|\nu_t - \mu_t|^2 \right) \\ &=\mathbb{E} \left(\underbrace{|\sqrt{\nu_t} - \sqrt{\mu_t}|^2}_{=Z_t^2} |\sqrt{\nu_t} + \sqrt{\mu_t}|^2 \right) \\ &\le \sqrt{\mathbb{E} \left(Z_t^4\right) \cdot \mathbb{E}\left( |\sqrt{\nu_t} + \sqrt{\mu_t}|^4 \right) } \hspace{1cm} \text{by cauchy-schwarz inequality} \tag{5} \end{align}$$

We have $$\mathbb{E}\left( |\sqrt{\nu_t} + \sqrt{\mu_t}|^4 \right) \le \mathbb{E}\left( (2(\nu_t + \mu_t))^2 \right) \le 8 \mathbb{E}\left( \nu_t^2+\mu_t^2 \right)$$ and as the expectation and the variance of CIR process are bounded, there exists a known, bounded function $M_t <+\infty$ such that $\mathbb{E}\left( |\sqrt{\nu_t} + \sqrt{\mu_t}|^4 \right) \le M_t$

From $(4),(5)$ we have $$\mathbb{E} \left(|\nu_t - \mu_t|^2 \right) \le \sqrt{M_t}\cdot \sqrt{e^{-2\kappa t}\cdot\left(\sqrt{\nu_0 + \epsilon}-\sqrt{\nu_0} \right)^4 } = \mathcal{O} (|\epsilon|)$$

We can conclude then $$\color{red}{\sup_{t\in[0,T]}\mathbb{E} \left(|\nu_t - \mu_t|^2 \right) = \mathcal{O} (|\epsilon|)}$$