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)?

  • $\begingroup$ Are the Brownian motions in these two processes jointly formed a two-dimensional Brownian motions? If they are the identical Brownian motion then the two processes are parallel and just differ by the constant shift $\epsilon$? $\endgroup$
    – BGM
    Dec 26, 2023 at 17:16
  • 1
    $\begingroup$ @BGM Yes, they are the identical Brownian motion, but it does not mean the two processes are parallel because their values at any time t are determined by their values on [0,t). What you said is right only when the right hand side of the differential equation is independent of the solution itself, which is not the case for CIR process. :) $\endgroup$
    – Shiningale
    Dec 27, 2023 at 2:08

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

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

  • 1
    $\begingroup$ Thank you very much! There is a small problem in your proof, that is, σ*sqrt(x) is not Lipschitz continuous with respect to x near x=0. However, uniqueness of strong solution to CIR equation has been proved by Yamada and Watanabe (see their article dx.doi.org/10.1215/kjm/1250523691 for more detailed information), so it doesn't matter. By the way, could you please give me some references related to your answer so that I can cite it in my paper? :) $\endgroup$
    – Shiningale
    Dec 27, 2023 at 2:03
  • $\begingroup$ @Shiningale I used only two references (the Lipschitz condition and the expectation, variance of CIR process). And you can use this answer in your paper, there is no problem. You don't need to cite it. $\endgroup$
    – NN2
    Dec 27, 2023 at 8:03
  • $\begingroup$ @NN2 I edited out the Lipschitz claim; I apologize in advance if you don't welcome it and you can undo or modify it further if you want. $\endgroup$
    – Snoop
    Dec 27, 2023 at 11:24
  • $\begingroup$ @Snoop thank you for the edit. The edit looks ok for me. $\endgroup$
    – NN2
    Dec 27, 2023 at 11:57
  • $\begingroup$ @NN2 thank you for your edit and answer :) $\endgroup$
    – Shiningale
    Dec 27, 2023 at 12:56

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