Deriving the Heston CIR Variance Process from an Ornstein–Uhlenbeck Model
Summary
This note derives the square-root variance process used in the Heston model from a process for the square root of variance. Applying Itô’s lemma to the square of that process adds a variance correction term to the drift and scales the diffusion by the square root of variance. Relabeling the parameters then puts the result in the standard Cox–Ingersoll–Ross form.
The note also works in reverse: applying Itô’s lemma to the square root of a CIR variance process recovers the Ornstein–Uhlenbeck form under the same parameter mapping. These calculations explain how the two specifications are related. They are an algebraic derivation, not empirical evidence about volatility behavior; the parameter correspondence depends on the stated equations and does not establish that either process is a suitable model in a given application.
Key ideas
- Applying Itô’s lemma to the squared square-root variance process produces a drift correction from its diffusion.
- The resulting variance process has mean-reverting drift and diffusion proportional to the square root of variance.
- A parameter mapping connects the Ornstein–Uhlenbeck and CIR specifications.
- Applying Itô’s lemma in the reverse direction recovers the square-root process.
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# How to get the CIR process in the Heston Model from the Ornstein-Uhlenbeck process modeling volatility
# How to get the CIR process in the Heston Model from the Ornstein-Uhlenbeck process modeling volatility
I am reading the Wikipedia page for the Heston Model and it said that with the Ornstein-Uhlenbeck process that models volatility, ${\displaystyle d{\sqrt {\nu _{t}}}=-\theta {\sqrt {\nu _{t}}}\,dt+\delta \,dW_{t}^{\nu }}$, using Ito's lemma you can get the CIR process ${\displaystyle d\nu _{t}=\kappa (\theta -\nu _{t})\,dt+\xi {\sqrt {\nu _{t}}}\,dW_{t}^{\nu }}$. I don't understand how you can get the second equation from the first, so I was was wondering if someone could explain that to me?
## Answer by Kevin (score 2, accepted)
https://quant.stackexchange.com/a/81688
There are two ways to go about your question. First, start from the process for $\sqrt{\nu_t}$ and get to the Heston SDE for the variance $\nu_t$. Conversely, you can start with the Heston SDE for the variance $\nu_t$ and find the process for $\sqrt{\nu_t}$. Both should give you the same results, of course.
Suppose $\text{d}X_t=\mu_t\text{d}t+\sigma_t\text{d}B_t$. Itô's Lemma tells us that $$\text{d}(X^2_t)=\left(2\mu_tX_t+\sigma^2_t\right)\text{d}t+2\sigma_tX_t\text{d}B_t.$$
In your example, $X_t=\sqrt{\nu_t}$, $\mu_t=-\theta\sqrt{\nu_t}$, and $\sigma_t=\delta$. Thus, $$\text{d}\nu_t=\left(-2\theta\nu_t+\delta^2\right)\text dt+2\delta\sqrt{\nu_t}\text{d}B_t.$$
Finally, define $\kappa=2\theta$, $\bar{\nu}=\frac{\delta^2}{2\theta}$, and $\xi=2\delta$ and you get $$\text d\nu_t=\kappa\left(\bar{\nu}-\nu_t\right)\text dt+\xi\sqrt{\nu_t}\text{d}B_t.$$
Conversely, suppose again $\text{d}X_t=\mu_t\text{d}t+\sigma_t\text{d}B_t$. Itô's Lemma also tells us that $$\text{d}\sqrt{X_t}=\left(\frac{1}{2\sqrt{X_t}}\mu_t-\frac{1}{8(X_t)^{3/2}}\sigma^2_t\right)\text{d}t+\frac{\sigma_t}{2\sqrt{X_t}}\text{d}B_t.$$ Now, $X_t=\nu_t$, $\mu_t=\kappa(\bar{\nu}-\nu_t)$, and $\sigma_t=\xi\sqrt{\nu_t}$. Thus, \begin{align} \text{d}\sqrt{\nu_t}&=\left(\frac{\kappa(\bar{\nu}-\nu_t)}{2\sqrt{\nu_t}}-\frac{\xi^2\nu_t}{8(\nu_t)^{3/2}}\right)\text{d}t+\frac{\xi\sqrt{\nu_t}}{2\sqrt{\nu_t}}\text{d}B_t \\ &= \left(\frac{4\kappa\bar{\nu}-\xi^2-4\kappa\nu_t}{8(\nu_t)^{1/2}}\right)\text{d}t+\frac{\xi}{2}\text{d}B_t. \end{align} Using again our definitions that $\kappa=2\theta$, $\bar{\nu}=\frac{\delta^2}{2\theta}$, and $\xi=2\delta$, we obtain \begin{align} \text{d}\sqrt{\nu_t}&=-\theta\sqrt{\nu_t}\text{d}t+\delta \text{d}B_t. \end{align}Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.