Approximating CIR Brownian Increments from Variance Endpoints
Summary
The document asks how to recover a Brownian increment over a time step when the starting and ending values of a Cox–Ingersoll–Ross variance process are known. The accepted answer applies a positivity-adjusted Euler–Maruyama approximation to the CIR equation and rearranges it to estimate the increment, provided the starting variance is nonzero. This is an approximation based on one discretized step, not an exact conditional distribution.
For a Heston-style simulation, the answer also gives an exact identity for the integral of the square root of variance against its driving Brownian motion, expressed using the variance endpoints and the integrated variance. This can remove that integral from a simulation calculation, although the integrated variance must still be handled. A second answer suggests reusing the normal or uniform random variable from a quadratic-exponential variance step to form a Brownian increment, but explicitly marks that suggestion as probably incorrect. The material therefore distinguishes a practical discretization estimate from a questionable sampling shortcut.
Key ideas
- A positivity-adjusted Euler step gives an approximate Brownian increment from two CIR variance values.
- The increment formula requires a nonzero starting variance and depends on the discretization approximation.
- In the Heston setting, the variance equation provides an identity for the integral of square-root variance against its driver.
- The integrated variance remains necessary when applying that identity.
- Reusing the quadratic-exponential scheme's sampling variable for the increment is presented as likely incorrect.
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Full text
# Sampling change in the driving brownian motion of a CIR process
# Sampling change in the driving brownian motion of a CIR process
I have volatility driven by a CIR process:
$$\mathrm{d}v_t = \kappa (\bar{v}-v_t)\mathrm{d}t + \omega \sqrt{v_t}\mathrm{d}W_v\text{.}\tag{1}$$
I am working with several (complicated) approximations of this process (for example, QE from the Andersen paper). Given $v(t)$, these approximations sample a $v(t+\epsilon)$. However, in addition to having a $v(t+\epsilon)$, I would like to sample
$$\int_t^{t+\epsilon} \mathrm{d}W_v = W_v(t+\epsilon)-W_v(t)\text{.}\tag{2}$$
Question: Given $v(t)$ and $v(t+\epsilon)$, how can I sample from the conditional distribution of (2)?
## Answer by ir7 (score 2, accepted)
https://quant.stackexchange.com/a/65845
One can use the Euler-Maruyama discretization scheme for CIR, 'fixed' for $v$ positivity, to get:
$$ v(t+\epsilon) -v(t)\approx \kappa (\bar{v} -v(t)^+)\epsilon + \omega \sqrt{v(t)^+} (W_v(t+\epsilon) - W_v(t)). $$
So, one approximation of the Brownian increment, when $v(t)$ and $v(t+\epsilon)$ are given, is:
$$ W_v(t+\epsilon) - W_v(t) \approx \frac{v(t+\epsilon) -v(t) - \kappa (\bar{v} -v(t)^+)\epsilon}{\omega \sqrt{v(t)^+} } \;\;\;\;\;({\rm when} \; v(t)\not= 0)$$
Note: In the Heston model context, one usually gets rid of the integral of $\sqrt{v(t)}dW_v(t) $ (integral against $W_v$) using the exact equality (equation (10), page 7 in Andersen's paper):
$$ \int_t^{t+\epsilon} \sqrt{v(u)}dW_v(u) = \omega^{-1} \left(v(t+\epsilon) -v(t) - \kappa \bar{v} \epsilon - \kappa\int_t^{t+\epsilon} v(u)du \right),$$
after employing Cholesky decomposition on $W_X$, leaving to compute an integral against a new Brownian motion $W$ that is independent of $v$, $\int_t^{t+\epsilon} \sqrt{v(u)}dW(u)$.
## Answer by mmencke (score 1)
https://quant.stackexchange.com/a/65832
Edit: This is probably incorrect.
The Quadratic Exponential scheme is the best one I have seen as it converges in distribution and is pretty fast, so nice choice there!
When $\eta$ is constant you can simplify the integral $$ \int_t^{t+\varepsilon}\eta dW(u)=\eta\int_t^{t+\varepsilon}dW(u)=\eta\left(W(t+\varepsilon)-W(t)\right) $$
In the QE scheme you either use a standard uniform variable or a standard normal variable. Denote them $U_V$ and $Z_V$, respectively. We know that changes in a Wiener process is normally distributed as follows $$ \Delta W\equiv W(t+\varepsilon)-W(t)\sim \mathcal{N}(0,\varepsilon) $$ where the second argument is the variance. So in the simulation we can find $$ \Delta W=\begin{cases} \sqrt{\varepsilon}\cdot Z_V&\text{if }\psi\leq\psi_c\\ \sqrt{\varepsilon}\cdot\Phi^{-1}(U_V)&\text{if }\psi>\psi_c \end{cases} $$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.