# 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)?

• One idea I had to solve my problem, that goes in a different direction of the OP (but I believe may still answer the OP due to Bayes' Theorem), is to sample $v(t+\epsilon)$ given (2). I'm imagining sampling a $W_v(t+\epsilon)$ from the normal distribution, and then using the brownian bridge to rewrite (1) and see if I can apply the QE algorithm to the resulting SDE. I haven't figured out how to get this to work either, but if you do, I can modify the OP to be more open to that type of solution (without you having to explain the Bayes' theorem step). Jul 2 at 10:40

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 $$v$$, $$\int_t^{t+\epsilon} \sqrt{v(u)}dW(u)$$.

• This solves the problem. Thanks! When $v(t)=0$, as often happens in QE, I plan to approximate $\int_t^{t+\epsilon} \mathrm{d}W_v=0$. Also: your "Note" will probably confuse future readers without additional context. I'm not trying to sample $W_v(t+\epsilon)$ to simulate $X$. Perhaps add a citation to equation (10) on page 7 of the Andersen paper? Jul 2 at 17:51
• Done. I'm glad it helped.
– ir7
Jul 2 at 17:54

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 $$\Delta W=\begin{cases} \sqrt{\varepsilon}\cdot Z_V&\text{if }\psi\leq\psi_c\\ \sqrt{\varepsilon}\cdot\Phi^{-1}(U_V)&\text{if }\psi>\psi_c \end{cases}$$

• No, I don't have $W_v(t+\epsilon)$. As far as I know, the process $W_v$ is not used explicitly or implicitly in the QE algorithm. If it were, that would also solve my problem. Jul 2 at 10:18
• My apologies - I have updated my answer with a description of how one can generate the Wiener path simultaneously. Jul 2 at 11:10
• I'm struggling to see why the $W_v(t+\epsilon)-W_v(t)$ you recommend is one that is implied in the QE scheme. My understanding is of QE is that we are approximating the distribution of $V(t+\epsilon)$ given $V(t)$ using two simpler distributions. I don't believe the random variables we sample have this relationship with $W_v$. Jul 2 at 11:32
• Importantly: $V(t+\epsilon)$ does not uniquely determine $W_v(t+\epsilon)$. This gives me more reason to doubt that this approximates $\Delta W_v$. Jul 2 at 11:46
• With no further explanation provided, I have to say that this incorrectly answers the question. Jul 2 at 13:33