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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)?

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  • $\begingroup$ 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). $\endgroup$
    – user54908
    Jul 2 at 10:40
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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 $v$, $\int_t^{t+\epsilon} \sqrt{v(u)}dW(u)$.

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  • $\begingroup$ 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? $\endgroup$
    – user54908
    Jul 2 at 17:51
  • $\begingroup$ Done. I'm glad it helped. $\endgroup$
    – ir7
    Jul 2 at 17:54
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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 $$ \Delta W=\begin{cases} \sqrt{\varepsilon}\cdot Z_V&\text{if }\psi\leq\psi_c\\ \sqrt{\varepsilon}\cdot\Phi^{-1}(U_V)&\text{if }\psi>\psi_c \end{cases} $$

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  • $\begingroup$ 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. $\endgroup$
    – user54908
    Jul 2 at 10:18
  • $\begingroup$ My apologies - I have updated my answer with a description of how one can generate the Wiener path simultaneously. $\endgroup$
    – mmencke
    Jul 2 at 11:10
  • $\begingroup$ 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 $W_v$. $\endgroup$
    – user54908
    Jul 2 at 11:32
  • $\begingroup$ Importantly: $V(t+\epsilon)$ does not uniquely determine $W_v(t+\epsilon)$. This gives me more reason to doubt that this approximates $\Delta W_v$. $\endgroup$
    – user54908
    Jul 2 at 11:46
  • $\begingroup$ With no further explanation provided, I have to say that this incorrectly answers the question. $\endgroup$
    – user54908
    Jul 2 at 13:33

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