I am trying to understand how quasi Monte Carlo (QMC) and the Brownian bridge (BB) can be combined to price an asset, but I am having a hard time understanding how. I am just considering a European option on a single asset modelled by a simple scalar SDE $S_t=a(S_t,t)\mathrm{d}t+b(S_t,t)\mathrm{d}W_t$. Here is what I understand so far.

  1. In the usual Euler-Maruyama approximation, we discretise using the $\hat{S}_{t+1}=a(\hat{S}_t,t)\Delta t+b(\hat{S}_t,t)\sqrt{\Delta t}\mathrm{d}Z_t$.

  2. In standard Monte Carlo (SMC), $\mathbb{E}f(S_t)\approx\frac1{L}\sum_{i=1}^Lf(S_t^{(L)})$, where $L$ is the simulation number.

  3. We can also use a single BB with $M=2^N$ uniform timesteps by generating the end point first, and using $W_t|W_T\sim\mathcal{N}\left(\frac{t}TW_T,\frac{t(T-t)}{T}\right)$ to generate midpoints recursively.

  4. In QMC, instead of pseudo-random numbers, we generate using low-discrepancy sequences, e.g. rank-1 lattice/Sobol points.

I therefore have the following questions:

  1. How do I combine the BB and QMC ideas? I haven't found a resource that really helps me understand how to use the two of them together.

  2. The variance of the BB is geometrically decreasing, isn't that a good thing? I read that the BB is only helpful for QMC estimation of the European option, but not for SMC. Where does this come from? The same resource mentions that they "have the same variance and thus convergence rate of $\mathcal{O}(L^{-1})$", which does not make sense to me.

  3. How does this extend to multiple assets? So far, I understand the Cholesky decomposition $LL^\mathrm{T}=\Sigma$, $L{\bf W}_t^\perp={\bf W}_t$ helps us capture the covariance, but I don't see how this works with QMC/BB.

Edit: the resource I was referring to in particular is this one on page 7, which says "Both schemes have the same variance, hence their MC convergence rates are the same, but QMC sampling shows different efficiencies for SD and BBD".


Combining Brownian bridges with quasi-Monte Carlo

  1. How do I combine the BB and QMC ideas?

Brownian bridges is just a means of producing a Brownian path $W$, and Monte Carlo is a means by which we can evaluate expectations and integrals. Regular Monte Carlo would use a pseudo-random point set $U$ to estimate such things, whereas quasi-Monte Carlo would use a low-discrepancy point set $\tilde{U}$. So what needs to be changed in the Monte Carlo scheme? The uniforms which generate the Gaussian increments $Z$ (aka the Normal random variables) should be switched for low-discrepancy quasi-random uniforms. To preserve the low-discrepancy property of the sequence, it is best to use a mapping which doesn't reject any samples. Arguably best for this is the inverse transform method, for which we usually have for regular Monte Carlo $Z=\Phi^{-1}(U)$, where $\Phi^{-1}$ is the inverse cumulative distribution function of the standard Gaussian distribution. For quasi-Monte Carlo you instead use quasi-Gaussian random variables $\tilde{Z}=\Phi^{-1}(\tilde{U})$.

Aside from this, you can largely proceed as normal, although for good estimates you really want to be doing randomised quasi-Monte Carlo.

I haven't found a resource that really helps me understand how to use the two of them together.

I think the best resource is Randomized Quasi-Monte Carlo: An Introduction for Practitioners by Pierre l’Ecuyer.

  1. How does this extend to multiple assets?

With this in mind, I think you just proceed as you usually would, but using the set of quasi-Gaussian random variables (and being mindful of randomising your uniform sequences).

Regarding your second question about SMC (Sequential Monte Carlo I presume), the variance of Brownian bridges, and your (unnamed) resource, I don't know enough to suggest an answer addressing this point.

  • $\begingroup$ Hi, thanks for the answer, I have updated my question with the link to where I got the quote from, although in general the idea seems to be that the Brownian bridge construction is only helpful for QMC in pricing a European option, but not for standard MC. I’ll read the link you have posted too before asking further questions! $\endgroup$
    – user107224
    Mar 5 at 20:04

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