Improving Monte Carlo: Control Variates

I’ve already discussed quite a lot about Monte Carlo in quantitative finance. MC can be used to value products for which an analytical price is not available in a given model, which includes most exotic derivatives in most models. However, two big problems are the time that it takes to run and the ‘Monte Carlo error’ in results.

One technique for improving MC is to use a ‘control variate’. The idea is to find a product whose price is strongly correlated to the product that we’re trying to price, but which is more easy to calculate or which we already know the price of. When we simulate a path in MC, it will almost surely give either an under-estimate or an over-estimate of the true price, but we don’t know which, and averaging all of these errors is what leads to the Monte Carlo error in the final result. The insight in the control variate technique is to use the knowledge given to us by the control variate to reduce this error. If the two prices are strongly correlated and a path produces an over-estimate of product price, it most likely also produces an over-estimate of the control variate and visa versa, which will allow us to improve our estimate of the product we’re trying to price.

The textbook example is the Asian Option. Although the arithmetic version of the asian option discussed in previous posts has no analytic expression in BS, a similar Geometric asian option does have an analytic price. So, for a given set of model parameters, we can calculate the price of the option. As a reminder, the payoff of an arithmetic asian option at expiry is

    \[C_{\rm arit}(T) = \Bigl({1\over N}\sum_{i=0}^{N-1} S(t_i) - K \Bigr)^+\]

and the payoff of the related geometric averaging asian is

    \[C_{\rm geo}(T) = \Bigl( \bigl(\prod_{i=0}^{N-1} S(t_i)\bigr)^{1\over N} - K \Bigr)^+\]

Denoting the price of the arithmetic option as X and the geometric option as Y, the traditional monte carlo approach is to generate N paths, and for each one calculate X_i (the realisation of the payoff along the path) and take the average over all paths, so that

    \[{\mathbb E}[X] = {1 \over N} \sum_{i=0}^{N-1} X_i\]

which will get closer to the true price as N \to \infty.

Using Y as a control variate, we instead calculate

    \[{\mathbb E}[X] = {1\over N} \sum_{i=0}^{N-1}\bigl( X_i - \lambda( Y_i - {\mathbb E}[Y] ) \bigr)\]

where {\mathbb E}[Y] is the price of the geometric option known from the analytical expression, and \lambda is a constant (in this case we will set it to 1).

What do we gain from this? Well, consider the variance of X_i - \lambda (Y_i - {\mathbb E } [Y])

    \[{\rm Var}\bigl( X_i - ( Y_i - {\mathbb E}[Y] ) \bigr) = {\rm Var}( X_i ) + \lambda^2 {\rm Var} ( Y_i ) - 2 \lambda {\rm Cov}( X_i Y_i )\]

(since {\mathbb E } [Y] is known, so has zero variance) which is minimal for \lambda = \sqrt{ {\rm Var}(X_i) \over {\rm Var}(Y_i) }\cdot \rho(X_i,Y_i) in which case

    \[{ {\rm Var}\bigl( X_i - ( Y_i - {\rm E}[Y] ) \bigr) \over {\rm Var}( X_i )} = 1 - \rho(X_i,Y_i)^2\]

that is, if the two prices are strongly correlated, the variance of the price calculated using the control variate will be a significantly smaller. I’ve plotted a sketch of the prices of the two types of average for 100 paths – the correlation is about 99.98%. Consequently, we expect to see a reduction in variance of about 2000 times for a given number of paths (although we now have to do a little more work on each path, as we need to calculate the geometric average as well as the arithmetic average of spots). This is roughly 45 times smaller standard error on pricing – well over an extra decimal place, which isn’t bad – and this is certainly much easier than running 2000 times as many paths to achieve the same result.

The relationship between payout for a geometric and an arithmetic asian option, which here demonstrate a 99.98% sample correlation
The relationship between payout for a geometric and an arithmetic asian option, which here demonstrate a 99.98% sample correlation. Parameters used were: r: 3%; vol: 10%; K: 105; S(0): 100; averaging dates: monthly intervals for a year

Leave a Reply

Your email address will not be published. Required fields are marked *