OSP-5 Control Variables

Techniques

Basic Idea

Suppose we want to estimate . If we have another variable with known expectation , then

and we can choose such that . If so, since is already known, can be simulated and computed.

Note that

As long as , it is possible to find such that .

In the case of multivariate

and thus

and

For uni-variate case

Sample Splitting and Cross-fitting

Note that the optimal

is unknown. We can estimate from an independent sample of .

We can use the 2-splitting (cross-fitting) method:

Generate independent observations

Split this samples into two parts each with observations. Call the first split and the second split .

Find

and store

Then find

and store

Return

Note that because is independent of , we have

and the same for . Therefore, is unbiased for .

Because are close in probability to , the asymptotic variance of can be estimated by , where

One template implementation using the cross-fitting method is as below


import numpy as np

def control_variable_cross_fitting(Y, X, E_X):
	"""
	to estimate E[Y], with X as control variable. E[X] = E_X is known
	"""
	n = len(X)
	I_1, I_2 = (X[:n//2], Y[:n//2]), (X[n//2:], Y[n//2:])
	splits = [(I_1, I_2), (I_2, I_1)]
	split_means = []
	split_ssq = []
	for (X_1, Y_1), (X_2, Y_2) in splits:
		# compute b_star on one split
		b_star = -np.sum((X_1 - E_X) * (Y_1 - np.mean(Y_1))) / np.sum((X_1 - E_X)**2)
		# estimate theta on the other split
		H_i = Y_2 + b_star * (X_2 - E_X)
		H_i_bar = np.mean(H_i)
		ssq = np.sum((H_i - H_i_bar) ** 2)
		split_means.append(H_i_bar)
		split_ssq.append(ssq)

	est_mean = np.mean(split_means)
	est_var = np.sum(split_ssq) / (n * (n - 1))
	
	return est_mean, est_var

Leave-one-out version

Instead of splitting into two parts (which can yield external randomness), we can use the leave-one-out method based on the fact that

for any computed using .

This yields the estimator

where

Note that here we do not compute for different to avoid the computation cost but use correction terms in the denominator (based on the Sherman-Morrison Formula, proof omitted).

The variance estimator is , where

Examples

Example (0)

Suppose we want to estimate where .

We can use as a control variable with known expectation of 3/2.

Example (1)

Consider the option pricing, the price is given by

Since under the risk-neutral measure, forms a martingale which implies

Hence, is a random variable with known mean and it’s reasonable to expect that and have a non-zero convariance.

For example,


S_0 = 50
K = 60
r = 0.05
sigma = 0.3
T = 1
n = 10000

f1 = (r - 0.5 * sigma ** 2) * T
f2 = sigma * np.sqrt(T)
z = rng.normal(size=n)
S_T = S_0 * np.exp(f1 + f2 * z)
pv = np.exp(-r * T)
C = pv * np.maximum(S_T - K, 0)
E_X = S_0 * np.exp(r * T) # known E[X]

est_mean, est_var = control_variable_cross_fitting(C, S_T, E_X)

Example (2)

An Asian arithmetic average call option has a payoff

which does not have a closed-form solution under the BS model.

A Geometric average Asian call option, however, has a closed-form solution under the BS model. Its payoff is

thus we can use (or just as a control variable for the .