Control variates

Using what you already know

Simulating a call, you also simulate the stock price at expiry S_T, whose true mean S0 * exp(r * T) you know exactly. In your sample the average of S_T came out a little off, say too high. Since the call payoff and S_T move together, the call's sample mean is probably too high as well. A control variate uses this: it subtracts a multiple of the observed error of the quantity you know from the estimate of the one you want.

import numpy as np

rng = np.random.default_rng(1)
z = rng.standard_normal(2000)
s_t = 100.0 * np.exp((0.05 - 0.5 * 0.2 ** 2) + 0.2 * z)             # terminal prices
call = np.maximum(s_t - 100.0, 0.0)                                  # the payoff we want

known = 100.0 * np.exp(0.05)                                         # E[S_T], exactly
print(round(float(s_t.mean()), 3), round(known, 3))                  # the sample mean of S_T vs its true mean
print(round(float(s_t.mean() - known), 3))                           # the observed error of the control

The best multiple

How much of the control's error should you subtract? The multiple b that minimises the variance of payoff - b * control is b = cov(payoff, control) / var(control), exactly the slope of a regression of the payoff on the control (the same sums as beta in the CAPM topic). The controlled estimate is mean(payoff) - b * (mean(control) - known_mean). It is unbiased, because the term being subtracted has expectation zero.

import numpy as np

def control_coefficient(payoff, control):
    dev = control - control.mean()
    return float(np.sum(dev * (payoff - payoff.mean())) / np.sum(dev ** 2))

rng = np.random.default_rng(1)
z = rng.standard_normal(2000)
s_t = 100.0 * np.exp((0.05 - 0.5 * 0.2 ** 2) + 0.2 * z)
call = np.maximum(s_t - 100.0, 0.0)
known = 100.0 * np.exp(0.05)

b = control_coefficient(call, s_t)
controlled = call.mean() - b * (s_t.mean() - known)
print(round(b, 4), round(float(call.mean()), 4), round(float(controlled), 4))       # the plain mean and the corrected one

# A payoff that is exactly proportional to the control is estimated without any error
import numpy as np

x = np.random.default_rng(2).standard_normal(50) + 5.0          # the control, known mean 5.0
y = 2.0 * x                                                      # the payoff
dev = x - x.mean()
b = np.sum(dev * (y - y.mean())) / np.sum(dev ** 2)
print(round(float(b), 6), round(float(y.mean() - b * (x.mean() - 5.0)), 6))      # 2.0 and exactly 10.0

How much error is saved

The adjusted samples payoff - b * control are what carry the noise, so the standard error of the controlled estimate is their standard deviation (ddof=1) divided by sqrt(n). With correlation rho between payoff and control, the variance falls by a factor 1 - rho^2: a correlation of 0.9 removes 81% of it, and 0.99 removes 98%. The control costs no extra simulation, only a few extra lines.

import numpy as np

def control_variate_error(payoff, control):
    dev = control - control.mean()
    b = np.sum(dev * (payoff - payoff.mean())) / np.sum(dev ** 2)
    return float((payoff - b * control).std(ddof=1) / np.sqrt(len(payoff)))

rng = np.random.default_rng(1)
z = rng.standard_normal(20000)
s_t = 100.0 * np.exp((0.05 - 0.5 * 0.2 ** 2) + 0.2 * z)
for strike in (80.0, 100.0, 120.0):
    call = np.maximum(s_t - strike, 0.0)
    plain = float(call.std(ddof=1) / np.sqrt(len(call)))
    print(strike, round(plain, 4), round(control_variate_error(call, s_t), 4), round(plain / control_variate_error(call, s_t), 2))

• A control variate is a simulated quantity with a known mean; subtract a multiple of its observed error from the estimate: mean(Y) - b * (mean(X) - E[X]).
• The best multiple is the regression slope b = cov(Y, X) / var(X); the error is the standard deviation of Y - b * X over sqrt(n).
• The variance drops by 1 - rho^2; a good control is highly correlated with the payoff and has an exactly known mean, and it costs no extra simulation.

Next up: scikit-learn pipelines for returns, fitting and evaluating predictive models on return data.

Lab · what's graded

• Choose the right multiple with control_coefficient25 pts
• Correct the estimate with control_variate_estimate25 pts
• Shrink the error bar with control_variate_error25 pts
• Count the gain with error_reduction25 pts