Comparing Delta-Hedge Control Variates with Antithetic Sampling
Summary
The discussion compares two ways to reduce Monte Carlo estimation error: a delta hedge used as a control variate and antithetic sampling. It explains that antithetic draws pair opposite samples, enforcing symmetry around the sampling mean. A delta hedge removes the payoff’s linear component, but does not ensure that any finite sample set has zero mean, so its variance reduction need not exceed the antithetic method’s.
The answer also compares computational efficiency rather than standard error alone. Antithetic sampling requires evaluating the payoff on paired draws, while a control variate requires estimating the hedge alongside the payoff. The relative cost depends on whether generating samples or evaluating the payoff and its delta dominates. The document gives no numerical comparison or general ranking; results depend on the payoff, the estimator, and implementation costs.
Key ideas
- Antithetic sampling pairs opposite draws to create symmetry in the sample.
- A delta hedge removes a payoff’s linear component but does not guarantee a zero-mean finite sample.
- Variance reduction from either method depends on the payoff and sampling setup.
- Efficiency should compare standard error with the computational cost of each method.
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# Answer by Brian B (score 3, accepted)
# In a Monte Carlo simulation, will a delta hedge control variate necessarily reduce the standard error more than an antithetic variate?
I have four Monte Carlo simulations and will list them in order of highest standard error to lowest.
- Plain MC
- MC with delta hedge control variate
- MC with antithetic variate
- MC with antithetic and delta variates
My textbook makes it seem as though the delta hedge should reduce SE more than the antithetic variate, but my results always show different. I am using 300 time steps and 100,000 simulations.
If it helps, my Black-Scholes $\Delta$ is computed as
```
(exp(-div*(M-t))*pnorm((log(St1/K)+(r-div+((sig^2)/2))*(M-t))/(sig*(sqrt(M-t))))
```
Where:
- `t = (i-1)*dt`
- `i` is the time step
## Answer by Brian B (score 3, accepted)
https://quant.stackexchange.com/a/30868
If your payoff is linear, then it is a little tough to see what's going on, so let's consider the quadratic case. Here's a generic quadratic to sample, centered at zero
Antithetic sampling introduces samples with a mean perfectly equal to zero, which effectively introduces perfect bilateral symmetry to the whole problem
In contrast, delta hedging will remove the linear component (just as antithetic sampling did) but will not force your samples to have zero mean, therefore any given set of samples will have some slight bias.
Finally, it's worth noting that for antithetic sampling with $N$ original samples, you have to compute $f(x)$ $2N$ times, rather than just $N$ times. For delta hedging you have to compute $\Delta_f(x)$ $N$ times in addition to the $N$ calculations of $f$.
It could be that the cost of computing $C(\{\Delta_f(x)\}, N)$ is quite cheap or essentially free, as when you are computing the Black-Scholes formula and need that component anyway, i.e.
$$ C(\{f(x), \Delta_f(x)\}, N) \approx C(\{f(x)\}, N) $$
or it could be that it costs quite a bit more, for example if you are using automatic differentiation on complex formulas
$$ C(\{f(x), \Delta_f(x)\}, N) \gg C(\{f(x)\}, N) $$
For the antithetic sampling, it may be that the calculation of $f(x)$ is so cheap that the main cost lies in forming the pseudorandom or quasirandom samples so that
$$ C(\{f(x)\}, 2N) \approx C(\{f(x)\}, N) $$
or it may be the case that the cost is all in calculation of $f(x)$ so that
$$ C(\{f(x)\}, 2N) \approx 2 C(\{f(x)\}, N) $$
Therefore from the point of view of efficiency, it is hard to say more about whether antithetics or delta hedging will achieve the best cost to standard error ratio without knowing details of $f$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.