# 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
• Do you want to calculate the delta by Monte Carlo method? – user16651 Nov 4 '16 at 8:49
• No. I am only using a delta hedge to obtain a smaller standard deviation in the MC simulation, which will lead me to a smaller standard error. I believe if I combine it with a gamma hedge, it will reduce the SE to smaller than the antithetic variate, but was curious if delta alone is generally supposed to be smaller than this. – Probability1 Nov 4 '16 at 11:55
• Which textbook are you using? – SRKX Nov 7 '16 at 2:39

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 $f$.

• Thanks! Great answer. I wish my textbook explained it this way haha. – Probability1 Nov 4 '16 at 22:23