I'm self studying for an actuarial exam and I am curious about a property of the antithetic variate method for increasing the Monte Carlo price accuracy (i.e. For every random draw of $z$, also include a draw of $-z$ in the simulation).


Assume the Black-Scholes framework and consider a European call option with strike $K$ expiring in $T$ years on a non-dividend paying stock currently priced at $S_0$ with an annual volatility $\sigma$. Suppose that a Monte Carlo simulation is used to estimate the expected value at expiration of the option.

The simulation was performed using $n$ draws $u_1, u_2, ..., u_n$ from a uniform distribution to generate the stock price. Suppose that each of these draws generates a stock price at expiration which gives a zero payoff for the call option and therefore $E(\text{Payoff}) = \frac{1}{n} \sum_{i = 1}^n C(S_T^i, K, T) = 0$, where $S_T^i$ is the stock price at expiration for the $i$th draw.

Using the same uniform draws, and applying the antithetic variate method, will $E(\text{Payoff}) = \frac{1}{2n} \sum_{i = 1}^{2n} C(S_T^i, K, T) > 0$ necessarily?

My intuition says yes, but I don't have a way of convincing myself why.


No, you can have

$$ \frac{1}{2n}\sum_{i=1}^{2n} C(S^i_T,K,T) = 0 $$

First off, there's the obvious case where $n=1$ and $u_1 = 0.5$

More generally, for options way out of the money it is common to have

$$ \frac{1}{n}\sum_{i=1}^{n} C(S^i_T,K,T) = 0 $$

even for very large $n$. Antithetic sampling does not change that.

  • $\begingroup$ Typically, using rare event sampling helps event these prices out $\endgroup$ – Drew Jul 13 '16 at 19:51

No. Antithetic variable method is usually for generating smaller standard error than your non-antithetic method, which is a direct result of the negative correlation between original variable and the antithetic variable.

For OTM option, there definitely will be a lot of path ending up with value 0. What may be a choice is to use importance sampling.

Write out the expectation under RN measure and manually extract another normal density with higher mean (in this case). Then use Monte Carlo method to get this new expectation. You can certainly apply antithetic method to it also to reduce the SE.


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