Antithetic Sampling with Second-Moment Matching

# Anti-thetic sampling and second moment matching


# Anti-thetic sampling and second moment matching












Background:

This is in reference to ch 7 problem 10 of Mark Joshi's concepts of mathematical finance.

Question:

> A normal random generator produces the following draws:

$$0.68, -0.31, -0.49, -0.19, -0.72, -0.16, -1.01, -1.60, 0.88, -0.97$$ What would these draws become after antithetic sampling and second moment matching.

Solution from Joshi - The sum-square of these and their negatives is $13.3482$. Divide by $20$ to get $0.6674$, whose square root is $\pm 0.81695$. Divide the twenty numbers by this quantity to get

0.83, -0.83,

-0.38, 0.38,

-0.60, 0.60,

-0.23, 0.23,

-0.88, 0.88,

-0.20, 0.20,

-1.24, 1.24,

-1.96, 1.96,

1.08, -1.08,

-1.19, 1.19.

I am confused by this solution as there is no formula for doing this calculation in the book. If someone could point out the formula or how he goes about getting this solution would be appreciated.

## Answer by Giogre (score 2, accepted)

https://quant.stackexchange.com/a/63048

I have just finished working on this exercise, it is rather simple and this is also an old question, but why not share my reasoning here anyway.

As far as the first point, performing antithetic sampling obviously speeds up convergence of the MonteCarlo method by picking two numbers for each draw, instead of one, $x$ and $-x$ with mean $= 0$.

Hence the given sample doubles to a 20-elements array $S$:

$$ S = \left[ \begin{align} 0.68,&-0.68,\\ −0.31,&\; 0.31,\\ −0.49,&\; 0.49,\\ −0.19,&\; 0.19,\\ −0.72,&\; 0.72,\\ −0.16,&\; 0.16,\\ −1.01,&\; 1.01,\\ −1.60,&\; 1.60,\\ 0.88,&\; -0.88,\\ −0.97,&\; 0.97 \end{align}\right] $$

Then in the second point we are asked to perform a second moment matching on $S$. We essentially need to rescale the draws in $S$ so that $$ \mathbb{E}[S^2] = \mathbb{E}[(N(0,1))^2] = \sigma^2_N = 1 $$

Compute $\mathbb{E}[S^2]$:

$$ \mathbb{E}[S^2] = \frac{\sum_{i=1}^{20} s_i^2}{20} = 0.66741 $$

with $s_i$ being the single draws in $S$.

In order to transform $\mathbb{E}[S^2]$ from $0.66741$ to $1$, each $s_i^2$ needs to be divided by $0.66741$. This way the denominator in the sum above will be $20 \cdot 0.66741$, and as a consequence $\mathbb{E}[S^2] = 1$. Thus the normalised set of draws $S$ will be, for $i = 1, 2, \dots, 20$:

$$ S = \frac{s_i}{\sqrt{0.66741}} = \left[ \begin{align} 0.83,& \, -0.83,\\ -0.38,& \, 0.38,\\ -0.60,& \, 0.60,\\ -0.23,& \, 0.23,\\ -0.88,& \, 0.88,\\ -0.20,& \, 0.20,\\ -1.24,& \, 1.24,\\ -1.96,& \, 1.96,\\ 1.08,& \, -1.08,\\ -1.19,& \, 1.19 \end{align} \right] $$