# 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.

• Anthithetic means given 10 numbers {x,y,z,...} you generate 20 numbers consisting of positive/negative pairs {x,-x,y,-y,z,-z,...} that's easy. But also you want the standard deviation to be exactly 1. Because the standard deviation of these 20 is a little two low (0.8) you divide by 0.8 to scale the numbers up somewhat to exactly hit $\sigma=1$ Feb 1, 2018 at 20:46
• This is quite clearly explained in the section "Variance Reduction". Quoting: "For example, if the variance of our sample is $V$ and we rescale every random number by $V^{-1/2}$, we obtain a sample with variance 1.". Feb 1, 2018 at 23:22

## 1 Answer

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]