# Solved – Variance reduction technique in Monte Carlo integration

I have some trouble understanding the variance reduction method called
"Antithetic variables":

Suppose that the integrand is \$g(x)=x^2\$ and the reference density \$f(x)=e^{-x}I_{[0,infty]}\$ is the Exp(1) density. The associated cdf is \$\$F_X(x)={1-e^{-x}}mathbb{I}_{[0,infty]}(x)\$\$

The antithetic variable method then uses the inverse cdf representation
begin{align*}
U&=1-e^{-X}\ X&=-log(1-U)
end{align*} where \$U~U[0,1]\$. To find an antithetic variable, I used begin{align*}U^*&=1-U\ 1-U&=1-e^{-X^*}\X^*&=-log(U)end{align*}

Therefore, given \$U\$,
\$\$Y=-log(1-U),Y^*=-log(U)\$\$ and \$Y\$ and \$Y^*\$ are Exp(1) variables. Both can be used in a Monte Carlo integration as e.g.
\$\$int_0^infty g(x)f(x)text(d)x approx frac{1}{2n}sum_{i=1}^n (Y_i+Y^*_i)\$\$

Please correct me where I'm wrong

Contents

All the theory you need

\$Zsim F\$, and you want to estimate \$mathrm{E}[Z]\$.

Let \$sigma^2=mathrm{Var}[Z]<infty\$.

Simple Monte Carlo

Construct \$X_1,X_2,dots\$ IID with \$X_1sim F\$.

Define \$bar{X}_n=frac{1}{n}sum_{i=1}^n X_i\$.

Result: \$mathrm{Var}[bar{X}_n]=sigma^2/n\$.

Strong Law: \$bar{X}_ntomathrm{E}[Z]\$ a.s.

Antithetic Variables

Construct \$X'_1,X'_2,dots\$ such that, for \$igeq 1\$,

1. \$X'_isim F\$;
2. \$mathrm{Cov}[X'_{2i-1},X'_{2i}]<0\$;
3. The pairs \$(X'_1,X'_2),(X'_3,X'_4) ,dots\$ are IID.

Define \$Y_i=(X'_{2i-1}+X'_{2i})/2\$ and \$bar{Y}_n=frac{1}{n}sum_{i=1}^n Y_i\$.

Result: \$\$mathrm{Var}[bar{Y}_n] =frac{sigma^2+mathrm{Cov}[X'_1,X'_2]}{2n} < frac{sigma^2}{2n}<mathrm{Var}[bar{X}_n].\$\$

Strong Law: \$bar{Y}_ntomathrm{E}[Z]\$ a.s.

Define \$U_1,U_2,dots\$ IID \$mathrm{U}[0,1]\$.

For \$igeq 1\$, define \$X'_{2i-1}=(log(1-U_i))^2\$ and \$X'_{2i}=(log(U_i))^2\$.

Prove 1, 2, 3 above.

Remember that \$U_isim 1-U_i\$, and \$mathrm{Cov}[U_i,1-U_i]<0\$.

Also, \$xmapsto (log x)^2\$ is monotonically decreasing for \$xin(0,1]\$.

Simulation

``n <- 10^6 u <- runif(n)  # simple x <- (log(1-u))^2 mean(x) sqrt(var(x)/n)  # antithetic y <- ((log(1-u))^2 + (log(u))^2)/2 mean(y) sqrt(var(y)/n) ``

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