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
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

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.

Your application

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 $xin(0,1]$.


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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