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
 A: All the theory you need
$Z\sim F$, 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 $X_1\sim F$.
Define $\bar{X}_n=\frac{1}{n}\sum_{i=1}^n X_i$.
Result: $\mathrm{Var}[\bar{X}_n]=\sigma^2/n$.
Strong Law: $\bar{X}_n\to\mathrm{E}[Z]$ a.s.
Antithetic Variables
Construct $X'_1,X'_2,\dots$ such that, for $i\geq 1$,


*

*$X'_i\sim F$;

*$\mathrm{Cov}[X'_{2i-1},X'_{2i}]<0$;

*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: $\bar{Y}_n\to\mathrm{E}[Z]$ a.s.
Your application
Define $U_1,U_2,\dots$ IID $\mathrm{U}[0,1]$.
For $i\geq 1$, define $X'_{2i-1}=(\log(1-U_i))^2$ and $X'_{2i}=(\log(U_i))^2$.
Prove 1, 2, 3 above. 
Remember that $U_i\sim 1-U_i$, and $\mathrm{Cov}[U_i,1-U_i]<0$.
Also, $x\mapsto (\log x)^2$ is monotonically decreasing for $x\in(0,1]$.
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)

