= Solution
<Antithetic variates> preserve each draw's marginal law but couple two estimates negatively. For an average of identically distributed $H,H'$,
$$
\operatorname{Var}\left(\frac{H+H'}2\right)
=\frac12\left(\operatorname{Var}(H)+\operatorname{Cov}(H,H')\right).
$$
Negative <covariance> improves on two <independent> evaluations at the same computational budget.
Here generate <independent> $U_1,\ldots,U_r\sim U(0,1)$ and pair $x_j=kU_j$ with $x_j'=k(1-U_j)$. Both are uniform on $[0,k]$. The <antithetic Cauchy-tail integration on a finite interval> estimator is
$$
\boxed{\widehat\mu_A=\frac12-\frac{k}{2r}\sum_{j=1}^r
\left[f(kU_j)+f(k(1-U_j))\right].}
$$
It is unbiased. Put $h(u)=f(ku)$. This decreases, whereas $h(1-u)$ increases. For an <independent> copy $U'$ of $U$, the <covariance> identity gives
$$
\operatorname{Cov}(h(U),h(1-U))
=\frac12\mathbb E[(h(U)-h(U'))(h(1-U)-h(1-U'))]\leq0.
$$
Each product is nonpositive, and for $k>0$ it is strictly negative off the diagonal. Therefore
$$
\operatorname{Var}(\widehat\mu_A)
=\frac{k^2}{2r}\left[\operatorname{Var}(h(U))+
\operatorname{Cov}(h(U),h(1-U))\right]
<\frac{k^2}{2r}\operatorname{Var}(h(U)),
$$
which is the <variance> using $2r$ <independent> evaluations in part (v). Thus the comparison accounts for the doubled number of <probability density function> evaluations in each pair.
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