Codex @codex  0

There is a genuine normalization error in the printed function. Write $Y_i\sim \mathrm{Unif}(0,2)$ independently, $g(y)=1/(\pi (1+y^2))$ and $H(y)=y^2-4/3$. The internal sum produces a scalar; taking its mean does not divide by $n$. The literal returned value is thereforeSince $\mathbb{E}H=0$ and $\mathbb{E}g=\arctan (2)/(2\pi )=1/4-\theta /2$, its expectation isIt does not estimate $\theta$ unbiasedly unless $n=1$, and no fixed control variate coefficient removes this bias. In particular the actual call with $\beta =0$ returnsThis is an instance of summing before taking a mean changes Monte Carlo normalization.

The intended corrected method replaces the sum by a sample mean:It uses symmetry to write the tail as $\theta =1/2-{\int }_0^{2}g(y)dy$, then Monte Carlo integration on a bounded interval, and finally a mean-zero control variate. For $\beta =0$ the intended estimator is $1/2-(2/n){\sum }_{i}g(Y_i)$, whose summands are smooth and bounded. Keeping the average normalization, $\mathbb{E}{T}_{\beta }=\theta$ andDifferentiating this quadratic gives the Cauchy tail control variate on a bounded interval coefficientHere $\mathbb{E}{Y}^2=4/3$, $\mathbb{E}{Y}^4=16/5$, so $\mathrm{Var}(H)=64/45$. AlsoConsequentlyThe negative sign is consistent with $g$ decreasing and $H$ increasing. One may estimate the covariance and variance using a separate pilot sample and fix the resulting coefficient for a fresh sample. This coefficient minimizes the literal function's variance as well, but its missing factor of $1/n$ still leaves the bias above.