Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 40 / 4

For independent standard Cauchy distribution draws $X_1,\dots ,X_n$, the code averages the indicators $I_i=1_{\{X_i>2\}}$. Each is a Bernoulli random variable with parameterHence it is an unbiased estimator of that tail probability, and independence givesThe variance is finite even though the underlying Cauchy draws have no finite mean: the variables being averaged are bounded indicators, not the Cauchy values themselves.

For $Y=2/(1-U)$ with $U$ uniform on $(0,1)$,Thus $Y$ has a Pareto distribution and is the importance sampling proposal, concentrated on the required tail. The target tail integrand is $f(y)=1/(\pi (1+y^2))$, so its importance weight isThe function returns $\tilde{\theta }=n^{-1}{\sum }_{i}w(Y_i)$. Since ${\mathbb{E}}_{q}w(Y)={\int }_2^{\infty }f(y)dy=\theta$, it is an unbiased estimator. For its second moment,An antiderivative is $\frac{1}{2}(\arctan y-y/(1+y^2))$, so the integral is $\frac{1}{2}(\pi \theta +2/5)$. Independence of the proposals givesThis is importance sampling of a Cauchy tail, with weights between $2/(5\pi )$ and $1/(2\pi )$. Their narrow range explains the much smaller variance than the indicator estimate, without appealing to nonexistent Cauchy moments.

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.