Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 40 / 2 / b

Take the indicator function $\theta (x)=1_{\{x\ge k\}}$. Its expectation under the standard Cauchy distribution is the desired upper-tail probability. To use the proposed tail probability density function in the remaining parts, assume $k>0$: a probability density function proportional to $x^{-2}$ on $[k,\infty )$ is not normalizable when that interval contains zero.

For $k>0$, ${\int }_k^{\infty }{x}^{-2}dx=1/k$. Thus the unnormalized integral is $1/k$ and the multiplicative normalization is $k$:Although this proposal does not cover the whole Cauchy distribution probability support, it covers the entire nonzero integrand $f\theta$, which is the required support condition for importance sampling.

For $x\ge k$, the proposal CDF is $G(x)=1-k/x$. Therefore inverse transform sampling with $V\sim U(0,1)$ givesEquivalently, put $U=1-V$, which is also uniform, and use $X=k/U$. This equivalent parametrization matches the uniforms used in the final estimator.

For $X_i=k/U_i$, the indicator is always one, and the importance weight reduces toAveraging these independent contributions givesAs a direct normalization check, its expectation is ${\int }_0^{1}k/[\pi (u^2+k^2)]du={\pi }^{-1}\arctan (1/k)$. For $k>0$ this equals $1/2-{\pi }^{-1}\arctan k$, the exact standard Cauchy distribution tail probability. The substitution $X=k/U$, rather than $k/(1-U)$ with the same symbol left unchanged, explains why the displayed denominator contains $U^2$.