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Assume independent samples $X_i\sim g$, $\int f|\theta |<\infty$, and $g>0$ wherever $f\theta$ can contribute. The importance weight is $w(x)=f(x)/g(x)$, and the ordinary importance sampling estimator isIts expectation is $\int \theta (x)f(x)dx=\mu$, by cancellation of $g$ in the expectation integral. The strong law of large numbers gives almost-sure convergence to $\mu$. Common probability support is sufficient; the precise support condition for importance sampling only requires covering the nonzero integrand for this particular expectation.

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 $U^2$.

Assume the draws are independent from $g$, that the target-weighted integrand is covered by its support, and that $\int |\theta (x)|f(x)dx<\infty$. Define the importance weight $w(x)=f(x)/g(x)$. The change-of-density identity isTherefore importance sampling uses the unbiased estimatorThe strong law of large numbers gives consistency. A finite second moment of the weighted integrand additionally gives finite variance and a usual independent-sample central limit theorem. A useful proposal places enough probability where $f|\theta |$ is large and avoids tiny proposal probability density function there; common support alone does not guarantee finite variance.

Let $w(\theta )=\pi (\theta )/\mu (\theta )$ be the importance weight. The two normalized densities giveSince $|F-Z|<C$, there is a finite constant $M$ such that $w(\theta )\le M$ everywhere. The Independence Metropolis–Hastings algorithm has an accepted transition density satisfyingConsequently the whole state space is a small set, with the one-step minorization condition $K(x,\cdot )\ge M^{-1}\pi (\cdot )$. Iterating this Doeblin condition gives uniform geometric convergence in total variation distance, so the chain is geometrically ergodic.