 Solution

For the new integrand, $\theta ={\mathbb{E}}_p[e^{-X_1}]=e^{1/2}$ and ${\mathrm{Var}}_p(e^{-X_1})=e^2-e$, independently of dimension. Chebyshev's inequality therefore yieldsA sufficient positive integer sample count isThe ceiling function ensures that the bound is at most $\delta$, so the strict-error event in the question has probability at least $1-\delta$. The required number of accepted draws is independent of $d$; their expected proposal count is $n{\sigma }^d$.