 Solution

The importance weight for the two normal densities isUse ${\hat{\mu }}_N=N^{-1}{\sum }_i{w}_{\sigma }(X_i)f(X_i)$ with $X_i\sim N(0,1)$. Its second moment before averaging isFor the source's general assumption $E_p{f}^2<\infty$, a guaranteed range isIndeed $w_{\sigma }\le 1/\sigma$ there, so $E_q(w_{\sigma }^2{f}^2)=E_p(w_{\sigma }{f}^2)\le {\sigma }^{-1}{E}_p{f}^2<\infty$.

This range cannot be enlarged uniformly over all such $f$. If $\sigma >1$, choose $f^2=e^{a{x}^2}$ withThen $E_p{f}^2<\infty$ but the displayed weighted second-moment integral diverges. Thus no variance range depending only on $\sigma$ can characterize every individual integrand beyond the guaranteed range.

If an additional boundedness assumption on $f$ is intended, the familiar sufficient range becomes $0<\sigma <\sqrt{2}$. For $f=1$ this condition is also necessary, but a compactly supported $f$ can have finite weighted variance for every $\sigma$. Distinguishing these assumptions prevents imposing an unstated restriction on the source integrand.