For arbitrary nonnegative raw weights $W_i$, let $\bar W=K^{-1}\sum_iW_i$. The empirical squared <coefficient of variation>, using variance divisor $K$, is
$$
\widehat{\operatorname{CV}}^2(W)
=\frac{K^{-1}\sum_i(W_i-\bar W)^2}{\bar W^2}
=\frac{K\sum_iW_i^2}{(\sum_iW_i)^2}-1.
$$
Substitution into the stated definition gives the usual <effective sample size of importance sampling>
$$
\boxed{\widehat{\operatorname{ESS}}
=\frac{(\sum_iW_i)^2}{\sum_iW_i^2}.}
$$
For normalized weights $w_i=W_i/\sum_jW_j$, this reduces to $\boxed{\widehat{\operatorname{ESS}}=1/\sum_iw_i^2}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 219 / 2 / c / Solution

{title2=$\operatorname{CV}(X)$}
{wiki}

For a random variable with nonzero mean, the coefficient of variation is the dimensionless ratio
$$
\operatorname{CV}(X)=\frac{\sqrt{\operatorname{Var}(X)}}{|\mathbb E X|}.
$$


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