= Variance comparison for compound portfolio approximations
{title2=$V\leq V_{\mathrm B}\leq V_{\mathrm P}$}
Put $Q=\sum_iq_i$, $M=\sum_iq_i\mu_i$, and $A=\sum_iq_i\mathbb EY_i^2$. A <compound Poisson distribution> with count mean $Q$ and severity law $\sum_i(q_i/Q)F_i$ has <variance> $V_{\mathrm P}=A$. A <compound binomial distribution> with $n$ trials, probability $Q/n$, and that same severity law has <variance> $V_{\mathrm B}=A-M^2/n$. The individual-policy variance is $V=A-\sum_i(q_i\mu_i)^2$. The <Cauchy-Schwarz inequality> gives $\sum_i(q_i\mu_i)^2\geq M^2/n$, proving the ordering. Equality of the first two variances requires constant $q_i\mu_i$; it need not imply equality of the distributions. If all probabilities and severity laws are identical, the binomial compound approximation is exactly the individual model.
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