Variance comparison for compound portfolio approximations ($V\le V_{\mathrm{B}}\le V_{\mathrm{P}}$)

Put $Q={\sum }_i{q}_i$, $M={\sum }_i{q}_i{\mu }_i$, and $A={\sum }_i{q}_i\mathbb{E}{Y}_i^2$. 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 ${\sum }_i(q_i{\mu }_i{)}^2\ge M^2/n$, 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.