Heterogeneous individual claims model ($T={\sum }_i{B}_i{Y}_i$)

Each policy has an independent Bernoulli random variable $B_i$ of probability $q_i$, and its independent severity $Y_i$ has expected value ${\mu }_i$ and variance ${\sigma }_i^2$. Independent policies give $\mathbb{E}T={\sum }_i{q}_i{\mu }_i$ and $\mathrm{Var}(T)={\sum }_i[q_i{\sigma }_i^2+q_i(1-q_i){\mu }_i^2]$. Conditioning on the claim indicator proves the formulas. Keeping the policy label preserves the dependence between the occurrence probability and the severity law; replacing all labels by a single mixture is an approximation unless the relevant laws agree.

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.