PROCEEDINGS May 15, 16, 17, 18, 2005 - Casualty Actuarial Society

104 WHY LARGER RISKS HAVE SMALLER INSURANCE CHARGESThis impliesUsingwe deriveG T¹+¢¹(t) ¡ G T¹(t)=E[T;t]=E[T ¹+¢¹ ;t] ¡ E[T ¹ ;t]=Z t0Z t0Z t0dxdF T¹(s) ¢ G T¢¹(t ¡ s):dsG T (s)Z xSwitching orders of integration, we haveE[T ¹+¢¹ ;t] ¡ E[T ¹ ;t]==Z t0Z t00dF T¹(y)dF T¹(y) ¢ G T¢¹(x ¡ y):Z tydxG T¢¹(x ¡ y)dF T¹(y)E[T ¢¹ ;t ¡ y]:Next, we integrate by parts and evaluate terms to obtainE[T ¹+¢¹ ;t] ¡ E[T ¹ ;t]=F T¹(0) ¢ E[T ¢¹ ;t]Z t+ dy F T¹(y) ¢ G T¢¹(t ¡ y):0Applying Equation (3.3a), we can conclude that this decreaseswith ¹, thus proving our result.Note that these results apply to discrete as well as continuousdistributions and that the proof does not require F T¹(0) toequal zero. Exhibit 3 shows how Poisson, Negative Binomial,and Gamma limited expected values vary as the mean changes.See Figure 3 for a graph of Poisson limited expected values atthe limit 3.0 that vary as a function of risk size.Now inequality 2.4 will be used to show that decomposablemodels must have charges that decrease with risk size.