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

114 WHY LARGER RISKS HAVE SMALLER INSURANCE CHARGESThen we perform integration by parts as follows to deriveZ 1=1d h()E[T();r¹]=¡(1 ¡ H())E[T();r¹]¯0The result follows since=+Z 10Z 101 ¡ H()= @E[£;] :@=0d(1 ¡ H()) @E[T();r¹]@d(1 ¡ H()) @E[(T();r¹] :@With this we can now show a smaller charge for the priorleads to a smaller charge for the unconditional distribution at acommon risk size.4.4. Smaller Charges for the Prior Lead to SmallerUnconditional Charges–Size FixedIfthen¹ =E[£ 1 ]=E[£ 2 ] and@ 2 E[T();r¹]@ 2 · 0' £2 · ' £1implies ' T(£2 ) · ' T(£ 1 ) : (4.4)ProofWe use Equation (4.5) to obtain' T(£1 )(r) ¡ ' T(£2 )(r)= ¡ 1 ¹Z 10Ã@E[£1 ;]d@!¡ @E[£2 ;] @E[T();r¹]:@ @