13.07.2015 Views

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

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

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

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

WHY LARGER RISKS HAVE SMALLER INSURANCE CHARGES 1<strong>15</strong>We integrate by parts to obtain¹(' T(£1 ) (r) ¡ ' T(£ 2 ) (r))= ¡(E[£ 1 ;] ¡ E[£ 2 ;]) @E[T();r¹]@¯=1Z 1+ d(E[£ 1 ;] ¡ E[£ 2 ;]) @2 E[T();r¹]0@ 2 :Since E[£ 1 ]=E[£ 2 ]=¹ and the first partial of the limited expectedvalue is bounded by unity as per Equation (3.3b), it followsthat the first term vanishes and we have¹ ¢ (' T(£1 ) (r) ¡ ' T(£ 1 ) (r))Z 1= d(E[£ 1 ;] ¡ E[£ 2 ;]) @2 E[T();r¹]0@ 2 :We use the expectation formula for the insurance charge, A1, toarrive at the formula E[£;]=¹ 1 ¡ ' T(£) :¹We then use this to substitute into the previous integral to yield' T(£1 )(r) ¡ ' T(£2 )(r)Z 1 @ 2 E[T();r¹]= d ' £2¡ '0¹£1¹ @ 2 :The result then follows immediately from the assumptions of theproposition.To gain a better understanding of the formulas, consider thefollowing example.EXAMPLE 5: Poisson Conditionals and Exponential PriorsLet T() be Poisson. We leave it as an exercise for the reader to=0

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!