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

98 WHY LARGER RISKS HAVE SMALLER INSURANCE CHARGESWhen n = 2, this formula says2 ¢ S 3 =2(T 1 + T 2 + T 3 )=(T 2 + T 3 )+(T 1 + T 3 )+(T 1 + T 2 )= S(» 1=3) + S(» 2=3) + S(» 3=3):Theformulaimplies· ;r¸Sn+1n(n +1)E =E[n ¢ Sn +1 n+1 ;n(n +1)r]23n+1 X=E4S(» k=n +1);n(n +1)r5:k=1Next, we apply the inequality min(A + B,C + D) ¸ min(A,B)+min(C,D) repeatedlytoget23n+1 Xn+1 XE4S(» k=n +1);n(n +1)r5 ¸ E[S(» k=n +1);nr]:k=1Since the T i are identically distributed and sample selection independent,it follows thatk=1E[S(» k=n +1);nr]=E[S n ;nr]and thus that23n+1 XE4S(» k=n +1);n(n +1)r5 ¸ (n +1)E[S n ;nr]:k=1Connecting inequalities and factoring out n from the right handexpectation, we obtain· ;r¸Sn+1·Sn;r¸n(n +1)E ¸ (n +1)nE :n +1 nAssuming without loss of generality for the purpose at handthat E[T] = 1, this inequality implies (1 ¡ ' Sn+1(r)) ¸ (1 ¡ ' Sn(r)).The result then follows.