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

WHY LARGER RISKS HAVE SMALLER INSURANCE CHARGES 97lated. Yet, even when correlation exists, the charge for the sumis less than or equal to the charge for T.To generalize Equation (2.2), suppose we take samples(T 1 ,T 2 ,:::) of a random variable, T. WedefineS(1,2,:::,n)=T 1 +T 2 + ¢¢¢+ T n . We make the assumption that such sums are sampleselection independent by which we mean that the distribution ofS(1,2,:::,n) is the same as the distribution of S(i 1 ,i 2 ,:::,i n )where(i 1 ,i 2 ,:::,i n ) is any ordered n-tuple of distinct positive integers.Note this does not require the T i to be independent of one another,but it does force the distribution of T 1 + T 2 , for example, tobethesameasT 1 + T 3 , T 2 + T 3 , T 21 + T 225 ,orthesumofanypairof distinct variables in our original sample. This would implythat there is a common correlation between any two of our samples.Under the assumption of sample selection independence,we will show the insurance charge for S n declines as n increases.While it might seem that there ought to be some simple inductionproof based on Equation (2.1), the only quick extension isthat the charge for S nm is less than or equal to the charge for S n .To arrive at the general proof, we will use a numeric groupingtrick and properties of the “min” operator.2.3. Insurance Charge for Sample Selection Independent SumsDeclines with Sample SizeProof' Sn+1(r) · ' Sn(r): (2.4)For k =1,2,:::,n +1, defineS(» k=n +1)=T 1 + T 2 + ¢¢¢+ T k¡1 + T k+1 + ¢¢¢+ T n+1 :In other words, S(» k=n +1)isthesumofthen out of the firstn + 1 trials obtained by excluding the kth trial. For example,S(» 2=3) = T 1 + T 3 . With this notation, we may write the followingformulan+1 Xn ¢ S n+1 = S(» k=n +1):k=1