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

WHY LARGER RISKS HAVE SMALLER INSURANCE CHARGES 133A.11. Variance Formula using the Integral of the InsuranceChargeIf r 2 G R (r) ! 0asr !1,thenVar(T)=¹ 2 Var(R)=¹ 2 2Z 10dr'(r) ¡ 1 : (A.11)Proof It suffices to prove the result when T is a continuousrandom variable. Integrate by parts twice and use the assumptionsto deriveZ 1r=1 Z 1dr'(r)=¡rG R (r)¯ + drrG R (r)0r=0=0+ r22 G R(r)¯r=1r=00Z 1+ dr r20 2 f R(r)= 1 2 E[R2 ]:Therefore, one has 2 R 10 dr'(r)=E[R2 ].Then, using the definition of the variance along with the factthat E[R] = 1, one can writeVar(R)=E[R 2 ] ¡ (E[R]) 2 =2Z 10dr'(r) ¡ 1:Note the coefficient of variation, CV, is given as the square rootof Var(R).The result, Equation (A.11), is intuitive since to have a largeinsurance charge a random variable must take on extreme valueswith some significant probability. This means it has a relativelylarge CV. The converse is not true. If CV(R 1 ) ¸ CV(R 2 ), then' 1 (r) must exceed ' 2 (r) on average, but not necessarily at everyentry ratio. Exhibit 1 shows a discrete counterexample in whichone random variable has a larger charge at some entry ratios,even though it has a smaller CV than another random variable.