Assuming inflation from time t to t' is a%, does the following hold for LEV function? LEV_t'(x) = (1+a%) ^2 * LEV_t(x/(1+a%)). I can provide with proof if needed. Thus we deduce for base limit b, we have LEV_t'(b) = (1+a%) ^2 * LEV_t(b/(1+a%)). Hence, ILF_t'(x) =ILF_t(x / (1+a%) ). This is given in the solution of Q&A bank Q4.2 and used for calculation of Q17(iv) in the second tutorial handout. Are the above steps correct? Thanks.
Nearly, but not quite, I think. In the proof that you sent me, in the first line, I think the LEV_t(b/(1+a)) term in the denominator of the RHS should instead be LEV_t(b), because the base limit will not change?
Thanks for your speedy reply! I convince myself that the following is true: LEV_t'(x) = (1+a%) ^2 * LEV_t(x/(1+a%)). If the base limit not inflated, how do we get "ILF_t'(x) =ILF_t(x / (1+a%) )"? Thanks again!
Hi Henrietta, Let Y = original loss distribution at time t'. If claims have inflated by a% between time t and t' then Y=(1+a)X, and: ILFt'(Y)=LEVY(y)/LEVY(b). However, we only have information about the original loss distribution X, so we need to write this expression in terms of X, ie where X=Y/(1+a). ILFt'(X)=LEVX(x/1+a)/LEVX(b) Hence ILFt'(X) = ILFt(x/1+a). Kind regards, Katherine.
Hi Katherine, I don't see how u get ILFt'(X)=LEVX(x/1+a)/LEVX(b) by substituting X=Y/(1+a) into ILFt'(Y)=LEVY(y)/LEVY(b). Thanks for help!
We know X=Y/(1+a),so substitute this into the numerator. The denominator is unchanged because the basic limit does not inflate.
