ILF curve and inflation

[henrietta]

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.

 

[Ian Senator]

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?

 

[henrietta]

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!

 

[Katherine Young]

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.

 

[lwang]

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)

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!

 

[Katherine Young]

We know X=Y/(1+a),so substitute this into the numerator. The denominator is unchanged because the basic limit does not inflate.