The distributions of some quantities for Erlang(2) risk models

for k = 0; 1; 2; : : : and m = 0; 1; 2; : : : so that the inverse of1X m x m mX m^q(r 1 ) m j ^q(r 2 ) jc (m + 1)! jm=1j=0is a function B x given by1X m x mB x (t) =c (m + 1)!This givesm=1mXj=0A x (t) = x h x (t) + x h x B x (t)x ex=c=x h x (t) + x h x B x (t) mjB mj;j (t):for t = x=c;for t > x=c:(3.6)From Dickson and Li (2010, 2012) we know how to …nd B k;m (t) for certainclaim size distributions.Example 1 Let p(x) = 2 xe x , so that q(x) = e x and ^q(r) = =(+r).Dickson and Li (2012) de…ne C n;m (t) be the inverse of 1=(r 1 + ) n+1 (r 2 +) m+1 ; and show thatwhere l;n;m =j=0C n;m (t) = c 2 t(ct) n+m e (+c)t1Xl=0(ct 2 ) l l;n;m(n + m + 2l + 2)lX n + 2j + 1 n + 1 m + 2(l j) + 1( 1) l j m + 1j n + 2j + 1 l j m + 2(l j) + 1 :Using these results, the inverse of1X m x mcism=1B x (t) =1X cm=1(m + 1)! mmXj=0x m(m + 1)! mj^q(r 1 ) mmXj=0j ^q(r 2 ) j mjC m j 1;j 1 (t):As q n (x) = n x n 1 e x = (a); it is straightforward to write down expressionsfor x (t), h x (t) and n(t), and to express these in terms of hypergeometricfunctions. Although the convolutions in formulae (3.5) and (3.6) lead torather messy formulae, it is not a di¢ cult task to evaluate these convolutionsby numerical integration. Of course, numerical implementation requires thatin…nite sums are truncated at an appropriate point.7