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

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4 The distribution of the duration of periodsof negative surplus4.1 Conditional distributionsIn this section, we study the density of the duration of a period of negativesurplus, given that the surplus process falls below 0. For the surplus processfU(t)g t0 ; if ruin occurs at time T , the process will cross level 0 at timeT + T 1 for the …rst time, where T 1 is the duration of the …rst period ofnegative surplus. Let T j , for j = 2; 3; : : : ; be the duration of the j-th periodof negative surplus. The distribution of T j depends on the initial surplus,as this a¤ects the phase of the Erlang(2) inter-arrival distribution when thesurplus upcrosses level 0 before the surplus drops below 0 for the j-th time.LethiD (u) = E e T 1j T < 1be the Laplace transform of T 1 with respect to ( 0): ThenD (u) ===Z 10Z 10Z 1Substituting (2.5) into (4.1) yields0Ehie T 1jY = y g(u; y) dyE[e Ty ] g(u; y) dyR (y) g(u; y)dy: (4.1)D (u) = 1 2 (^g(u; r 1 ) + ^g(u; r 2 )) + 2c (2 ^q(r 1) + ^q(r 2 )) ^g(u; r 1 ) ^g(u; r 2 )r 2 r 1;(4.2)where ^g(u; r i ) = R 1e riy g(u; y)dy, for i = 1; 2; are the Laplace transforms0of g(u; y) with respect to r i .Let i (u) and g i (u; y) be the probability of ruin and the defective densityof the de…cit at ruin for a modi…ed Erlang(2) surplus process in which thedistribution of the time to the …rst claim is Erlang(i), for i = 1; 2: Clearly2(u) = (u) and g 2 (u; y) = g(u; y): Then g i (u; y) = g i (u; y)= i (u); i = 1; 2;are the densities of the de…cit at ruin given that ruin has occurred for thesurplus process with the distribution of the time to the …rst claim beingErlang(i), for i = 1; 2, respectively.Let D (i); i = 1; 2; be the Laplace transform of T j for j = 2; 3; : : : ; giventhat the distribution of the time to the next claim from the last recovery time8
prior to the occurrence of the j-th period of negative surplus is Erlang(i).ThenZ 1 hiD (i)= E e T jjY = y g i (0; y) dy; i = 1; 2:0Clearly D (2)= D (0), i.e. the density of T j for j = 2; 3; : : : ; is the same asthat of T 1 for u = 0 if the distribution of the time to the next claim fromthe last recovery time prior to the occurrence of the j-th period of negativesurplus is Erlang(2).In what follows, we show how to invert D (u) and D (1)with respect to to obtain the conditional density function of T 1 and T j for j = 2; 3; : : : intwo examples. We revisit these examples in Section 4.3.Example 2 Let p(x) = e x so that q(x) = p =(x)e x ; x > 0, and^q(s) = (=(s + )) 1=2 : It is well known that g(u; y) = e y , leading toD (u) = 1 2 r 1 + + + r 2 + 2c (2 ^q(r 1) + ^q(r 2 ))(r 1 + )(r 2 + ) :(4.3)It follows from Dickson and Li (2010) that the inverse of the …rst term withrespect to isX 1ce (+c)t (c 2 t 3 ) m1(2m)!(m + 1)! = ce (+c)t 0F 22 ; 2; c2 t 3;4wherem=0pF q (B 1 ; B 2 ; : : : ; B p ; C 1 ; C 2 ; : : : ; C q ; Z) =1Xm=0is the generalised hypergeometric function, with (a) n =the inverse of c(r 1 + )(r 2 + )is(B 1 ) m (B 2 ) m : : : (B p ) m Z m(C 1 ) m (C 2 ) m : : : (C q ) m m!(a + n)= (a), and 3ct e (+c)t 0F 22 ; 2; c2 t 3: (4.4)4Last,2c^q(r 1 ) ^q(r 2 )(r 1 + )(r 2 + )= 3=22c(r 1 + )(r 2 + )911:(r 1 + ) 1=2 (r 2 + ) 1=2(4.5)
Page 1 and 2: The distributions of some quantitie
Page 3 and 4: where r 1 and r 2 are the two posit
Page 5 and 6: is a compound Poisson density funct
Page 7: for k = 0; 1; 2; : : : and m = 0; 1
Page 11 and 12: = l+2 2l+2 c l+22lX 3m + 2m=0m 23m
Page 13: Then we can express the density of
Page 16 and 17: with probability (a j ) 2 2 (0) it
Page 18: and satisfyingB t (z) r =1X tk + r

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

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