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

The distributions of some quantities forErlang(2) risk modelsDavid C.M. Dickson and Shuanming LiAbstractWe study the distributions of [1] the …rst time that the surplusreaches a given level and [2] the duration of negative surplus in a SparreAndersen risk process with the inter-claim times being Erlang(2) distributed.These distributions can be obtained through the inversion ofLaplace transforms using the inversion relationship for the Erlang(2)risk model given by Dickson and Li (2010).Keywords: Sparre Andersen risk model; Erlang(2) inter-claim times;Generalised Lundberg equation; Duration of negative surplus; Firsthitting time; Laplace transform.1 IntroductionConsider a Sparre Andersen insurance surplus processU(t) = u + c tN(t)XX i ; t 0 ; (1.1)i=1where u 0 is the initial surplus, fX i g 1 i=1 are i.i.d. random variables withcommon distribution function (d.f.) P = 1 P and density function p;representing claim amounts. We denote by k = E[X k 1 ] the k-th moment ofX 1 . The counting process fN(t); t 0g denotes the number of claims up totime t and is de…ned as N(t) = maxfk : W 1 + W 2 + + W k tg; wherethe inter-claim times {W i g 1 i=1 are assumed to be i.i.d. random variables withcommon density function k. Further, we assume that fW i g 1 i=1 and fX i g 1 i=1are independent.In this paper, we consider a risk model in which the inter-claim times areErlang(2) distributed, i.e. k(t) = 2 te t ; t > 0; > 0: We also assume thatc E [W 1 ] > E [X 1 ] ; so 2c > 1 , providing a positive safety loading factor.1

De…ne T to be the time of ruin from initial surplus u so that T = infft :U(t) < 0 j U(0) = ug, with T = 1 if U(t) 0 for all t 0: We de…ne(u) = P (T < 1) to be the ultimate ruin probability.LetG(u; y) = P (T < 1; jU(T )j y j U(0) = u)so that G(u; y) represents the probability that ruin occurs from initial surplusu with a de…cit at ruin of no more than y. We denote its (defective) densityas g(u; y): Let Y denote the de…cit at ruin given that ruin has occurred.Then G(u; y) = G(u; y)= (u) and g(u; y) = g(u; y)= (u) are the distributionfunction and the density function of Y , respectively.In this paper we consider the distribution of the …rst hitting time of thesurplus process through level x > 0, starting from u = 0, and the relatedproblem of the distribution of the duration of periods of negative surplus.This leads to the problem of calculating the distribution of the total durationof negative surplus. By a symmetry argument, the distribution of the …rsthitting time for our risk model is the same as the distribution of the ruintime in a dual risk model given byN(t)XU D (t) = x c t + X i ; t 0 ;where the condition 2c > 1 for the insurance surplus process would bereplaced by 2c < 1 : See Cheung (2010), for example, for a discussion ofdual risk models.Let ^a(s) = R 10e sx a(x) dx denote the Laplace transform of a function a.i=12 The …rst hitting timeDe…neT x = infft > 0 : U(0) = 0; U(t) xg; x > 0;to be the …rst time that the surplus process upcrosses through level x, andde…ne for > 0R (x) = E e Tx U(0) = 0to be the Laplace transform of T x . Li (2008) shows thatR (x) = r 2 =ce r1x + r 1 =ce r2x ; x > 0; (2.1)r 2 r 1 r 1 r 22

where r 1 and r 2 are the two positive solutions of Lundberg’s fundamentalequation: 2 + s= 2^p(s): (2.2)c c2 We know from Dickson and Hipp (2001) that 0 < r 1 < ( + )=c < r 2 andr 1 = + cr 2 = + cc ^q(r 1);+ c ^q(r 2);where ^q(s) = p^p(s):Let us rewrite equation (2.1) in a di¤erent form:Similarly, we can writeR (x) = e r 1x r 1 =cr 1 r 2e r 1xe r 2x = e r 1x c (1 ^q(r 1)) e r 1xe r 2xr 1 r 2: (2.3)R (x) = e r 2xThese lead to a third formulation, namelyc (1 + ^q(r 2)) e r 1xe r 2x: (2.4)r 1 r 2R (x) = 1 2 e r1x + e r 2x+ 2c (2 ^q(r 1) + ^q(r 2 )) e r 1xe r 2x: (2.5)r 2 r 1This is our preferred equation for R (x) since it is symmetric in r 1 and r 2 ,unlike equations (2.3) and (2.4). As we shall see, the main task in invertingR (x) is dealing with (e r 1xe r2x )=(r 2 r 1 ). This term arises in relatedLaplace transforms for the Erlang(2) risk model, as shown by Sun (2005).We can apply the method in Dickson and Li (2010) to invert the Laplacetransform in (2.5) to obtain the distribution of T x . Dickson and Li (2010)show that ifthen^f(r 1 ) =Z 1g(t) = ce t f(ct) +0e r 1t f(t)dt = ^g() =1X n t nn=11 e t Z ctn!30Z 10e t g(t)dt; (2.6)yq n (ct y)f(y)dy; (2.7)

where q n denotes the n-fold convolution of q. Further, if^f(r 2 ) =thenZ 1e r 2t f(t)dt = ^h() =Z 1001Xh(t) = ce t ( ) n t n 1 e tf(ct) +n!n=1Z ct0e t h(t)dt; (2.8)yq n (ct y)f(y)dy: (2.9)From these results, Dickson and Li (2012) observe that ^q(r 1 )inverse of a function n given by^q(r 2 ) is then(t) = 2c1X (t) 2mm=1(2m)!1 e tq 2m (ct):Similarly, ^q(r 1 ) + ^q(r 2 ) is the inverse of a function m given bym(t) = 2c1Xn=0(t) 2n e t(2n + 1)! q(2n+1) (ct):3 The density of the …rst hitting timeIn this section we …rst …nd general expressions for the density of the …rsthitting time T x by inverting its Laplace transform, then we simplify ourresults for Erlang(2) claims. Consider …rst inversion of e r1x . The equationsatis…ed by r 1 is + cr 1 = ^q(r 1 );which is of exactly the same form as Lundberg’s fundamental equation for theclassical risk model. It therefore follows from Dickson and Willmot (2005)thate r 1x= e (+)x=c += e (+)x=c +1X (=c) nxn!n=1Z 1x=cZ 10e t x t e(ct(y + x) n 1 e (+)(y+x)=c q n (y)dyx; t)dtwhere1Xe(x; t) = en=1t (t)nq n (x)n!4

is a compound Poisson density function. Thus, the inverse of e r1x isex=cfor t = x=c;h x+ (t) =e(ct x; t) for t > x=c:xtThis is just the well-known result from the classical risk model where theLaplace transform of T x is e x where is the unique positive solution ofLundberg’s fundamental equation for that model. (See Gerber and Shiu(1998).)Similarly, the inverse of e r2x is h x (t) where ex=ch x (t) = P 1n=1 e t (xtfor t = x=c;1) n (t) nq n (ctn!x) = h x (t) for t > x=c:Hence the inverse of (e r1x + e r2x ) =2 is(ex=cfor t = x=c; x (t) = Px 1t n=1 e t (t) 2n(2n)! q2n (ct x) = x (t) for t > x=c;(3.1)(3.2)and the inverse of (e r 1x x (t) = x t1Xen=1Now let A x (t) be such thate r 2x ) =2 ist (t)2n1(2n 1)! q(2n 1) (ct x) for t > x=c. (3.3)Z 1To …nd A x (t) we note thate r 1xSimilarly,e r 2x0e t A x (t)dt = e r 1xe r 2xr 2 r 1:= e r 1x e (r 2 r 1 )x1r 2 r 1 r 2 r 11X= e r ( 1)1xn x n(r 2 r 1 ) n 1n!n=11X = e r ( 1)1xn x n n 1 (^q(r 1 ) + ^q(r 2 )) n 1n! cn=11X = e r ( 1)1xn x nn 1 ^m():n! ce r 1xe r 2xn=1r 2 r 1= e r 2x1Xn=15x nn! n 1 ^m(); (3.4)c

givinge r 1xe r 2x= 1 1X x nn 1 ^m()( 1) n 1 e r1x + e r 2xr 2 r 1 2 n! cn=1= x 2 e r1x + e r 2x+ 1 1Xx 2n+1 ^m()2 (2n + 1)! cn=111X x 2n2n 1 ^m()e r 1xe r 2x:2 (2n)! cn=1 2ne r 1x + e r 2x Hence by formulae (3.2) and (3.3), for t > x=cA x (t) = x x (t) +1Xn=1(x=c) 2n(2n + 1)! x m 2n (t)1X (x=c) 2n 1(2n)! x m (2nn=11) (t)!;with A x (x=c) = x e x=c . Finally, let f Txinversion of (2.5) givesbe the density function of T x : Thenfor t > x=c, withf Tx (t) = x (t) + c A x(t)2c n A x(t) (3.5)Pr(T x = x=c) = x (x=c) + c A x(x=c) = e x=c (1 + x=c) = Pr(W 1 > x=c):Although formula (3.5) is a general result, it is a formula that is not easilyimplemented, even for simple claim size distributions like the exponentialdistribution. As an alternative approach to …nding A x (t), we can use thefact that ^m() = ^q(r 1 ) + ^q(r 2 ) to write formula (3.4) as1X !me r 2x x m+1x +c (m + 1)! (^q(r 1) + ^q(r 2 )) m= xe r 2x1 +m=11X m x mcm=1Now let B k;m (t) be such thatZ 10(m + 1)!mX !m^q(r 1 ) m j ^q(r 2 ) j :jj=0e t B k;m (t) dt = ^q(r 1 ) k ^q(r 2 ) m6

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

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)

From Dickson and Li (2010), the inverse of (r 1 + ) 1=2 is a function V 1 (t)given byV 1 (t) = e (+c)t 1X m t m 1 m=2 (ct) (m+1)=2m+12 m!+ 1 2m=0and the inverse of (r 2 + ) 1=2 is a function W 1 (t) given byW 1 (t) = e (+c)t21X ( ) m t m 1 m=2 (ct) (m+1)=2m+1m!+ 1 2m=0meaning that the inverse of (r 1 + ) 1=2 (r 2 + ) 1=2 is a function, say g(t),given by1Xg(t) = n e (+c)t t 3n+1wheren=0 n = 2n+1 n+1=2 c n+1(2n + 1)! (n + 2) :Hence the inverse of (4.5) is the convolution of g with a function h, wherefrom (4.4)1Xh(t) = h m e (+c)t t 3m+1wherem=0h m = 2m+1 m+3=2 c m+1:(2m + 2)! m!The Laplace transform (with parameter s) of the product of the m-th term ofh(t) and the n-th term of g(t) isgivingwhere l =h m n(3m + 2) (3n + 2)( + c + s) 3(m+n)+4h g(t) =1Xl=0 le (+c)t t 3l+3(3l + 4)lXh m l m (3m + 2) (3 (l m) + 2)m=0lX= l+2 2l+2 c l+2 (3m + 1)! (3(l m) + 1)!(2m + 2)! m! (2(l m) + 1)!(l m + 1)!m=010

= l+2 2l+2 c l+22lX 3m + 2m=0m 23m + 2 3(l m) + 2l m + 113(l m) + 2 :(4.6)In order to express our …nal answer in terms of hypergeometric functions(and hence make computation straightforward), it is necessary to express thesum in (4.6) in closed form. As shown in the Appendix,givinglX 3m + 2m=0m 23m + 2h g(t) = 2 e (+c)t 3(l m) + 2l m + 11Xl=013(l m) + 2 l+2 2l+2 c l+2 t 3l+3l! (2l + 4)!=4 (3l + 3)!l! (2l + 4)!Simpli…cation givesh g(t) = 1 512 2 2 c 2 t 3 e (+c)t 0F 22 ; 3; c2 t 3and hence the inverse of (4.3) gives the density of T 1 as 1f T1 (u; t) = ce (+c)t 0F 22 ; 2; c2 t 33+ t 0 F 242 ; 2; c2 t 34 1512 2 2 c 2 t 3 e (+c)t 0F 22 ; 3; c2 t 3:4Remarks1. The distribution of the duration of the …rst period of negative surplusis independent of the initial surplus u due to the memoryless propertyof the exponential distribution.2. As g i (u; y) = g(u; y) = e y ; i = 1; 2; the distribution of the durationof other periods of negative surplus is the same as that of the durationof the …rst period of negative surplus.4Example 3 We now consider the case when p(x) = 2 xe x .from Li and Garrido (2004) or Sun (2005) thatIt followsg(u; y) = e y a 11 e R 1u + a 12 e R 2u + 2 ye y a 21 e R 1u + a 22 e R 2u ;11

where R 1 and R 2 are the negative solutions of 2 2s = 2 ; (4.7)c c 2 s + anda 11 =a 12 =a 21 =a 22 = 2c 2 (r + ) 2 (3 + 2r) R 1 (r + 2)R 2 R 1; 2c 2 (r + ) 2 (3 + 2r) R 2 (r + 2)R 1 R 2; R 1;c 2 (r + ) R 2 R 1 2 2 R 2;c 2 (r + ) R 1 R 2with r > 0 being the positive solution of equation (4.7). The ultimate ruinprobability can be obtained asThenwhere(u) =FurtherZ 10g(u; y)dy = (a 11 + a 21 ) e R 1u + (a 12 + a 22 ) e R 2u ; u 0:g(u; y) =g(u; y)(u)= (u) e y + (1 (u)) 2 ye y ;(u) = a 11e R1u + a 12 e R 2u; u 0:(u) 2^g(u; r i ) = (u)r i + + (1 (u)) ; i = 1; 2;r i + so that formula (4.2) simpli…es toD (u) = (u) 12 r 1 + + 1 + 2 (1 (u)) 1r 2 + 2 (r 1 + ) + 12 (r 2 + ) 2 (u)+c (r 1 + )(r 2 + ) + 2 (1 (u))c (r 1 + )(r 2 + ) + 2 (1 (u))2 c (r 1 + ) 2 (r 2 + ) 2 (u) 2 (u)+2c (r 1 + ) 2 (r 2 + ) 2c (r 1 + )(r 2 + ) 2+ 3 (1 (u)) 3 (1 (u))2c (r 1 + ) 3 (r 2 + ) 2c (r 1 + )(r 2 + ) : 3 (4.8)12

Then we can express the density of T 1 asf T1(u; t) = (u)2(C 0; 1 (t) + C 1;0 (t)) + 2 (1 (u))(C 1; 1 (t) + C 1;1 (t))2(u)=2)C 1;0 (t) + 2 (1c+ (u) C 0;0 (t) + 2 (1c+ 3 (1 (u))2c(C 2;0 (t) C 0;2 (t)) :As shown in Dickson and Li (2010, 2012) each of these C n;m functions canbe expressed in terms of hypergeometric functions.To …nd the density of the duration of other periods of negative surplus, weneed only …nd g 1 (0; y) as g 2 (0; y) = g(0; y): Dickson and Li (2012) give thebivariate Laplace transform of the time of ruin and the de…cit at ruin for themodi…ed surplus process in which the initial surplus is 0 and the distributionof the time to the …rst claim is exponential with parameter : Setting = 0and inverting this Laplace transform givesThenandwhereg 1 (0; y) = c(r + )2 2 (r + ) 2 e yc 2 (r + ) 21(0) =+Z 10c(r + ) + 2 2 ye y ; y > 0:c 2 (r + )g 1 (0; y)dy =2c(r + )2 2 ;c 2 (r + ) 2g 1 (0; y) = g 1(0; y)1(0) = 1e y + (1 1 ) 2 ye y ; y > 0; 1 = c(r + )2 (r + ) :2c(r + ) 2 3(u)=2)C 0;1 (t)cThen D (1)= R 1E[e T jjY = y]g0 1 (0; y)dy = R 1R0 (y)g 1 (0; y)dy and we can…nd an expression for D (1)by replacing (u) by 1 in (4.8). Let f (1)T 2bethe conditional density of the duration of other periods of negative surplus,conditioning on there being one phase of the Erlang distribution until thenext claim following the previous upcrossing of the surplus process through 0.13

where (0) = ( 1 (0); 2 (0)) with i (0) = 1 i (0):Now let u = e 2 (u) = ( (u) ^g(u; r)) ; (1 ) (u) + ^g(u; r)cr cr crand let I be the 2 2 identity matrix. Note that > (0) = [I (0)]1 > ;where 1 = (1; 1): Then the distribution of N can be re-expressed asPr(N = 0) = 1 u 1 > ;Pr(N = n) = u [ (0)] n 1 [I (0)] 1 > ; n = 1; 2; : : : :This shows that N follows a discrete phase-type distribution with representation( u ; (0)) :Example 4 Let p(x) = e x . Then it is well-known, e.g. Grandell (1991),that (u) = (1 R=)e Ru where R is the unique negative solution ofand froms 1 (u) = 2 (u) 2= 2 c c 2 s + c d du 2(u)(see, for example, Dickson and Li (2012)) we obtain 1 (u) = (1+cR=) 2 (u).As g i (u; y) = i (u) e y , we have8 < cr cr(+r) i(u); j = 1; ij (u) = : 1i(u); j = 2:cr + cr(+r)When = 2; = 1 and c = 1:2 we have Pr(N = 0) = 1 (u), and forn = 1; 2; 3; : : :Pr(N = n) = 0:1698(0:8302 n 1 ) (u).4.3 Unconditional distributionsWe can use the same ideas as in Section 4.2 to …nd the distribution of thej-th period of negative surplus for j = 2; 3; 4; : : :. Let a j be a vector given bya j = e 2 (u) (0) j 2 :Then for i = 1; 2 the i-th element of a j is the probability that the (j 1)-thupcrossing of the surplus process through 0 occurs with i phases until thenext claim. Thus, with probability (a j ) 1 1 (0) the density of T j is f (1)T 2(t) and15

with probability (a j ) 2 2 (0) it is f T1(0; t). Thus, given that there is a j-thperiod of negative surplus, its density iswhereg j (t) = j f (1)T 2(t) + (1 j ) f T1(0; t) (4.9) j =(a j ) 1 1 (0)(a j ) 1 2 (0) + (a j ) 2 2 (0) :In principle, this allows us to specify the distribution of the total duration ofnegative surplus. The following examples illustrate contrasting situations.Example 5 Let p(x) = e x . Then we have noted that f T j g 1 j=1 are i.i.d.random variables. Consequently, the distribution of the total duration of negativesurplus is compound phase-type, meaning that we can use the recursiveformula proposed by Wu and Li (2010) to calculate this distribution by discretisingthe distribution of T 1 .Example 6 Let p(x) = 2 xe x . In this case we can calculate the elementsof (u) usingc dg 1 (u; y) = g(u; y) g(u; y): duTable 4.1 shows the weight j for di¤erent values of j when u = 0 and 10;with = 2 and c = 1:1. As j increases, the value of j for each value ofj j (u = 0) j (u = 10)2 0491300 0.4895983 0.527392 0.5273424 0.528454 0.5284535 0.528486 0.5284866 0.528487 0.5284877 0.528487 0.528487Table 4.1: Values of ju is unchanged (to the number of decimal places shown), suggesting that thee¤ect of the initial surplus on the distribution of T j quickly diminishes as jincreases. The total duration of negative surplus is now a random sum wherethe quantities being summed are not i.i.d. random variables. In such a situationconvolutions must be calculated numerically and there does not appearto be a neat approach to calculating the distribution of the total duration ofnegative surplus.16

References[1] Cheung, E.C.K. (2010) A unifying approach to the analysis of businesswith random gains. Scandinavian Actuarial Journal. In press.[2] Dickson, D.C.M. and Hipp, C. (2001) On the time to ruin for Erlang(2)risk processes. Insurance: Mathematics & Economics 29, 333–344.[3] Dickson, D.C.M. and Li, S. (2010) Finite time ruin problems for theErlang(2) risk model. Insurance: Mathematics & Economics 46, 12–18.[4] Dickson, D.C.M. and Li, S. (2012) Erlang risk models and …nite timeruin problems. Scandinavian Actuarial Journal. In press.[5] Dickson, D.C.M. and Willmot, G.E. (2005) The density of the time toruin in the classical Poisson risk model. ASTIN Bulletin 35, 45–60.[6] Gerber, H.U. and Shiu, E.S.W. (1998) On the time value of ruin. NorthAmerican Actuarial Journal 2 (1), 48–78.[7] Graham, R.L., Knuth, D.E. and Patashnik, O. (1994) Concrete Mathematics,2nd edition. Addison-Wesley, Upper Saddle River, NJ.[8] Grandell, J. (1991) Aspects of Risk Theory. Springer-Verlag, New York.[9] Li, S. (2008) The time of recovery and the maximum severity of ruinin a Sparre Andersen model. North American Actuarial Journal 12 (4),413–424.[10] Li, S. and Garrido, J. (2004) On ruin for the Erlang( n) risk process.Insurance: Mathematics & Economics 34, 391–408.[11] Sun, L.-J. (2005) The expected discounted penalty at ruin in the Erlang(2)risk process. Statistics & Probability Letters 72, 205–217.[12] Wu, X. and Li, S. (2010) Matrix-form recursions for a family of compounddistributions. ASTIN Bulletin 40, 351–368.APPENDIXIn order evaluate the sum in (4.6), we make use of the generalised binomialseries B t (z) described in Graham et al (1994), de…ned by1X tk + 1 zkB t (z) =k tk + 1k=017

and satisfyingB t (z) r =1X tk + r r zkk tk + r :k=0We can write the sum in (4.6) in an obvious notation aslX 3m + 2m=0m 23m + 2 3(l m) + 2l m + 11lX3(l m) + 2 =m=0a m b l m :De…ne A(z) = P 1m=0 a m z m and B(z) = P 1m=0 b m z m . Then A(z) = B 3 (z) 2and1X 3n + 2 znB(z) =n + 1 3n + 2 = X 1 3m 1 zm 1m 3m 1n=0m=1X 1 3m 1 z= z 1 mm 3m 1 + z 1m=0= z 1 (1 B 3 (z) 1 ):Thus, if C(z) = A(z)B(z) = P 1m=0 c l z l , then c l = P lm=0 a m b l m , but wealso haveC(z) = B 3 (z) 2 z 1 (1 B 3 (z) 1 ) = z 1 B 3 (z) 2 z 1 B 3 (z):The coe¢ cient of z l in C(z) is thus the coe¢ cient of z l+1 in B 3 (z) 2giving 3l + 5 2 3l + 4 1 4 (3l + 3)!c l ==l + 1 3l + 5 l + 1 3l + 4 l! (2l + 4)! :B 3 (z),David C M Dickson, Shuanming LiCentre for Actuarial StudiesDepartment of EconomicsUniversity of MelbourneVictoria 3010Australiaemail: dcmd@unimelb.edu.au, shli@unimelb.edu.au18