The distributions of some quantities for Erlang(2) risk models
The distributions of some quantities for Erlang(2) risk models
The distributions of some quantities for Erlang(2) risk models
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<strong>The</strong> <strong>distributions</strong> <strong>of</strong> <strong>some</strong> <strong>quantities</strong> <strong>for</strong><strong>Erlang</strong>(2) <strong>risk</strong> <strong>models</strong>David C.M. Dickson and Shuanming LiAbstractWe study the <strong>distributions</strong> <strong>of</strong> [1] the …rst time that the surplusreaches a given level and [2] the duration <strong>of</strong> negative surplus in a SparreAndersen <strong>risk</strong> process with the inter-claim times being <strong>Erlang</strong>(2) distributed.<strong>The</strong>se <strong>distributions</strong> can be obtained through the inversion <strong>of</strong>Laplace trans<strong>for</strong>ms using the inversion relationship <strong>for</strong> the <strong>Erlang</strong>(2)<strong>risk</strong> model given by Dickson and Li (2010).Keywords: Sparre Andersen <strong>risk</strong> model; <strong>Erlang</strong>(2) inter-claim times;Generalised Lundberg equation; Duration <strong>of</strong> negative surplus; Firsthitting time; Laplace trans<strong>for</strong>m.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 <strong>of</strong>X 1 . <strong>The</strong> counting process fN(t); t 0g denotes the number <strong>of</strong> 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 <strong>risk</strong> model in which the inter-claim times are<strong>Erlang</strong>(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 <strong>of</strong> ruin from initial surplus u so that T = infft :U(t) < 0 j U(0) = ug, with T = 1 if U(t) 0 <strong>for</strong> 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 <strong>of</strong> 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.<strong>The</strong>n G(u; y) = G(u; y)= (u) and g(u; y) = g(u; y)= (u) are the distributionfunction and the density function <strong>of</strong> Y , respectively.In this paper we consider the distribution <strong>of</strong> the …rst hitting time <strong>of</strong> thesurplus process through level x > 0, starting from u = 0, and the relatedproblem <strong>of</strong> the distribution <strong>of</strong> the duration <strong>of</strong> periods <strong>of</strong> negative surplus.This leads to the problem <strong>of</strong> calculating the distribution <strong>of</strong> the total duration<strong>of</strong> negative surplus. By a symmetry argument, the distribution <strong>of</strong> the …rsthitting time <strong>for</strong> our <strong>risk</strong> model is the same as the distribution <strong>of</strong> the ruintime in a dual <strong>risk</strong> model given byN(t)XU D (t) = x c t + X i ; t 0 ;where the condition 2c > 1 <strong>for</strong> the insurance surplus process would bereplaced by 2c < 1 : See Cheung (2010), <strong>for</strong> example, <strong>for</strong> a discussion <strong>of</strong>dual <strong>risk</strong> <strong>models</strong>.Let ^a(s) = R 10e sx a(x) dx denote the Laplace trans<strong>for</strong>m <strong>of</strong> a function a.i=12 <strong>The</strong> …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 <strong>for</strong> > 0R (x) = E e Tx U(0) = 0to be the Laplace trans<strong>for</strong>m <strong>of</strong> 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 <strong>of</strong> 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 <strong>for</strong>m: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 2x<strong>The</strong>se lead to a third <strong>for</strong>mulation, 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 <strong>for</strong> 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 trans<strong>for</strong>ms <strong>for</strong> the <strong>Erlang</strong>(2) <strong>risk</strong> model, as shown by Sun (2005).We can apply the method in Dickson and Li (2010) to invert the Laplacetrans<strong>for</strong>m in (2.5) to obtain the distribution <strong>of</strong> 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 <strong>of</strong> 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 <strong>of</strong> 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 <strong>of</strong> a function m given bym(t) = 2c1Xn=0(t) 2n e t(2n + 1)! q(2n+1) (ct):3 <strong>The</strong> density <strong>of</strong> the …rst hitting timeIn this section we …rst …nd general expressions <strong>for</strong> the density <strong>of</strong> the …rsthitting time T x by inverting its Laplace trans<strong>for</strong>m, then we simplify ourresults <strong>for</strong> <strong>Erlang</strong>(2) claims. Consider …rst inversion <strong>of</strong> e r1x . <strong>The</strong> equationsatis…ed by r 1 is + cr 1 = ^q(r 1 );which is <strong>of</strong> exactly the same <strong>for</strong>m as Lundberg’s fundamental equation <strong>for</strong> theclassical <strong>risk</strong> model. It there<strong>for</strong>e 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 <strong>of</strong> e r1x isex=c<strong>for</strong> t = x=c;h x+ (t) =e(ct x; t) <strong>for</strong> t > x=c:xtThis is just the well-known result from the classical <strong>risk</strong> model where theLaplace trans<strong>for</strong>m <strong>of</strong> T x is e x where is the unique positive solution <strong>of</strong>Lundberg’s fundamental equation <strong>for</strong> that model. (See Gerber and Shiu(1998).)Similarly, the inverse <strong>of</strong> e r2x is h x (t) where ex=ch x (t) = P 1n=1 e t (xt<strong>for</strong> t = x=c;1) n (t) nq n (ctn!x) = h x (t) <strong>for</strong> t > x=c:Hence the inverse <strong>of</strong> (e r1x + e r2x ) =2 is(ex=c<strong>for</strong> t = x=c; x (t) = Px 1t n=1 e t (t) 2n(2n)! q2n (ct x) = x (t) <strong>for</strong> t > x=c;(3.1)(3.2)and the inverse <strong>of</strong> (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) <strong>for</strong> 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 <strong>for</strong>mulae (3.2) and (3.3), <strong>for</strong> 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 <strong>of</strong> (2.5) givesbe the density function <strong>of</strong> T x : <strong>The</strong>n<strong>for</strong> 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 <strong>for</strong>mula (3.5) is a general result, it is a <strong>for</strong>mula that is not easilyimplemented, even <strong>for</strong> simple claim size <strong>distributions</strong> 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 <strong>for</strong>mula (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
<strong>for</strong> k = 0; 1; 2; : : : and m = 0; 1; 2; : : : so that the inverse <strong>of</strong>1X 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):<strong>for</strong> t = x=c;<strong>for</strong> t > x=c:(3.6)From Dickson and Li (2010, 2012) we know how to …nd B k;m (t) <strong>for</strong> certainclaim size <strong>distributions</strong>.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 <strong>of</strong> 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 <strong>of</strong>1X 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 straight<strong>for</strong>ward to write down expressions<strong>for</strong> x (t), h x (t) and n(t), and to express these in terms <strong>of</strong> hypergeometricfunctions. Although the convolutions in <strong>for</strong>mulae (3.5) and (3.6) lead torather messy <strong>for</strong>mulae, 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 <strong>The</strong> distribution <strong>of</strong> the duration <strong>of</strong> periods<strong>of</strong> negative surplus4.1 Conditional <strong>distributions</strong>In this section, we study the density <strong>of</strong> the duration <strong>of</strong> a period <strong>of</strong> 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 <strong>for</strong> the …rst time, where T 1 is the duration <strong>of</strong> the …rst period <strong>of</strong>negative surplus. Let T j , <strong>for</strong> j = 2; 3; : : : ; be the duration <strong>of</strong> the j-th period<strong>of</strong> negative surplus. <strong>The</strong> distribution <strong>of</strong> T j depends on the initial surplus,as this a¤ects the phase <strong>of</strong> the <strong>Erlang</strong>(2) inter-arrival distribution when thesurplus upcrosses level 0 be<strong>for</strong>e the surplus drops below 0 <strong>for</strong> the j-th time.LethiD (u) = E e T 1j T < 1be the Laplace trans<strong>for</strong>m <strong>of</strong> T 1 with respect to ( 0): <strong>The</strong>nD (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, <strong>for</strong> i = 1; 2; are the Laplace trans<strong>for</strong>ms0<strong>of</strong> g(u; y) with respect to r i .Let i (u) and g i (u; y) be the probability <strong>of</strong> ruin and the defective density<strong>of</strong> the de…cit at ruin <strong>for</strong> a modi…ed <strong>Erlang</strong>(2) surplus process in which thedistribution <strong>of</strong> the time to the …rst claim is <strong>Erlang</strong>(i), <strong>for</strong> i = 1; 2: Clearly2(u) = (u) and g 2 (u; y) = g(u; y): <strong>The</strong>n g i (u; y) = g i (u; y)= i (u); i = 1; 2;are the densities <strong>of</strong> the de…cit at ruin given that ruin has occurred <strong>for</strong> thesurplus process with the distribution <strong>of</strong> the time to the …rst claim being<strong>Erlang</strong>(i), <strong>for</strong> i = 1; 2, respectively.Let D (i); i = 1; 2; be the Laplace trans<strong>for</strong>m <strong>of</strong> T j <strong>for</strong> j = 2; 3; : : : ; giventhat the distribution <strong>of</strong> the time to the next claim from the last recovery time8
prior to the occurrence <strong>of</strong> the j-th period <strong>of</strong> negative surplus is <strong>Erlang</strong>(i).<strong>The</strong>nZ 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 <strong>of</strong> T j <strong>for</strong> j = 2; 3; : : : ; is the same asthat <strong>of</strong> T 1 <strong>for</strong> u = 0 if the distribution <strong>of</strong> the time to the next claim fromthe last recovery time prior to the occurrence <strong>of</strong> the j-th period <strong>of</strong> negativesurplus is <strong>Erlang</strong>(2).In what follows, we show how to invert D (u) and D (1)with respect to to obtain the conditional density function <strong>of</strong> T 1 and T j <strong>for</strong> 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 <strong>of</strong> 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 <strong>of</strong> 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 <strong>of</strong> (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 <strong>of</strong> (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 <strong>of</strong> (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 <strong>of</strong> (4.5) is the convolution <strong>of</strong> 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!<strong>The</strong> Laplace trans<strong>for</strong>m (with parameter s) <strong>of</strong> the product <strong>of</strong> the m-th term <strong>of</strong>h(t) and the n-th term <strong>of</strong> 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 <strong>of</strong> hypergeometric functions(and hence make computation straight<strong>for</strong>ward), it is necessary to express thesum in (4.6) in closed <strong>for</strong>m. 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 <strong>of</strong> (4.3) gives the density <strong>of</strong> 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. <strong>The</strong> distribution <strong>of</strong> the duration <strong>of</strong> the …rst period <strong>of</strong> negative surplusis independent <strong>of</strong> the initial surplus u due to the memoryless property<strong>of</strong> the exponential distribution.2. As g i (u; y) = g(u; y) = e y ; i = 1; 2; the distribution <strong>of</strong> the duration<strong>of</strong> other periods <strong>of</strong> negative surplus is the same as that <strong>of</strong> the duration<strong>of</strong> the …rst period <strong>of</strong> 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 <strong>of</strong> 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 <strong>of</strong> equation (4.7). <strong>The</strong> ultimate ruinprobability can be obtained as<strong>The</strong>nwhere(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 <strong>for</strong>mula (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
<strong>The</strong>n we can express the density <strong>of</strong> 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 <strong>of</strong> these C n;m functions canbe expressed in terms <strong>of</strong> hypergeometric functions.To …nd the density <strong>of</strong> the duration <strong>of</strong> other periods <strong>of</strong> negative surplus, weneed only …nd g 1 (0; y) as g 2 (0; y) = g(0; y): Dickson and Li (2012) give thebivariate Laplace trans<strong>for</strong>m <strong>of</strong> the time <strong>of</strong> ruin and the de…cit at ruin <strong>for</strong> themodi…ed surplus process in which the initial surplus is 0 and the distribution<strong>of</strong> the time to the …rst claim is exponential with parameter : Setting = 0and inverting this Laplace trans<strong>for</strong>m gives<strong>The</strong>nandwhereg 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)c<strong>The</strong>n 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 <strong>for</strong> D (1)by replacing (u) by 1 in (4.8). Let f (1)T 2bethe conditional density <strong>of</strong> the duration <strong>of</strong> other periods <strong>of</strong> negative surplus,conditioning on there being one phase <strong>of</strong> the <strong>Erlang</strong> distribution until thenext claim following the previous upcrossing <strong>of</strong> 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): <strong>The</strong>n the distribution <strong>of</strong> 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 . <strong>The</strong>n it is well-known, e.g. Grandell (1991),that (u) = (1 R=)e Ru where R is the unique negative solution <strong>of</strong>and froms 1 (u) = 2 (u) 2= 2 c c 2 s + c d du 2(u)(see, <strong>for</strong> 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 <strong>for</strong>n = 1; 2; 3; : : :Pr(N = n) = 0:1698(0:8302 n 1 ) (u).4.3 Unconditional <strong>distributions</strong>We can use the same ideas as in Section 4.2 to …nd the distribution <strong>of</strong> thej-th period <strong>of</strong> negative surplus <strong>for</strong> j = 2; 3; 4; : : :. Let a j be a vector given bya j = e 2 (u) (0) j 2 :<strong>The</strong>n <strong>for</strong> i = 1; 2 the i-th element <strong>of</strong> a j is the probability that the (j 1)-thupcrossing <strong>of</strong> the surplus process through 0 occurs with i phases until thenext claim. Thus, with probability (a j ) 1 1 (0) the density <strong>of</strong> 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 <strong>of</strong> 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 <strong>of</strong> the total duration <strong>of</strong>negative surplus. <strong>The</strong> following examples illustrate contrasting situations.Example 5 Let p(x) = e x . <strong>The</strong>n we have noted that f T j g 1 j=1 are i.i.d.random variables. Consequently, the distribution <strong>of</strong> the total duration <strong>of</strong> negativesurplus is compound phase-type, meaning that we can use the recursive<strong>for</strong>mula proposed by Wu and Li (2010) to calculate this distribution by discretisingthe distribution <strong>of</strong> T 1 .Example 6 Let p(x) = 2 xe x . In this case we can calculate the elements<strong>of</strong> (u) usingc dg 1 (u; y) = g(u; y) g(u; y): duTable 4.1 shows the weight j <strong>for</strong> di¤erent values <strong>of</strong> j when u = 0 and 10;with = 2 and c = 1:1. As j increases, the value <strong>of</strong> j <strong>for</strong> each value <strong>of</strong>j 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 <strong>of</strong> ju is unchanged (to the number <strong>of</strong> decimal places shown), suggesting that thee¤ect <strong>of</strong> the initial surplus on the distribution <strong>of</strong> T j quickly diminishes as jincreases. <strong>The</strong> total duration <strong>of</strong> negative surplus is now a random sum wherethe <strong>quantities</strong> 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 <strong>of</strong> the total duration <strong>of</strong>negative surplus.16
References[1] Cheung, E.C.K. (2010) A unifying approach to the analysis <strong>of</strong> businesswith random gains. Scandinavian Actuarial Journal. In press.[2] Dickson, D.C.M. and Hipp, C. (2001) On the time to ruin <strong>for</strong> <strong>Erlang</strong>(2)<strong>risk</strong> processes. Insurance: Mathematics & Economics 29, 333–344.[3] Dickson, D.C.M. and Li, S. (2010) Finite time ruin problems <strong>for</strong> the<strong>Erlang</strong>(2) <strong>risk</strong> model. Insurance: Mathematics & Economics 46, 12–18.[4] Dickson, D.C.M. and Li, S. (2012) <strong>Erlang</strong> <strong>risk</strong> <strong>models</strong> and …nite timeruin problems. Scandinavian Actuarial Journal. In press.[5] Dickson, D.C.M. and Willmot, G.E. (2005) <strong>The</strong> density <strong>of</strong> the time toruin in the classical Poisson <strong>risk</strong> model. ASTIN Bulletin 35, 45–60.[6] Gerber, H.U. and Shiu, E.S.W. (1998) On the time value <strong>of</strong> 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 <strong>of</strong> Risk <strong>The</strong>ory. Springer-Verlag, New York.[9] Li, S. (2008) <strong>The</strong> time <strong>of</strong> recovery and the maximum severity <strong>of</strong> ruinin a Sparre Andersen model. North American Actuarial Journal 12 (4),413–424.[10] Li, S. and Garrido, J. (2004) On ruin <strong>for</strong> the <strong>Erlang</strong>( n) <strong>risk</strong> process.Insurance: Mathematics & Economics 34, 391–408.[11] Sun, L.-J. (2005) <strong>The</strong> expected discounted penalty at ruin in the <strong>Erlang</strong>(2)<strong>risk</strong> process. Statistics & Probability Letters 72, 205–217.[12] Wu, X. and Li, S. (2010) Matrix-<strong>for</strong>m recursions <strong>for</strong> a family <strong>of</strong> compound<strong>distributions</strong>. ASTIN Bulletin 40, 351–368.APPENDIXIn order evaluate the sum in (4.6), we make use <strong>of</strong> 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 . <strong>The</strong>n 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):<strong>The</strong> coe¢ cient <strong>of</strong> z l in C(z) is thus the coe¢ cient <strong>of</strong> 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 <strong>for</strong> Actuarial StudiesDepartment <strong>of</strong> EconomicsUniversity <strong>of</strong> MelbourneVictoria 3010Australiaemail: dcmd@unimelb.edu.au, shli@unimelb.edu.au18