AA STATE-DEPENDENT DUAL RISK MODEL
LINGJIONG ZHU
Abstract.
In a dual risk model, the premiums are considered as the costs andthe claims are regarded as the profits. The surplus can be interpreted as thewealth of a venture capital, whose profits depend on research and development.In most of the existing literature of dual risk models, the profits follow thecompound Poisson model and the cost is constant. In this paper, we develop astate-dependent dual risk model, in which the arrival rate of the profits and thecosts depend on the current state of the wealth process. Ruin probabilities areobtained in closed-forms. Further properties and results will also be discussed. Introduction
The classic risk model is based on the surplus process U t = u + ρt − (cid:80) N t − i =1 C i ,where the insurer starts with the initial reserve u and receives the premium ata constant rate ρ and C i are the claims. A central problem is to study the ruinprobability, i.e., the probability that the surplus process will ever hit zero. In recentyears, a dual risk model has attracted many attentions, in which the surplus processis modeled as(1.1) U t = u − ρt + N t − (cid:88) i =1 C i where C i are i.i.d. positive random variables distributed according to Q ( dc ) inde-pendent of N t , which is a Poisson process with intensity λ . We assume λ E [ C ] > ρ .The surplus can be interpreted as the wealth of a venture capital, whose prof-its depend on research and development. The profits are uncertain and modeledas a jump process and the costs are more predictable and are modeled as a de-terministic process. The company pays expenses continuously over time for theresearch and development and gets profits at random discrete times in the future.Many properties have been studied for the dual risk model. The ruin probability ψ ( u ) = P ( τ < ∞| U = u ), where(1.2) τ = inf { t > U t ≤ } , satisfies the equation, see e.g. Afonso et al. [2](1.3) ψ ( u ) = e − λ uρ + (cid:90) uρ λe − λt (cid:90) ∞ ψ ( u − ρt + c ) Q ( dc ) dt. Date : 10 September 2015.
Revised:
13 October 2015.2000
Mathematics Subject Classification.
Key words and phrases. dual risk model, state-dependent, ruin probability. a r X i v : . [ q -f i n . R M ] O c t LINGJIONG ZHU
It is well known that ψ ( u ) = e − αu where α is the unique positive solution to theequation:(1.4) λ (cid:18)(cid:90) ∞ e − αx Q ( dx ) − (cid:19) = − ρα. Avanzi et al. [5] worked on optimal dividends in the dual risk model where thewealth process follows a L´evy process and the optimal strategy is a barrier strat-egy. Albrecher et al. [3] studied a dual risk model with tax payments. For generalinterclaim time dsitributions and exponentially distributed C i ’s, an expression forthe ruin probability with tax is obtained in terms of the ruin probability withouttaxation. When the interclaim times are exponential or mixture of exponentials,explicit expressions are obtained. Ng [18] considered a dual model with a thresholddividend strategy, with exponential interclaim times. Afonso et al. [2] worked ondividend problem in the dual risk model, assuming exponential interclaim times.They presented a new approach for the calculation of expected discounted divi-dends. and studied ruin and dividend probabilities, number of dividends, time to adividend. and the distribution for the amount of single dividends. Avanzi et al. [4]studied a dividend barrier strategy for the dual risk model whereby dividend deci-sions are made only periodically, but still allow ruin to occur at any time. Cheung[10] studied the Laplace transform of the time of recovery after default, amongstother concepts for a dual risk model. Cheung and Drekic [11] studied dividendmoments in the dual risk model. They derived integro-differential equations for themoments of the total discounted dividends which can be solved explicitly assumingthe jump size distribution has a rational Laplace transform. Rodr´ıguez et al. [21]worked on a dual risk model with Erlang interclaim times, studied the ruin proba-bility, the Laplace transform of the time of ruin for generally distributed C i ’s. Theyalso studied the expected discounted dividends assuming the profits follow a PhaseType distribution. When the profits are Phase Type distributed, Ng [19] also stud-ied the cross probabilities. Yang and Sendova [23] studied the Laplace transform ofthe ruin time, expected discounted dividends for the Sparre-Andersen dual model.The dual risk model has also been used in the context of venture capital invest-ments and some optimization problems have been studied, see e.g. Bayraktar andEgami [7]. In Fahim and Zhu [13], they studied the optimal control problem for thedual risk model, which is the minimization of the ruin probability of the underlyingcompany by optimizing over the investment in research and development.In this paper, we develop a state-dependent dual risk model. The innovations ofa company may have self-exciting phenomena, i.e., an innovation or breakthroughwill increase the chance of the next innovation and breakthrough. Also, when thewealth process increases, the company will be in a better shape to innovate andhence the arrival rate of the profits, may depend on the state of the wealth ratherthan simply being Poisson. Also, the expenses that a company pays for researchand develop may also increase after the company receive more profits. For the hightech and fast-growing companies, the running cost and the revenues of a companygrow in line with the size of the company, see e.g. Table 1, where we consideredthe annual total revenues, cost of total revenues and the gross profits in the years2011-2014 . We can see the upward trend of growth for Google. Therefore, for a Gross profit is the difference between the revenue and the cost of the revenue. Available on Google Finance
DUAL RISK MODEL 3 high tech company for Google, the usual constant assumption for running cost, theintensity of profits arrivals in the dual risk model might be too simplistic. On theother hand, for a traditional company like Coca-Cola, the annual total revenues,cost of total revenues and the gross profits do not vary too much year over year,see e.g. Table 2, where we considered the annual total revenues, cost of totalrevenues and the gross profits in the years 2011-2014 . That might also be thepattern for a high tech company that has already matured and no longer has stellargrowth. Therefore, the dual risk model in the existing literature might be a goodmodel when the financials of a company do not change too much over time. Astate-dependent dual risk model might be more appropriate when the underlyingcompany has phenomenal growth.Full Year 2011 2012 2013 2014Revenue (millions) $37,905 $46,039 $55,519 $66,001Cost of Revenue (millions) $13,188 $17,176 $21,993 $25,313Gross Profit (millions) $24,717 $28,863 $33,526 $40,688 Table 1.
Revenue and Cost by Google during 2011-2014.Full Year 2011 2012 2013 2014Revenue (millions) $46,542 $48,017 $46,854 $45,998Cost of Revenue (millions) $18,215 $19,053 $18,421 $17,889Gross Profit (millions) $28,327 $28,964 $28,433 $28,109
Table 2.
Revenue and Cost by Coca-Cola during 2011-2014.So it will be reasonable to assume that the costs depend on the state of thewealth of the company. Indeed, it is not only possible that the company spendsmore capital on research and development when the profits increase, it is also quitecommon in the technology industry to increase the capital spending on researchwhen the company is lagging behind its pairs so that it is fighting for survival andcatch-up. When we assume that the cost is constant, the wealth process of thecompany is illustrated Figure 1 till the ruin time. If we allow the cost to dependlinearly on the wealth, the wealth process of the company is illustrated in Figure2. When the dual risk model uses the classical compound Poisson as the wealthprocess, the probability that the company eventually ruins decays exponentiallyin terms of the initial wealth of the company. As we will see later in the paper,e.g. Figure 3 and Figure 4, by allowing the costs and arrival rates of the profitsdepending on the state of the wealth process, the model becomes much more robust,and the ruin probability can decay superexponentially in terms of the initial wealth,i.e., Figure 3, Table 3, and it can also decay polynomially in terms of the initialwealth, i.e., Figure 4, Table 4.We are interested to develop a state-dependent dual risk model, which still leadsto closed-form solutions to the ruin probabilities. Let us assume that the wealthprocess U t satisfies the dynamics(1.5) dU t = − η ( U t ) dt + dJ t , U = u, Available on Google Finance
LINGJIONG ZHU where J t = (cid:80) N t i =1 C i and N t is a simple point process with intensity λ ( U t − ) attime t . Here, η ( · ) : R ≥ → R ≥ and λ ( · ) : R ≥ → R ≥ are both continuouslydifferentiable. Throughout the paper, unless specified otherwise, we assume that C i are i.i.d. exponentially distributed with parameter γ >
0. While allowing η ( · )and λ ( · ) to be general, the drawback of our model is that we restrict C i ’s to beexponentially distributed for the paper. It will be an interesting future researchproject to investigate generally distributed C i ’s. For the wealth process U t in (1.5),we will obtain closed-form expressions for the ruin probability and further propertieswill also be studied.It is worth noting that the U t process in (1.5) is an extension of the Hawkesprocess with exponential kernel and exponentially distributed jump sizes. that is,a simple point process N t with intensity λ (cid:0) ue − βt + (cid:80) i : τ i Ruin Probability.Theorem 1. Assume that (cid:82) ∞ λ ( v ) η ( v ) e γv − (cid:82) v λ ( w ) η ( w ) dw dv exists and is finite. Then, theruin probability ψ ( u ) = P ( τ < ∞| U = u ) is given by (2.1) ψ ( u ) = P ( τ < ∞ ) = (cid:82) ∞ u λ ( v ) η ( v ) e γv − (cid:82) v λ ( w ) η ( w ) dw dv (cid:82) ∞ λ ( v ) η ( v ) e γv − (cid:82) v λ ( w ) η ( w ) dw dv . Figure 1 and Figure 2 are illustrations of the wealth process again time till thetime when the company is ruined. In Figure 1, η ( z ) is a constant and we cansee that the wealth process always decays with the constant rate. In Figure 2, η ( z ) is linear in z , i.e. η ( z ) = α + βz , for some α, β > η ( z ) gives us more flexibility. DUAL RISK MODEL 5 t U t Figure 1. An illustration of the wealth process against time tillthe company is ruined. t U t Figure 2. An illustration of the wealth process against time tillthe company is ruined. Example 2. Assume that U t = u − ρt + (cid:80) N t i =1 C i , where for any t ≤ τ , N t isa simple point process whose intensity depends linearly on the wealth process, i.e.with intensity α + βU t − for some α, β > . That is η ( v ) = ρ and λ ( v ) = α + βv for LINGJIONG ZHU any v ≥ . Hence, the ruin probability is given by ψ ( u ) = (cid:82) ∞ u α + βvρ e γv − (cid:82) v α + βwρ dw dv (cid:82) ∞ α + βvρ e γv − (cid:82) v α + βwρ dw dv (2.2) = (cid:82) ∞ u ( α + βv ) e γv − ρβ ( α + βv ) dv (cid:82) ∞ ( α + βv ) e γv − ρβ ( α + βv ) dv = − ρe − α ρβ − αx − β ρ x + γx + √ πγ β (2 ρβ ) / e γ ( ργ − α )2 β erf (cid:16) α + βx − γρ √ βρ (cid:17) (cid:12)(cid:12)(cid:12)(cid:12) ∞ u − ρe − α ρβ − αx − β ρ x + γx + √ πγ β (2 ρβ ) / e γ ( ργ − α )2 β erf (cid:16) α + βx − γρ √ βρ (cid:17) (cid:12)(cid:12)(cid:12)(cid:12) ∞ = √ πγ β (2 ρβ ) / e γ ( ργ − α )2 β (cid:104) − erf (cid:16) α + βu − γρ √ βρ (cid:17)(cid:105) + ρe − α ρβ − αu − β ρ u + γu √ πγ β (2 ρβ ) / e γ ( ργ − α )2 β (cid:104) − erf (cid:16) α − γρ √ βρ (cid:17)(cid:105) + ρe − α ρβ , where erf ( x ) := √ π (cid:82) x e − t dt is the error function. Example 3. Assume that λ ( v ) = µη ( v ) for some constant µ > γ . Then, the ruinprobability is explicitly given by (2.3) ψ ( u ) = (cid:82) ∞ u e − ( µ − γ ) v dv (cid:82) ∞ e − ( µ − γ ) v dv = e − ( µ − γ ) u . In general, if we assume that the profits C i are i.i.d. with probability distribution Q ( dc ) , then, (2.4) A f ( u ) = − η ( u ) ∂f∂u + λ ( u ) (cid:90) ∞ [ f ( u + c ) − f ( u )] Q ( dc ) = 0 . We aim to find f such that A f = 0 . In general, this may not yield closed formsolutions. In the special case that λ ( u ) = µη ( u ) , for some µ > E Q [ c ] , then, A f ( u ) =0 reduces to (2.5) − f (cid:48) ( u ) + µ (cid:90) ∞ [ f ( u + c ) − f ( u )] Q ( dc ) = 0 . Let us try the Ansatz f ( u ) = e θu . Then, we get (2.6) − θ + µ ( E Q [ e θc ] − 1) = 0 . The function F ( θ ) := − θ + µ ( E Q [ e θc ] − is convex in θ and F (0) = 0 . Since F (cid:48) (0) = − µ E Q [ c ] > , we conclude that F ( θ ) = 0 has a unique negative solution θ ∗ . Then, f ( u ) = e θ ∗ u and f ( ∞ ) = 0 . Therefore, (2.7) ψ ( u ) = P ( τ < ∞ ) = e θ ∗ u . DUAL RISK MODEL 7 Example 4. Assume that λ ( v ) = ( α + β √ v ) η ( v ) for some constant α > γ and β > .Then, the ruin probability is given by ψ ( u ) = (cid:82) ∞ u ( α + β √ v ) e γv − αv − β √ v dv (cid:82) ∞ ( α + β √ v ) e γv − αv − β √ v dv (2.8) = − √ πβγ ( α − γ ) / e β α − γ erf ( √ x ( α − γ )+ β √ α − γ ) − αα − γ e − αx − β √ x + γx (cid:12)(cid:12)(cid:12)(cid:12) ∞ u − √ πβγ ( α − γ ) / e β α − γ erf ( √ x ( α − γ )+ β √ α − γ ) − αα − γ e − αx − β √ x + γx (cid:12)(cid:12)(cid:12)(cid:12) ∞ = √ πβγ ( α − γ ) / e β α − γ (cid:104) erf ( √ u ( α − γ )+ β √ α − γ ) − (cid:105) + αα − γ e − αu − β √ u + γu √ πβγ ( α − γ ) / e β α − γ (cid:104) erf ( β √ α − γ ) − (cid:105) + αα − γ , where erf ( x ) := √ π (cid:82) x e − t dt is the error function. Example 5. Assume that λ ( v ) = ( αv β + γ ) η ( v ) , for some constants α, β > .Then, the ruin probability is given by ψ ( u ) = (cid:82) ∞ u ( αv β + γ ) e − αβ +1 v β +1 dv (cid:82) ∞ ( αv β + γ ) e − αβ +1 v β +1 dv (2.9) = − γβ +1 ( αβ +1 ) − β +1 Γ( β +1 , αx β +1 β +1 ) − e − αβ +1 x β +1 (cid:12)(cid:12)(cid:12)(cid:12) ∞ u − γβ +1 ( αβ +1 ) − β +1 Γ( β +1 , αx β +1 β +1 ) − e − αβ +1 x β +1 (cid:12)(cid:12)(cid:12)(cid:12) ∞ = e − αβ +1 u β +1 + γβ +1 ( αβ +1 ) − β +1 Γ( β +1 , αu β +1 β +1 )1 + γβ +1 ( αβ +1 ) − β +1 Γ( β +1 , , where Γ( s, x ) := (cid:82) ∞ x t s − e − t dt is the incomplete gamma function. Example 6. Assume that λ ( v ) = ( α − β v ) η ( v ) , for some constants α > γ and α > β > . Then, the ruin probability is given by ψ ( u ) = (cid:82) ∞ u ( α − β v ) e − ( α − γ ) v + β log( v +1) dv (cid:82) ∞ ( α − β v ) e − ( α − γ ) v + β log( v +1) dv (2.10) = (cid:82) ∞ u [ α ( v + 1) β − β ( v + 1) β − ] e − ( α − γ ) v dv (cid:82) ∞ [ α ( v + 1) β − β ( v + 1) β − ] e − ( α − γ ) v dv = β ( α − γ )Γ( β, ( α − γ )( x + 1)) − α Γ( β + 1 , ( α − γ )( x + 1)) (cid:12)(cid:12)(cid:12)(cid:12) ∞ u β ( α − γ )Γ( β, ( α − γ )( x + 1)) − α Γ( β + 1 , ( α − γ )( x + 1)) (cid:12)(cid:12)(cid:12)(cid:12) ∞ = − β ( α − γ )Γ( β, ( α − γ )( u + 1)) + α Γ( β + 1 , ( α − γ )( u + 1)) − β ( α − γ )Γ( β, ( α − γ )) + α Γ( β + 1 , ( α − γ )) , where Γ( s, x ) := (cid:82) ∞ x t s − e − t dt is the incomplete gamma function. LINGJIONG ZHU Example 7. Assume that λ ( v ) = ( γ + β v ) η ( v ) , for some constant β > . Then,the ruin probability is given by ψ ( u ) = (cid:82) ∞ u ( γ + β v ) e − β log( v +1) dv (cid:82) ∞ ( γ + β v ) e − β log( v +1) dv (2.11) = γγ + β − u ) β − + β − γ + β − u ) β . Remark 8. One way to interpretate the formula for the ruin probability is to writeit as (2.12) ψ ( u ) = E [ e γV V ≥ u ] E [ e γV ] , where V is a positive random variable with probability density function λ ( v ) η ( v ) e − (cid:82) v λ ( w ) η ( w ) dw . Expected Dividends. One can also study the single dividend payment prob-lem. Let b > U be the barrier of the dividend. For the first time that the wealthprocess U t goes above the barrier b , say at the first-passage time τ b := inf { t > U t ≥ b } , a dividend of the amount D = U τ b − b is paid out. No dividend is paidout if the company is ruined before ever hitting the barrier b . We are interested tocompute that expected value of the dividend to be paid out E [ D τ b <τ ].Note that under the assumption that the C i ’s are i.i.d. exponentially distributedwith parameter γ > 0, from the memoryless property of exponential distribution, U τ b − b is also exponentially distributed with parameter γ > 0. Therefore,(2.13) E [ D τ b <τ ] = 1 γ P ( τ b < τ ) . Hence, the problem reduces to compute the probability that the dividend will bepaid out before the company is ruined. Theorem 9. Assume that (cid:82) ∞ λ ( v ) η ( v ) e γv − (cid:82) v λ ( w ) η ( w ) dw dv exists and is finite. Then, theprobability φ ( u, b ) := P ( τ b < τ | U = u ) is given by (2.14) φ ( u, b ) = (cid:82) u λ ( v ) η ( v ) e γv − (cid:82) v λ ( w ) η ( w ) dw dv (cid:82) ∞ γe − γc (cid:82) b + c λ ( v ) η ( v ) e γv − (cid:82) v λ ( w ) η ( w ) dw dvdc , and the expected dividend is given by (2.15) E [ D τ b <τ ] = 1 γ (cid:82) u λ ( v ) η ( v ) e γv − (cid:82) v λ ( w ) η ( w ) dw dv (cid:82) ∞ γe − γc (cid:82) b + c λ ( v ) η ( v ) e γv − (cid:82) v λ ( w ) η ( w ) dw dvdc . One can consider multiple dividend payments as follows:(2.16) τ (1) b := inf { t > U t > b } , τ ( i ) b := inf { t > τ ( i ) b : U t > b } , i ≥ . Then τ ( i ) b is the i th payment of the dividend if τ ( i ) b < τ .Let N be the total number of dividends to be paid out before the ruin occursand (cid:80) Ni =1 D i be the total value of dividends to be paid out before the ruin occurs. DUAL RISK MODEL 9 Recall that φ ( u, b ) = P ( τ b < τ | U = u ), was computed in Theorem 9 withclosed-form formulas. It is easy to see that P ( N = 0) = 1 − φ ( u, b ) , (2.17) P ( N = n ) = φ ( u, b ) φ ( b, b ) n − (1 − φ ( b, b )) , n ≥ . (2.18)Therefore,(2.19) E (cid:34) N (cid:88) i =1 D i (cid:35) = 1 γ E [ N ] = 1 γ φ ( u, b )1 − φ ( b, b ) . One can also compute the Laplace transform of the total amount of dividendsto be paid out. For any θ > E (cid:104) e − θ (cid:80) Ni =1 D i (cid:105) = E (cid:104) E (cid:104) e − θ (cid:80) Ni =1 D i (cid:12)(cid:12) N (cid:105)(cid:105) (2.20) = E (cid:34)(cid:18) γγ + θ (cid:19) N (cid:35) = 1 − φ ( u, b ) + φ ( u, b ) ∞ (cid:88) n =1 (cid:18) γγ + θ (cid:19) n φ ( b, b ) n − (1 − φ ( b, b ))= 1 − φ ( u, b ) + φ ( u, b )(1 − φ ( b, b )) γγ + θ − γγ + θ φ ( b, b ) . Example 10. Assume that λ ( v ) = µη ( v ) for some constant µ > γ . Then, we cantake f ( x ) = e − ( µ − γ ) x and (2.21) (cid:90) ∞ f ( b + c ) γe − γc dc = γµ e − µb . Example 11. Assume that η ( v ) = ρ and λ ( v ) = α + βv . Then, we can take (2.22) f ( x ) = − ρe − α ρβ − αx − β ρ x + γx + √ πγ β (2 ρβ ) / e γ ( ργ − α )2 β erf (cid:18) α + βx − γρ √ βρ (cid:19) . We can compute that (2.23) (cid:90) ∞ f ( b + c ) γe − γc dc = √ πγ √ β ρ / e γ ( ργ − α )2 β erf (cid:18) α + βb − γρ √ βρ (cid:19) . Example 12. Assume that λ ( v ) = ( α − β v ) η ( v ) , for some constants α > γ and α > β > . Then, we can take (2.24) f ( x ) = β ( α − γ )Γ( β, ( α − γ )( x + 1)) − α Γ( β + 1 , ( α − γ )( x + 1)) , and we can compute that (cid:90) ∞ f ( b + c ) γe − γc dc (2.25)= − β ( α − γ ) e ( b +1) γ ( α − γ ) β ( β − α − β Γ( β − , ( b + 1) α ) − β ( α − γ ) e ( b +1) γ ( α − γ ) β ( b + 1) β − α − e − α ( b +1) + β ( α − γ )Γ( β, ( b + 1)( α − γ ))+ α − β e ( b +1) γ ( α − γ ) β +1 Γ( β + 1 , ( b + 1) α ) − α Γ( β + 1 , ( b + 1)( α − γ )) . Example 13. Assume that λ ( v ) = ( α + β √ v ) η ( v ) for some α > γ and β > . Then,we can take (2.26) f ( x ) = √ πβγ ( α − γ ) / e β α − γ erf (cid:18) √ x ( α − γ ) + β √ α − γ (cid:19) − αα − γ e − αx − β √ x + γx . And we can compute that (cid:90) ∞ f ( b + c ) γe − γc dc (2.27)= √ πβγ ( α − γ ) / e β α − γ e bγ e − β γ ( α − γ )( γ +1) √ γ + 1 erfc (cid:32) √ b ( γ + 1) + β √ α − γ √ γ + 1 (cid:33) + erf (cid:18) √ b + β √ α − γ (cid:19) − αα − γ γ (cid:34) − √ πβe β α + bγ α / erfc (cid:32) α √ b + β √ α (cid:33) + e bγ − αb − β √ b α (cid:35) . where erfc ( x ) := 1 − erf ( x ) is the complementary error function. First and Second Moments of the Wealth Process. We are also inter-ested to study the first and second moments of the wealth process U t . Note thatsince the wealth process is defined only up to the ruin time τ , we should evaluate E [ U t ∧ τ ] and E [ U t ∧ τ ], which in general is a challenge to compute since it will requireus to know explicitly the distribution of the ruin time. We derive the first and sec-ond moments of the wealth process U t for a special case instead. Let η ( u ) ≡ ρ + µu and λ ( u ) = α + βu , for some α, β ≥ 0, i.e.(2.28) dU t = − ( ρ + µU t ) dt + dJ t where J t = (cid:80) N t i =1 C i , where N t is a simple point process with intensity λ ( U t − ) = α + βU t − and C i are i.i.d. with distribution Q ( dc ).In this case, τ ≥ T , where T is the time that the ODE(2.29) du t = − ( ρ + µu t ) dt, u t = u, hits zero. It is easy to solve the above ODE and get(2.30) u t = (cid:18) ρµ + u (cid:19) e − µt − ρµ , T = 1 µ log (cid:18) µuρ (cid:19) . Then, for any t < T , t ∧ τ = t . DUAL RISK MODEL 11 Proposition 14. For any t < µ log (cid:16) µuρ (cid:17) , (2.31) E [ U t ] = (cid:18) ρ − α E Q [ c ] µ − β E Q [ c ] + u (cid:19) e − ( µ − β E Q [ c ]) t − ρ − α E Q [ c ] µ − β E Q [ c ] , and E [ U t ] = ue − β ( E Q [ c ] − µ ) t + α E Q [ c ] 1 − e − β ( E Q [ c ] − µ ) t β ( E Q [ c ] − µ ) − (2( α E Q [ c ] − ρ ) + β E [ c ]) ρ − α E Q [ c ] µ − β E Q [ c ] 1 − e − β ( E Q [ c ] − µ ) t β ( E Q [ c ] − µ )+ (2( α E Q [ c ] − ρ ) + β E [ c ]) (cid:18) ρ − α E Q [ c ] µ − β E Q [ c ] + u (cid:19) e − β ( E Q [ c ] − µ ) t − e − β ( E Q [ c ] − µ ) t β ( E Q [ c ] − µ ) . Laplace Transform of Ruin Time. In the ruin theory of the dual riskmodels, it is of great interest to study the Laplace transform of the ruin time,(2.32) ψ ( u, δ ) = E (cid:2) e − δτ τ< ∞ (cid:3) , where δ > 0. Note that ψ ( u, δ ) can also be interpreted as a perpetual digit option,with payoff 1 dollar at the time of ruin, with discount coefficient δ > 0, which canbe taken as the risk-free rate. Theorem 15. Assume that the equation λ ( u ) η ( u ) f (cid:48)(cid:48) ( u ) + [ λ ( u ) η (cid:48) ( u ) + λ ( u ) − γη ( u ) λ ( u ) − λ (cid:48) ( u ) η ( u ) + δλ ( u )] f (cid:48) ( u )(2.33) − ( γλ ( u ) + λ (cid:48) ( u )) δf ( u ) = 0 has a uniformly bounded positive solution f ( u ) that satisfies f ( ∞ ) = 0 . Then, wehave ψ ( u, δ ) = f ( u ) /f (0) . In general, a second-order linear ODE with non-constant coefficients do not yieldclosed-form solutions. Nevertheless, there are a wide range of special cases that doyield analytical solutions. Example 16. λ ( u ) ≡ λ , η ( u ) = ρ + µu . Then, we have (2.34) [ λρ + λµu ] f (cid:48)(cid:48) ( u ) + [( λµ + λ − γλρ + δλ ) − γλµuλ ] f (cid:48) ( u ) − γλδf ( u ) = 0 . This is a special 2nd order ODE that has a solution, see e.g. Polyanin and Zaitsev [20](2.35) f ( u ) = e γu J (cid:18) µ + λµ , µ + λ + δµ ; − γu − ργµ (cid:19) , where J ( a, b ; x ) is a solution to the degenerate hypergeometric equation (2.36) xy (cid:48)(cid:48) ( x ) + ( b − x ) y (cid:48) ( x ) − ay ( x ) = 0 , which has the solution in the case b > a > : (2.37) J ( a, b ; x ) = Γ( b )Γ( a )Γ( b − a ) (cid:90) e xt t a − (1 − t ) b − a − dt, where Γ( · ) is the gamma function. Example 17. Assume that λ ( u ) = µe − γu . Then, (2.38) η ( u ) f (cid:48)(cid:48) ( u ) + [ η (cid:48) ( u ) + µe − γu + δ ] f (cid:48) ( u ) = 0 , which yields that (2.39) f ( u ) = (cid:90) u η ( v ) e − (cid:82) v µe − γw + δη ( w ) dw dv. Example 18. Assume that λ ( u ) = γη ( u ) . Then, (2.40) η ( u ) f (cid:48)(cid:48) ( u ) + δη ( u ) f (cid:48) ( u ) − ( γη ( u ) + η (cid:48) ( u )) δf ( u ) = 0 . Further assume that η ( u ) = α + βu , where α, β > . Then, (2.41) f (cid:48)(cid:48) ( u ) + ( δβu + δα ) f (cid:48) ( u ) + ( − δγβu + δβ − δγα ) f ( u ) = 0 , which yields the solution (2.42) f ( u ) = e γu J (cid:32) γ + δβ δβ , 12 ; − δβ (cid:18) u + 2 γ + δαδβ (cid:19) (cid:33) , where J was defined in (2.37) . Example 19. Assume that η ( u ) ≡ η is a constant and λ ( u ) = µe λu for some µ, λ > . Then, we get (2.43) f (cid:48)(cid:48) ( u ) + (cid:20) µη e λu − γ − λ + δη (cid:21) f (cid:48) ( u ) − (cid:20) γη + λη (cid:21) δf ( u ) = 0 , and it is equivalent to (2.44) f (cid:48)(cid:48) ( u ) + ( ae λu + b ) f (cid:48) ( u ) + cf ( u ) = 0 , where (2.45) a := µη , b := − γ − λ + δη , c := − (cid:20) γη + λη (cid:21) . By letting ξ = e u , (2.44) reduces to (2.46) ξ f ξξ + ( aξ λ + b + 1) ξf ξ + cf ξ = 0 , and by letting z = ξ λ , w = f z − k , where k satisfies the quadratic equation (2.47) λ k + λbk + c = 0 , we have that (2.46) reduces to (2.48) λ zw zz + [ λaz + 2 kλ + λ ( λ + b )] w z + kλa = 0 , which has solution, see e.g. [20](2.49) w ( z ) = J (cid:18) k, k + 1 + bλ ; − aλ z (cid:19) , where J was defined in (2.37) . DUAL RISK MODEL 13 Expected Ruin Time. We have already computed the ruin probability P ( τ < ∞ ) in Theorem 1 under certain assumptions. Note that when P ( τ < ∞ ) < E [ τ ] = ∞ . In the case that the ruin occurs with probability one, i.e., P ( τ < ∞ ) = 1,we can also compute that expected time that the ruin occurs. Theorem 20. Assume that P ( τ < ∞ ) = 1 and let us define f ( u ) := (cid:90) u λ ( v ) η ( v ) (cid:90) v [ − λ (cid:48) ( w ) − γλ ( w )] 1 λ ( w ) e γ ( v − w ) − (cid:82) vw λ ( r ) η ( r ) dr dwdv (2.50) + g (0) (cid:90) u η (0) η ( v ) λ ( v ) λ (0) e γv e − (cid:82) v λ ( w ) η ( w ) dw dv, where (2.51) g (0) := 1 + λ (0) (cid:82) ∞ (cid:82) c λ ( v ) η ( v ) (cid:82) v [ − λ (cid:48) ( w ) − γλ ( w )] λ ( w ) e γ ( v − w ) − (cid:82) vw λ ( r ) η ( r ) dr γe − γ ( c − u ) dwdvdcη (0) − λ (0) (cid:82) ∞ (cid:82) c η (0) η ( v ) λ ( v ) λ (0) e γv e − (cid:82) v λ ( w ) η ( w ) dw γe − γ ( c − u ) dvdc . Assume that sup
In this section, we illustrate the ruin probability ψ ( u )obtained in Theorem 1 by some numerical examples. The summary statistics ofthe ruin probability ψ ( u ) for the case λ ( u ) = ( αu β + γ ) η ( u ) with fixed α = γ = 1 . β = 0 . 0, 0 . 5, 1 . 0, 1 . . ψ ( u ) for the case λ ( u ) = ( γ + β u ) η ( u ) with fixed γ = 1 . β = 1 . 5, 2 . 0, 2 . 5, 3 . . ψ ( u ) interms of the initial wealth u is not necessarily exponential. It exhibits a rich classof behaviors as we vary the parameter β . Therefore, the state-dependent dual riskmodel we have built is much more flexible and robust than many of the classicaldual risk models in the literature. ψ ( u ) u = 1 u = 2 u = 3 u = 4 u = 5 β = 0 . β = 0 . β = 1 . β = 1 . β = 2 . Table 3. Illustration of the ruin probability ψ ( u ) when λ ( u ) =( αu β + γ ) η ( u ) for fixed α = γ = 1.3. Appendix: Proofs Proof of Theorem 1. From (1.5), the infinitesimal generator of the wealth process U t can be written as(3.1) A f ( u ) = − η ( u ) ∂f∂u + λ ( u ) (cid:90) ∞ [ f ( u + c ) − f ( u )] γe − γc dc. Let us find a function f such that A f = 0, which is equivalent to(3.2) − η ( u ) f (cid:48) ( u ) − λ ( u ) f ( u ) + λ ( u ) (cid:90) ∞ u f ( c ) γe − γ ( c − u ) dc = 0 . u ψ ( u ) Figure 3. Illustration of the ruin probability ψ ( u ) against theinitial wealth u when λ ( u ) = ( αu β + γ ) η ( u ). The black, blue,green, red and cion lines denote the cases when β = 0 . 0, 0 . 5, 1 . . . 0. The α and γ are fixed to be 1 . 0. We can see from theplot that when β = 0 . 0, the ruin probability exponentially decaysin the initial wealth. Otherwise, the shape of decay is not expo-nential. ψ ( u ) u = 1 u = 2 u = 3 u = 4 u = 5 β = 1 . β = 2 . β = 2 . β = 3 . β = 3 . Table 4. Illustration of the ruin probability ψ ( u ) when λ ( u ) =( γ + β u ) η ( u ) for fixed γ = 1.Differentiating (3.2) with respect to u , we get − η (cid:48) ( u ) f (cid:48) ( u ) − η ( u ) f (cid:48)(cid:48) ( u ) − λ (cid:48) ( u ) f ( u ) − λ ( u ) f (cid:48) ( u ) + λ (cid:48) ( u ) (cid:90) ∞ u f ( c ) γe − γ ( c − u ) dc (3.3) − λ ( u ) γf ( u ) + λ ( u ) γ (cid:90) ∞ u f ( c ) γe − γ ( c − u ) dc = 0 . Substituting (3.2) into (3.3), we get(3.4) η ( u ) f (cid:48)(cid:48) ( u ) = (cid:20) λ (cid:48) ( u ) λ ( u ) η ( u ) + γη ( u ) − η (cid:48) ( u ) − λ ( u ) (cid:21) f (cid:48) ( u ) . DUAL RISK MODEL 15 u ψ ( u ) Figure 4. Illustration of the ruin probability ψ ( u ) against theinitial wealth u when λ ( u ) = ( γ + β u ) η ( u ). The black, blue,green, red and cion lines denote the cases when β = 1 . 5, 2 . 0, 2 . . . 5. The γ is fixed to be 1 . 0. The ruin probability decayspolynomially against the initial wealth.By letting f (cid:48) ( u ) = g ( u ), we have(3.5) dgg = (cid:18) λ (cid:48) ( u ) λ ( u ) + γ − η (cid:48) ( u ) η ( u ) − λ ( u ) η ( u ) (cid:19) du, which implies that(3.6) g ( u ) = λ ( u ) η ( u ) e γu − (cid:82) u λ ( v ) η ( v ) dv is a particular solution to (3.5). Hence, A f ( u ) = 0 for(3.7) f ( u ) := (cid:90) u λ ( v ) η ( v ) e γv − (cid:82) v λ ( w ) η ( w ) dw dv. By our assumption, f ( ∞ ) exists and is finite and it is also clear that for any0 ≤ u ≤ ∞ , 0 ≤ f ( u ) ≤ f ( ∞ ) < ∞ . Hence, by optional stopping theorem,(3.8) f ( u ) = E u [ f ( U τ )] = f (0) P ( τ < ∞ ) + f ( ∞ ) P ( τ = ∞ ) , which implies that(3.9) ψ ( u ) = P ( τ < ∞ ) = (cid:82) ∞ u λ ( v ) η ( v ) e γv − (cid:82) v λ ( w ) η ( w ) dw dv (cid:82) ∞ λ ( v ) η ( v ) e γv − (cid:82) v λ ( w ) η ( w ) dw dv . (cid:3) Proof of Theorem 9. Let us recall that for(3.10) f ( u ) = (cid:90) u λ ( v ) η ( v ) e γv − (cid:82) v λ ( w ) η ( w ) dw dv, we have A f = 0 and by our assumption f is uniformly bounded. By optionalstopping theorem, f ( u ) = E u [ f ( U τ ∧ τ b )](3.11) = f (0) P ( τ < τ b ) + (cid:90) ∞ f ( b + c ) γe − γc dc P ( τ b < τ ) , which implies that E [ D τ b <τ ] = 1 γ P ( τ b < τ )(3.12) = 1 γ f ( u ) − f (0) (cid:82) ∞ f ( b + c ) γe − γc dc − f (0)= 1 γ (cid:82) u λ ( v ) η ( v ) e γv − (cid:82) v λ ( w ) η ( w ) dw dv (cid:82) ∞ γe − γc (cid:82) b + c λ ( v ) η ( v ) e γv − (cid:82) v λ ( w ) η ( w ) dw dvdc . (cid:3) Proof of Proposition 14. The infinitesimal generator of U t process is given by(3.13) A f ( u ) = − ( ρ + µu ) f (cid:48) ( u ) + ( α + βu ) (cid:90) ∞ [ f ( u + c ) − f ( u )] Q ( dc ) . Let f ( u ) = u , we get A u = − ρ − µu + E Q [ c ]( α + βu ). By Dynkin’s formula, E [ U t ] = u + (cid:90) t E (cid:2) − ρ − µU s + E Q [ c ]( α + βU s ) (cid:3) ds (3.14) = u + ( − ρ + α E Q [ c ]) t + ( β E Q [ c ] − µ ) (cid:90) t E [ U s ] ds, which yields that(3.15) E [ U t ] = (cid:18) ρ − α E Q [ c ] µ − β E Q [ c ] + u (cid:19) e − ( µ − β E Q [ c ]) t − ρ − α E Q [ c ] µ − β E Q [ c ] . Let f ( u ) = u , we get A u = α E Q [ c ] + (2( α E Q [ c ] − ρ ) + β E [ c ]) u + 2( β E [ c ] − µ ) u .By Dynkin’s formula,(3.16) E [ U t ] = u + α E Q [ c ] t +(2( α E Q [ c ] − ρ )+ β E [ c ]) (cid:90) t E [ U s ] ds +2( β E [ c ] − µ ) (cid:90) t E [ U s ] ds, which implies the ODE:(3.17) ddt E [ U t ] = α E Q [ c ] + (2( α E Q [ c ] − ρ ) + β E [ c ]) E [ U t ] + 2( β E [ c ] − µ ) E [ U t ] . DUAL RISK MODEL 17 This is a first-order linear ODE. Together with (3.15), we get E [ U t ] = ue − β ( E Q [ c ] − µ ) t + (cid:90) t e − β ( E Q [ c ] − µ )( t − s ) α E Q [ c ] ds (3.18) + (2( α E Q [ c ] − ρ ) + β E [ c ]) (cid:90) t e − β ( E Q [ c ] − µ )( t − s ) (cid:18) ρ − α E Q [ c ] µ − β E Q [ c ] + u (cid:19) e − ( µ − β E Q [ c ]) s ds − (2( α E Q [ c ] − ρ ) + β E [ c ]) (cid:90) t e − β ( E Q [ c ] − µ )( t − s ) ρ − α E Q [ c ] µ − β E Q [ c ] ds = ue − β ( E Q [ c ] − µ ) t + α E Q [ c ] 1 − e − β ( E Q [ c ] − µ ) t β ( E Q [ c ] − µ ) − (2( α E Q [ c ] − ρ ) + β E [ c ]) ρ − α E Q [ c ] µ − β E Q [ c ] 1 − e − β ( E Q [ c ] − µ ) t β ( E Q [ c ] − µ )+ (2( α E Q [ c ] − ρ ) + β E [ c ]) (cid:18) ρ − α E Q [ c ] µ − β E Q [ c ] + u (cid:19) e − β ( E Q [ c ] − µ ) t − e − β ( E Q [ c ] − µ ) t β ( E Q [ c ] − µ ) . (cid:3) Proof of Theorem 15. Assume that we have a uniformly bounded positive function f such that A f = δf . Note that f ( U t ) f ( U ) e − (cid:82) t A ff ( U s ) ds = f ( U t ) f ( U ) e − δt is a martingale.By optional stopping theorem,(3.19) 1 = E (cid:20) f ( U τ ) f ( U ) e − (cid:82) τ A ff ( U s ) ds (cid:21) = f (0) f ( u ) E (cid:2) e − δτ τ< ∞ (cid:3) . Therefore,(3.20) ψ ( u, δ ) = f ( u ) /f (0) . Let us now try to find a function f such that A f = δf . Note that A f = δf isequivalent to(3.21) − η ( u ) f (cid:48) ( u ) − λ ( u ) f ( u ) + λ ( u ) (cid:90) ∞ f ( u + c ) γe − γc dc = δf, which implies that λ ( u ) η ( u ) f (cid:48)(cid:48) ( u ) + [ λ ( u ) η (cid:48) ( u ) + λ ( u ) − γη ( u ) λ ( u ) − λ (cid:48) ( u ) η ( u ) + δλ ( u )] f (cid:48) ( u )(3.22) − ( γλ ( u ) + λ (cid:48) ( u )) δf ( u ) = 0 . (cid:3) Proof of Theorem 20. Recall that the infinitesimal generator of U t process is givenby(3.23) A f ( u ) = − η ( u ) ∂f∂u + λ ( u ) (cid:90) ∞ [ f ( u + c ) − f ( u )] γe − γc dc. Let us find a function f such that A f = − 1. That is,(3.24) − η ( u ) f (cid:48) ( u ) − λ ( u ) f ( u ) + λ ( u ) (cid:90) ∞ u f ( c ) γe − γ ( c − u ) dc = − . Differentiating the equation (3.24) with respect to u , we get − η (cid:48) ( u ) f (cid:48) ( u ) − η ( u ) f (cid:48)(cid:48) ( u ) − λ (cid:48) ( u ) f ( u ) − λ ( u ) f (cid:48) ( u ) + λ (cid:48) ( u ) (cid:90) ∞ u f ( c ) γe − γ ( c − u ) dc (3.25) − λ ( u ) γf ( u ) + λ ( u ) γ (cid:90) ∞ u f ( c ) γe − γ ( c − u ) dc = 0 . Substituting (3.24) into (3.25), we get(3.26) f (cid:48)(cid:48) ( u ) + (cid:20) η (cid:48) ( u ) η ( u ) + λ ( u ) η ( u ) − λ (cid:48) ( u ) λ ( u ) − γ (cid:21) f (cid:48) ( u ) = − (cid:20) λ (cid:48) ( u ) λ ( u ) + γ (cid:21) η ( u ) . Let g ( u ) = f (cid:48) ( u ), then g ( u ) satisfies a first-order linear ODE:(3.27) g (cid:48) ( u ) + (cid:20) η (cid:48) ( u ) η ( u ) + λ ( u ) η ( u ) − λ (cid:48) ( u ) λ ( u ) − γ (cid:21) g ( u ) = − (cid:20) λ (cid:48) ( u ) λ ( u ) + γ (cid:21) η ( u ) , which yields the general solution g ( u ) = (cid:90) u (cid:20) − λ (cid:48) ( v ) λ ( v ) − γ (cid:21) η ( v ) e − (cid:82) uv (cid:104) η (cid:48) ( w ) η ( w ) + λ ( w ) η ( w ) − λ (cid:48) ( w ) λ ( w ) − γ (cid:105) dw dv (3.28) + g (0) e − (cid:82) u (cid:104) η (cid:48) ( v ) η ( v ) + λ ( v ) η ( v ) − λ (cid:48) ( v ) λ ( v ) − γ (cid:105) dv = λ ( u ) η ( u ) (cid:90) u [ − λ (cid:48) ( v ) − γλ ( v )] 1 λ ( v ) e γ ( u − v ) − (cid:82) uv λ ( w ) η ( w ) dw dv + g (0) η (0) η ( u ) λ ( u ) λ (0) e γu e − (cid:82) u λ ( v ) η ( v ) dv . Hence, we can choose f ( u ) = (cid:82) u g ( v ) dv , which gives f ( u ) = (cid:90) u λ ( v ) η ( v ) (cid:90) v [ − λ (cid:48) ( w ) − γλ ( w )] 1 λ ( w ) e γ ( v − w ) − (cid:82) vw λ ( r ) η ( r ) dr dwdv (3.29) + g (0) (cid:90) u η (0) η ( v ) λ ( v ) λ (0) e γv e − (cid:82) v λ ( w ) η ( w ) dw dv. Next, let us determine g (0). Recall that g (0) = f (cid:48) (0) and also notice that f (0) = 0.Note that by letting u = 0 in (3.24), we get(3.30) − η (0) g (0) + λ (0) (cid:90) ∞ f ( c ) γe − γ ( c − u ) dc = − , which implies that − − η (0) g (0) + λ (0) (cid:90) ∞ (cid:90) c λ ( v ) η ( v ) (cid:90) v [ − λ (cid:48) ( w ) − γλ ( w )] 1 λ ( w ) (3.31) · e γ ( v − w ) − (cid:82) vw λ ( r ) η ( r ) dr γe − γ ( c − u ) dwdvdc + g (0) λ (0) (cid:90) ∞ (cid:90) c η (0) η ( v ) λ ( v ) λ (0) e γv e − (cid:82) v λ ( w ) η ( w ) dw γe − γ ( c − u ) dvdc. Therefore,(3.32) g (0) = 1 + λ (0) (cid:82) ∞ (cid:82) c λ ( v ) η ( v ) (cid:82) v [ − λ (cid:48) ( w ) − γλ ( w )] λ ( w ) e γ ( v − w ) − (cid:82) vw λ ( r ) η ( r ) dr γe − γ ( c − u ) dwdvdcη (0) − λ (0) (cid:82) ∞ (cid:82) c η (0) η ( v ) λ ( v ) λ (0) e γv e − (cid:82) v λ ( w ) η ( w ) dw γe − γ ( c − u ) dvdc . DUAL RISK MODEL 19 By Dynkin’s formula, for any K > E [ f ( U τ ∧ K )] = f ( u ) + E (cid:34)(cid:90) τ ∧ K A f ( U s ) ds (cid:35) = f ( u ) − E [ τ ∧ K ] . By our assumption sup
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