Original article Biomath 3 (2014), 1404211, 1–6 B f Volume ░, Number ░, 20░░ BIOMATH ISSN 1314-684X Editor–in–Chief: Roumen Anguelov B f BIOMATH h t t p : / / w w w . b i o m a t h f o r u m . o r g / b i o m a t h / i n d e x . p h p / b i o m a t h / Biomath Forum On a Bivariate Poisson Negative Binomial Risk Process Krasimira Kostadinova1, Leda Minkova2 1Faculty of Mathematics and Informatics, Shumen University ”K. Preslavsky” Shumen, Bulgaria, Email: kostadinova@shu-bg.net 2Faculty of Mathematics and Informatics, Sofia University ”St. Kl.Ohridski” Sofia, Bulgaria, Email: leda@fmi.uni-sofia.bg Received: 19 February 2014, accepted: 21 April 2014, published: 30 May 2014 Abstract—In this paper we define a bivariate counting process as a compound Poisson process with bivariate negative binomial compounding dis- tribution. We investigate some of its basic properties, recursion formulas and probability mass function. Then we consider a risk model in which the claim counting process is the defined bivariate Poisson negative binomial process. For the defined risk model we derive the distribution of the time to ruin in two cases and the corresponding Laplace transforms. We discuss in detail the particular case of exponentially distributed claims. Keywords-bivariate negative binomial distribu- tion; compound birth process; ruin probability I. Introduction We consider the stochastic process N(t), t > 0 defined on a fixed probability space (Ω,F , P) and given by N(t) = X1 + X2 + . . . + XN1 (t), (1) where Xi, i = 1, 2, . . . are independent, identically distributed (iid) as X random variables, indepen- dent of N1(t). We suppose that the counting pro- cess N1(t) is a Poisson process with intensity λ > 0 (N1(t) ∼ Po(λt)). In this case N(t) is a compound Poisson process. The probability mass function (PMF) and probability generating function (PGF) of N1(t) are given by P(N1(t) = i) = (λt)ie−λt i! , i = 0, 1, . . . (2) and ψN1 (t)(s) = e −λt(1−s). (3) The compound Poisson distribution is analyzed by many authors; see Johnson et al. [3], Grandell, [2], Minkova [8]. The corresponding compound Poisson process is commonly used as a counting process in risk models; see for example Klugman et al. [5], Minkova [9]. In this paper we suppose that the compounding random variable X has a bivariate negative bino- mial distribution, given in the next Section II. In Section III we define a counting process with the Bivariate Poisson Negative Binomial distribution (BPNB). We derive the moments and the joint PMF. Then, in Section IV, two types of ruin probability are considered for the risk model with BPNB distributed counting process. We derive the Laplace transforms and analyze the case of exponentially distributed claims. Citation: Krasimira Kostadinova, Leda Minkova, On a Bivariate Poisson Negative Binomial Risk Process, Biomath 3 (2014), 1404211, http://dx.doi.org/10.11145/j.biomath.2014.04.211 Page 1 of 6 http://www.biomathforum.org/biomath/index.php/biomath http://dx.doi.org/10.11145/j.biomath.2014.04.211 K Kostadinova et al., On a Bivariate Poisson Negative Binomial... II. Bivariate Negative Binomial Distribution Let us consider the bivariate negative binomial distribution, defined by the following PGF, given in Kocherlakota and Kocherlakota [6] ψ1(s1, s2) = ( γ 1 −αs1 −βs2 )r , (4) where γ = 1 − α − β and r ≥ 1 is a given integer number. We use the notation (X, Y ) ∼ BN B(r,α,β). The PMF of (X, Y ) is given by P(X = k, Y = l) = ( k + l l )( r + k + l − 1 k + l ) αkβlγr (5) for k, l = 0, 1, . . . , (k, l) , (0, 0), and P(X = 0, Y = 0) = γr. The marginal distributions are again negative binomial with PGFs ψ1(s1) = ψ1(s1, 1) = ( γ 1 −β−αs1 )r (6) and ψ1(s2) = ψ1(1, s2) = ( γ 1 −α−βs2 )r . (7) Denote by ρ1 = α 1−β and ρ2 = β 1−α the corre- sponding parameters of X and Y. In terms of ρ1 and ρ2, the PGFs (6) and (7) have the form ψ1(s1) = ( 1 −ρ1 1 −ρ1 s1 )r and ψ1(s2) = ( 1 −ρ2 1 −ρ2 s2 )r . The PGF of the sum X + Y is given by ψ1(s, s) = ( γ 1 − (α + β)s )r = ( γ 1 − (1 −γ)s )r . (8) From (8) it follows that X + Y has a negative binomial distribution with parameters r and γ, say X + Y ∼ N B(r,γ), and for j = 0, 1, . . . , PMF P(X + Y = j) = ( r + j − 1 j ) γr(1 −γ) j. Denote the PGF in (8) by ψ1(s) = ψ1(s, s). III. The bivariate counting process In this section we consider a compound Pois- son process with bivariate negative binomial com- pounding distribution. The resulting process is a bivariate counting process (N1(t), N2(t)), defined by the PGF ψ(s1, s2) = e −λt(1−ψ1 (s1,s2 )), (9) where ψ1(s1, s2) is the PGF of the compounding distribution, given in (4). We say that the counting process defined by (9) has a bivariate Poisson Negative binomial distribution with parameters λt,α and β, and use the notation (N1(t), N2(t)) ∼ BPN B(λt,α,β). The marginal distributions are de- fined by the following PGFs ψ(s1) = e −λt(1−ψ1 (s1 )) and ψ(s2) = e −λt(1−ψ1 (s2 )), where ψ1(s1) and ψ1(s2) are given by (6) and (7). The means are given by E(N1(t)) = rαλt γ and E(N2(t)) = rβλt γ , while the variances are V ar(N1(t)) = αr γ2 [1 + rα − β]λt and V ar(N2(t)) = βr γ2 [1 −α + rβ]λt. From (9) we obtain ∂2ψ(s1, s2) ∂s1∂s2 = ψ(s1, s2)rαβλt × [ rγ2rλt (1 −αs1 −βs2)2r+2 + (r + 1)γr (1 −αs1 −βs2)r+2 ] . Upon setting s1 = s2 = 1 we obtain the product moment of N1(t) and N2(t) to be E(N1(t)N2(t)) = rαβ γ2 (rλt + r + 1)λt, which yields the covariance between N1(t) and N2(t) as Cov(N1(t), N2(t)) = r(r + 1)αβ γ2 λt. For the correlation coefficient we have Corr(N1(t), N2(t)) = (r + 1) √ αβ (1+rα−β)(1+rβ−α). Biomath 3 (2014), 1404211, http://dx.doi.org/10.11145/j.biomath.2014.04.211 Page 2 of 6 http://dx.doi.org/10.11145/j.biomath.2014.04.211 K Kostadinova et al., On a Bivariate Poisson Negative Binomial... In terms of ρ1 and ρ2, the covariance and the correlation coefficient have the forms Cov(N1(t), N2(t)) = r(r + 1)ρ1ρ2 (1 −ρ1)(1 −ρ2) λt and Corr(N1(t), N2(t)) = (r + 1) √ ρ1ρ2 (1 + rρ1)(1 + rρ2) . A. Joint Probability Mass Function The probability function of the joint distribu- tion of (N1(t), N2(t)) is given by expanding the PGF ψ(s1, s2) in powers of s1 and s2. Denote by f (i, j) = P(N1(t) = i, N2(t) = j), i, j = 0, 1, 2, . . . , the joint probability mass function of (N1(t), N2(t)). We rewrite the PGF of (9) in the form ψ(s1, s2) = e−λt ∞∑ m=0 (λt)m m! ψm1 (s1, s2) = e−λt ∞∑ m=0 (λtγr)m m! 1 (1 −αs1 −βs2)rm . (10) Denote by ψ(i, j)(s1, s2) = ∂i+ jψ(s1, s2) ∂si1∂s j 2 , for i, j = 0, 1, . . . , and (i, j) , (0, 0), the derivatives of ψ(s1, s2). From (10) we get the following: ψ(i, j)(s1, s2) = e−λtαiβ j ∑ ∞ m=1 (λtγr )m m! × rm(rm+1)...(rm+i−1)(rm+i)...(rm+i+ j−1) (1−αs1−βs2 )rm+i+ j . (11) From Johnson et al. [4], it is known that for i, j = 0, 1, . . . , (i, j) , (0, 0), f (i, j) = ψ(i, j)(s1, s2) i! j! ∣∣∣∣∣∣ s1=s2=0. (12) The result is given in the next theorem. Theorem 1. The probability mass function of (N1(t), N2(t)) is given by f (i, j) = ( i + j j ) αiβ j × ∞∑ m=1 ( rm + i + j − 1 i + j ) (λtγr)m m! e−λt, i, j = 0, 1, . . . , (i, j) , (0, 0), (13) and f (0, 0) = e−λt(1−γ r ). Proof. The initial value f (0, 0) = e−λt(1−γ r ) follows simply from the PGF ψ(0, 0) = f (0, 0). Then (13) follows from (11) and (12). � IV. Bivariate riskmodel Consider the following bivariate surplus process U1(t) = u1 + c1t − ∑N1 (t) j=1 Z 1 j U2(t) = u2 + c2t − ∑N2 (t) j=1 Z 2 j for two lines of business. Here u1 and u2 are the initial capitals, c1, c2 represent the premium incomes per unit time and Z1, Z11, Z 1 2, . . . , and Z2, Z21, Z 2 2, . . . are two independent sequences of independent random variables, independent of the counting processes N1(t) and N2(t), representing the corresponding claim sizes. The univariate case of this model was analyzed in Kostadinova [7]. Let µ1 = E(Z1) and µ2 = E(Z2) be the means of the claims. Denote by S 1(t) = ∑N1 (t) j=1 Z 1 j and S 2(t) = ∑N2 (t) j=1 Z 2 j the corresponding accumulated claim processes. The model, analyzed in Chan et al., [1] is the case when N1(t) = N2(t) = N(t). Here we consider two possible times to ruin τmax = inf{t|max(U1(t), U2(t)) < 0} and τsum = inf{t|U1(t) + U2(t) < 0}, and the corresponding ruin probabilities Ψmax(u1, u2) = P(τmax < ∞) and Ψsum(u1, u2) = P(τsum < ∞). For the event of τmax we have the following: Biomath 3 (2014), 1404211, http://dx.doi.org/10.11145/j.biomath.2014.04.211 Page 3 of 6 http://dx.doi.org/10.11145/j.biomath.2014.04.211 K Kostadinova et al., On a Bivariate Poisson Negative Binomial... {max(U1(t), U2(t)) < 0} = {U1(t) < 0, U2(t) < 0} = {u1 + c1t − S 1(t) < 0, u2 + c2t − S 2(t) < 0} = {S 1(t) > u1 + c1t, S 2(t) > u2 + c2t}. It follows that the ruin probability ψmax(u1, u2) is the joint survival function of (S 1(t), S 2(t)). In a similar way, we obtain the event for the τsum : {U1(t) + U2(t) < 0} = {S 1(t) + S 2(t) > u1 + u2 + (c1 + c2)t}, i.e., the ruin probability ψsum(u1, u2) is the survival function of the sum S 1(t) + S 2(t). According to the definition of Ψmax(u1, u2), for the no initial capitals, we have the following Ψ1(0)Ψ2(0) ≤ Ψmax(0, 0) ≤ min{Ψ1(0), Ψ2(0)}, where Ψ1(0) and Ψ2(0) are the ruin probabilities of the models U1(t) and U2(t) with no initial capitals. The univariate Poisson Negative binomial risk model is analyzed in [7], where it is given that Ψ1(0) = ρ1 c1(1 −ρ1) λµ1r and Ψ2(0) = ρ2 c2(1 −ρ2) λµ2r. It follows that the upper bound of the ruin probability is given by Ψmax(0, 0) ≤ min{ ρ1 c1(1 −ρ1) λµ1r, ρ2 c2(1 −ρ2) λµ2r}. The lower bound has the form Ψmax(0, 0) ≥ ρ1ρ2 c1c2(1 −ρ1)(1 −ρ2) λ2µ1µ2r 2. A. Laplace transforms Denote by LTZ1 (s1) and LTZ2 (s2) the Laplace transforms of the random variables Z1 and Z2. Then, the Laplace transform of (S 1(t), S 2(t)) is given by LT(S 1 (t),S 2 (t))(s1, s2) = E[e −s1 S 1 (t)−s2 S 2 (t)] = ∑ ∞ i=0 ∑ ∞ j=0 E[e −s1 (Z11 +...+Z 1 i )−s2 (Z 2 1 +...+Z 2 j )] ×P(N1(t) = i, N2(t) = j). According to the construction of the counting process, for the Laplace transform of (S 1(t), S 2(t)) we have LT(S 1 (t),S 2 (t))(s1, s2) = ψ(N1 (t),N2 (t))(LTZ1 (s1), LTZ2 (s2)) = e −λt [ 1− ( γ 1−αLT Z1 (s1 )−βLTZ2 (s2 ) )r] . Using the parameters γ 1 −β = 1 −ρ1 and γ 1 −α = 1 −ρ2 of the marginal compounding distributions, we obtain the Laplace transforms of the marginal compound distributions to be LTS 1 (t)(s1) = LT(S 1 (t),S 2 (t))(s1, 0) = e −λt [ 1− ( 1−ρ1 1−ρ1 LTZ1 (s1 ) )r] and LTS 2 (t)(s2) = LT(S 1 (t),S 2 (t))(0, s2) = e −λt [ 1− ( 1−ρ2 1−ρ2 LTZ2 (s2 ) )r] . We need the following result about Laplace transforms, given in Omey and Minkova ([10]) Lemma 1. For the joint survival function P(S 1(t) > x, S 2(t) > y) we have∫ ∞ 0 ∫ ∞ 0 e−s1 x−s2 y P(S 1(t) > x, S 2(t) > y)d xdy = 1 − LTS 1 (t)(s1) − LTS 2 (t)(s2) + LT(S 1 (t),S 2 (t))(s1, s2) s1 s2 . (14) Biomath 3 (2014), 1404211, http://dx.doi.org/10.11145/j.biomath.2014.04.211 Page 4 of 6 http://dx.doi.org/10.11145/j.biomath.2014.04.211 K Kostadinova et al., On a Bivariate Poisson Negative Binomial... In our case we have ∫ ∞ 0 ∫ ∞ 0 e−s1 x−s2 y P(S 1(t) > x, S 2(t) > y)d xdy = 1s1 s2 [ 1 − e −λt [ 1− ( 1−ρ1 1−ρ1 LTZ1 (s1 ) )r] −e −λt [ 1− ( 1−ρ2 1−ρ2 LTZ2 (s2 ) )r] + e −λt [ 1− ( γ 1−αLT Z1 (s1 )−βLTZ2 (s2 ) )r]] . Lemma 2. For the survival function P(S 1(t) + S 2(t) > x) we have∫ ∞ 0 e−sx P(S 1(t) + S 2(t) > x)d x = 1 s [1 − LTS 1 (t)+S 2 (t)(s)]. (15) Then, for the ruin probability ψsum we have: LTS 1 (t)+S 2 (t)(s) = LTS 1 (t),S 2 (t)(s, s) = e −λt [ 1− ( γ 1−αLT Z1 (s)−βLT Z2 (s) )r] . and hence∫ ∞ 0 e−sx P(S 1(t) + S 2(t) > x)d x = 1 s [ 1 − e −λt [ 1− ( γ 1−αLT Z1 (s)−βLT Z2 (s) )r]] . B. Exponentially distributed claims Let us consider the case of exponentially dis- tributed claim sizes, i.e. FZ1 (x) = 1 − e − x µ1 , x ≥ 0 and GZ2 (y) = 1 − e − y µ2 , y ≥ 0, and µ1,µ2 > 0. Denote by e(n, x) = n∑ k=0 xk k! = exΓ(n + 1, x) Γ(n + 1) , where Γ(n) is a Gamma function and Γ(a, x) =∫ ∞ x ta−1e−tdt is the incomplete Gamma function, the truncated exponential sum function. For the ruin probability ψmax we have P(S 1(t) > x, S 2(t) > y) = ∑ ∞ i, j=0 F ∗i (x)G ∗ j (y)P(N1(t) = i, N2(t) = j), where F ∗i (x) = e− x µ1 e(i − 1, x µ1 ), i = 1, 2, . . . is the tail distribution of Z11 + . . . + Z 1 i and G ∗ j (y) = e− y µ2 e( j − 1, y µ2 ), j = 1, 2, . . . is the tail distribution of Z21 + . . . + Z 2 j . In this case we have P(S 1(t) > x, S 2(t) > y) = e−λt(1−γ r ) + ∑ ∞ i=1 α ie(i − 1, x µ1 ) × ∑ ∞ m=1 ( rm+i−1 i ) (λtγr )m m! e − x µ1 e−λt + ∑ ∞ j=1 β je( j − 1, y µ2 ) × ∑ ∞ m=1 ( rm+ j−1 j ) (λtγr )m m! e − y µ2 e−λt + ∑ ∞ i=1 ∑ ∞ j=1 α iβ je(i − 1, x µ1 )e( j − 1, y µ2 ) ( i+ j j ) × ∑ ∞ m=1 ( rm+i+ j−1 i+ j ) (λtγr )m m! e − x µ1 − y µ2 e−λt. Substituting x = u1 + c1t and y = u2 + c2t in the last expression, we obtain the ruin probability ψmax(u1, u2), as it was shown in the previous section. For the ruin probability ψsum(u1, u2) in the case Z1 = Z2 = Z, and hence µ1 = µ2 = µ, we obtain for the Laplace transform of the sum S 1(t)+S 2(t) : LTS 1 (t)+S 2 (t)(s) = e −λt [ 1− ( γ 1−(α+β)LTZ (s) )r] . This means that S 1(t) + S 2(t) = Z1 + . . . + ZN(t), where N(t) = X1 +. . .+ XN1 (t), N1(t) ∼ Po(λt), and Xi ∼ N B(r,γ). The survival function of the sum S 1(t) + S 2(t) is the survival function of the sum of claims, i.e. P(S 1(t) + S 2(t) > x) = P(Z1 + . . . + ZN(t) > x) = ∑ ∞ i=0 F ∗i (x)P(N(t) = i), where F ∗i (x) = e− x µ e(i − 1, x µ ), i = 1, 2, . . . is the Biomath 3 (2014), 1404211, http://dx.doi.org/10.11145/j.biomath.2014.04.211 Page 5 of 6 http://dx.doi.org/10.11145/j.biomath.2014.04.211 K Kostadinova et al., On a Bivariate Poisson Negative Binomial... tail distribution of Z1 + . . . + Zi. Hence we have P(S 1(t) + S 2(t) > x) = e−λt(1−γ r ) + ∑ ∞ i=1(1 −γ) ie(i − 1, x µ ) × ∑ ∞ m=1 ( rm+i−1 i ) (λtγr )m m! e − x µ e−λt = e−λt(1−γ r ) + ∑ ∞ i=1(α + β) ie(i − 1, x µ ) × ∑ ∞ m=1 ( rm+i−1 i ) (λtγr )m m! e − x µ e−λt and for x = u1 + u2 + (c1 + c2)t, we obtained the ruin probabilities ψsum. V. Conclusion In this study we introduce a compound Pois- son process with bivariate negative binomial com- pounding distribution. Also, we find the moments and the joint probability mass function. Then we define the bivariate risk model with bivariate Poisson negative binomial counting process. We find the Laplace transform of the ruin probability and investigate a special case of exponentially distributed claims. Acknowledgment The authors are thankful to the anonymous reviewers and the editor for making some use- ful comments and suggestions. This work was supported by the European Social Fund through the Human Resource Development Operational Programme under contract BG051PO001-3.3.06- 0052 (2012/2014) and by grant RD-08-255/2013 of Shumen University, Bulgaria. References [1] Chan W–S., Yang H. and Zhang L. (2003). Some results on ruin probabilities in a two–dimensional risk model, Insurance Mathematics & Economics, 33, 345–358. [2] Grandell J.(1997). Mixed Poisson Processes, Chapman & Hall, London. [3] Johnson N.L., Kemp A.W. and Kotz S. (2005). Univari- ate Discrete Distributions, Wiley Series in Probability and Mathematical Statistics. 3th edition. [4] Johnson N.L., Kotz S. and Balakrishnan N. (1997). Discrete Multivariate Distributions, John Wiley & Sons, New York. [5] Klugman S. A., Panjer H. and Willmot G. (1998) Loss Models. From Data to Decisions, John Wiley & Sons,Inc. [6] Kocherlakota S. and Kocherlakota K. (1992). Bivariate Discrete Distributions, Marcel Dekker, New York. [7] Kostadinova K.Y. (2013). On a Poisson Negative Bino- mial Process, in: Advanced Research in Mathematics, and Computer Science, Doctoral Conference in Math- ematics, Informatics and Education, September, 19–21, Sofia, 25–33. [8] Minkova L.D. (2002). A Generalization of the Classical Discrete Distributions, Commun.Statist. - Theory and Methods, 31, 871–888. [9] Minkova L.D. (2004). The Pólya-Aeppli process and ruin problems, J. Appl. Math. Stoch. Anal., 3, 221–234. [10] Omey E. and Minkova L.D. (2013). Bivariate Geometric Distributions, (submitted). Biomath 3 (2014), 1404211, http://dx.doi.org/10.11145/j.biomath.2014.04.211 Page 6 of 6 http://dx.doi.org/10.11145/j.biomath.2014.04.211 Introduction Bivariate Negative Binomial Distribution The bivariate counting process Joint Probability Mass Function Bivariate risk model Laplace transforms Exponentially distributed claims Conclusion References