A Renewal Generated Geometric Catastrophe Model with Discrete-Time Markovian Arrival Process

Abstract

Any event that results in sudden change of the normal functioning of a system may be thought of as a catastrophe. Stochastic processes involving catastrophes have very rich application in modeling of a dynamic population in areas of ecology, marketing, queueing theory, etc. When the size of the population reduces abruptly as a whole, due to a catastrophe, it is termed as the total catastrophe. However, in many real-life circumstances the catastrophes have a mild influence on the population and have a sequential effect on the individuals. This paper presents a discrete-time catastrophic model in which the catastrophes occur according to renewal process, and it eliminates each individual of the population in sequential order with probability p until the one individual survives or the entire population wipes out. The individuals arrive according to the discrete-time Markovian arrival process. Using the supplementary variable technique, we obtain the steady-state vector generating function (VGF) of the population size at various epochs. Further using the inversion method of VGF, the population size distribution is expressed in terms of the roots of the associated characteristic equation. We further give a detailed computational procedure by considering inter-catastrophe time distributions as discrete phase-type as well as arbitrary. Finally, a few numerical results in form of tables and graphs are presented. Moreover, the impact of the correlation of arrival process on the mean population size is also investigated.

Appendix

In order to prove that S(s) ≡|(s − p)I − qsA(s)| = 0 has m roots in |s|≤ 1, we follow the procedure of Dudin and Klimenok (1996). For this, we first introduce some notations and prove some lemmas.

$$ \begin{array}{@{}rcl@{}} \text{Let} S_{(n)}(s)&\triangleq&|(s-p)\mathbf{I}_n- qs\mathbf{A}_{(n)}(s)|,~~~~n=1,...,m, \end{array} $$

(1)

where the matrix A_(n)(s) is derived from the matrix A(s) by deleting last (m − n) rows and columns, and I_n is the n × n identity matrix. One may note that S_(m)(s) ≡ S(s).

Lemma 1

Show that |qsA_{i, j}(s)|_|s|= 1 ≤ qA_{i, j}(1), for 1 ≤ i, j ≤ m.

Proof

For 1 ≤ i, j ≤ m, we have

$$ \begin{array}{@{}rcl@{}} |qsA_{i,j}(s)|_{|s|=1} & \le & q |s|_{|s|=1} A_{i,j}(|s|)_{|s|=1},\\ & = & q A_{i,j}(1). \end{array} $$

Hence Lemma 1 is proved. □

Lemma 2

The following inequalities hold:

$$ \begin{array}{@{}rcl@{}} |s-p-qsA_{i,i}(s)|_{|s|=1} & > & \sum\limits_{j=1, j \ne i}^{m-1} |q s A_{i,j}(s)|_{|s|=1},~~ ~ 1 \le i \le m-1, \end{array} $$

(93)

$$ \begin{array}{@{}rcl@{}} \text{and~} |s-p-qsA_{m,m}(s)|_{|s|=1} & \ge & \sum\limits_{j=1}^{m-1} |q s A_{m,j}(s)|_{|s|=1}. \end{array} $$

(94)

Proof

$$ \begin{array}{@{}rcl@{}} |s-p-qsA_{i,i}(s)|_{|s|=1} & \ge & |s|_{|s|=1} - p - |q s A_{i,i}(s)|_{|s|=1}, \\ & \ge & 1 - p - q A_{i,i}(1), \\ & = & q \sum\limits_{j=1, j \ne i}^{m} A_{i,j}(1),~~~~1 \le i \le m. \end{array} $$

(95)

Now using the fact that qA_{m, m}(1) > 0 and using Lemma 1, we obtain

$$ \begin{array}{@{}rcl@{}} q \sum\limits_{j=1, j \ne i}^{m} A_{i,j}(1) & > & q \sum\limits_{j=1, j \ne i}^{m-1} A_{i,j}(1) \ge \sum\limits_{j=1, j \ne i}^{m-1} |q s A_{i,j}(s)|_{|s|=1}. \end{array} $$

(96)

Hence inequality (93) follows from Eqs. 95 and 96. Now for i = m, from inequality (95) and using Lemma 1 we obtain

$$ \begin{array}{@{}rcl@{}} |s-p-qsA_{m,m}(s)|_{|s|=1} & \ge & q \sum\limits_{j=1}^{m-1} A_{m,j}(1) \ge \sum\limits_{j=1}^{m-1} |q s A_{m,j}(s)|_{|s|=1}. \end{array} $$

(97)

Hence inequality (94) is proved. □

Lemma 3

For 1 ≤ i ≤ m, (s − p) − qsA_{i, i}(s) has exactly one zero in the region |s| < 1.

Proof

From Lemma 1 we have

$$ \begin{array}{@{}rcl@{}} |-qsA_{i,i}(s)|_{|s|=1} &\le& q A_{i,i}(1) < q = 1-p = |s|_{|s|=1} - p \le |s-p|_{|s|=1}. \\ \Rightarrow ~~~~~~~~~~ |-qsA_{i,i}(s)|_{|s|=1} & < & |s-p|_{|s|=1}. \end{array} $$

From Rouch$\acute {e}$’s theorem, (s − p) and (s − p) − qsA_{i, i}(s) have the same number of zeros in the region |s| < 1. Hence (s − p) − qsA_{i, i}(s), 1 ≤ i ≤ m has exactly one zero in the region |s| < 1. □

Lemma 4

The determinant S_(n)(s), 1 ≤ n ≤ m − 1 has exactly nzeros in the region |s| < 1.

Proof

We use the induction method to prove this Lemma. From Lemma 3 it is true for n = 1. Let the determinant S_(k− 1)(s) has exactly (k − 1) zeros in |s| < 1. Now we prove that S_(k)(s) has exactly k zeros in |s| < 1.

$$ \begin{array}{@{}rcl@{}} S^{(k)}(s)&=&-\sum\limits_{j=1}^{k-1} q s A_{k,j}(s)U_{k,j}(s)+\left( s-p-qsA_{k,k}(s)\right)S^{(k-1)}(s), \end{array} $$

where U_{k, j}(s) is the cofactor of − qsA_{k, j}(s).

$$ \begin{array}{@{}rcl@{}} \Big|\frac{S_{(k)}(s)-\left( s-p-qsA_{k,k}(s)\right)S_{(k-1)}(s)}{\left( s-p-qsA_{k,k}(s)\right)S_{(k-1)}(s)}\Big| &=& \Big|\frac{-{\sum}_{j=1}^{k-1} q s A_{k,j}(s)U_{k,j}(s)}{\left( s-p-qsA_{k,k}(s)\right)S_{(k-1)}(s)}\Big| \\ & \le & \frac{{\sum}_{j=1}^{k-1} |q s A_{k,j}(s)||U^{*}_{k,j}(s)|}{|s-p-qsA_{k,k}(s)|}, \end{array} $$

where $|U^{*}_{k,j}(s)|={\frac {|U_{k,j}(s)|}{|S_{(k-1)}(s)|}}$ is the unique solution (by Cramer’s rule, provided S_(k− 1)(s)≠ 0) of the following system of equations

$$ \left[\begin{array}{lllll} \!s-p-qs A_{1,1}(s) & -qsA_{1,2}(s) & {\dots} & -qsA_{1,k-1}(s) \\ \!-qsA_{2,1}(s) & s-p-qs A_{2,2}(s) & {\dots} & -qsA_{2,k-1}(s) \\ \!{\vdots} & {\vdots} & {\vdots} & {\vdots} \\ \!-qsA_{k-1,1}(s) & -qsA_{k-1,2}(s) & {\dots} & s-p-qs A_{k-1,k-1}(s) \end{array}\!\right] \!\left[\begin{array}{lllll} U^{*}_{k,1}(s) \\ U^{*}_{k,2}(s) \\ {\vdots} \\ U^{*}_{k,k-1}(s) \end{array}\right] \!= \!\left[\begin{array}{lllll} \!-qsA_{1,k}(s) \\ \!-qsA_{2,k}(s) \\ \!{\vdots} \\ \!-qsA_{k-1,k}(s) \end{array}\!\right] $$

The i th (1 ≤ i ≤ k − 1) equation of above system of equations is given by

$$ \begin{array}{@{}rcl@{}} \left( s-p- q s A_{i,i}(s)\right)U^{*}_{k,i}(s) - \sum\limits_{j=1,~j\ne i}^{k-1} q s A_{i,j}(s) U^{*}_{k,j}(s) &=&-qsA_{i,k}(s). \end{array} $$

(98)

Now, we assume the contrary that

$$\max_{j}|U^{*}_{k,j}(s)|=|U^{*}_{k,i}(s)| \ge 1.$$

$$\Rightarrow~~~~ \left|\frac{U^{*}_{k,j}(s)}{U^{*}_{k,i}(s)}\right| \le 1 ~~~~ \& ~~~~\Big|\frac{1}{U^{*}_{k,i}(s)}\Big| \le 1.$$

Now Eq. 98 can be written as

$$ \begin{array}{@{}rcl@{}} |s-p- q s A_{i,i}(s)| &\le& \sum\limits_{j=1,~j\ne i}^{k-1} |q s A_{i,j}(s)| \left|\frac{U^{*}_{k,j}(s)}{U^{*}_{k,i}(s)}\right| + |qsA_{i,k}(s)|\left|\frac{1}{U^{*}_{k,i}(s)}\right|, \\ &\le& \sum\limits_{j=1,~j\ne i}^{k} |q s A_{i,j}(s)|, \end{array} $$

(99)

which is a contradiction to Lemma 2. Thus we have $|U^{*}_{k,i}(s)| < 1.$

Hence

$$ |S_{(k)}(s)- \left( s-p-qsA_{k,k}(s) \right) S_{(k-1)}(s)|<|\left( s-p-qsA_{k,k}(s)\right)S_{(k-1)}(s)|. $$

Thus by Rouch$\acute {e}$’s theorem S_(k)(s) and $\left (s-p-qsA_{k,k}(s)\right )S_{(k-1)}(s)$ have same number of zeros inside |s| = 1. Now by our assumption S_(k− 1)(s) has (k − 1) zeros and using Lemma 3, $\left (s-p-qsA_{k,k}(s)\right )$ has one zero inside |s| = 1. This implies $\left (s-p-qsA_{k,k}(s)\right )S_{(k-1)}(s)$ has k zeros inside the unit circle. Hence S_(k)(s) has k zeros inside |s| = 1. □

Lemma 5

The function$\left (s-p-qsA_{m,m}(s)\right )S_{(m-1)}(s)$has exactly mzeros in the region |s| < 1.

Proof

Using Lemma 4 it is easy to see that S_(m− 1)(s) has (m − 1) zeros in the region |s| < 1, and from Lemma 3 $\left (s-p-qsA_{m,m}(s)\right )$ has one zero inside |s| = 1. Also s = 1 is not a zero of the function under consideration. Therefore, all the zeros of $\left (s-p-qsA_{m,m}(s)\right )S_{(m-1)}(s)$ are concentrated in the region |s| < 1. □

Theorem 1

S(s) ≡|(s − p)I − qsA(s)| = 0 has mroots in |s|≤ 1.

Proof

The point s = 1 is a root of the function S_(m)(s) = 0 because the determinant S_(m)(1) can be reduced to a determinant that has the zero column. Hence, we need to prove that the function S_(m)(s) = 0 has exactly m roots in the region |s|≤ 1. Now we first prove the following inequality on the boundary of the curve $\{|s|<1,|s-1|<\epsilon \}\bigcap \{Re(s)>1\}$:

$$ \begin{array}{@{}rcl@{}} |S_{(m)}(s)-\left( s-p-qsA_{m,m}(s)\right)S_{(m-1)}(s)| &<& |\left( s-p-qsA_{m,m}(s)\right)S_{(m-1)}(s)|.~~ \end{array} $$

(100)

In order to prove the above inequality it is sufficient to prove the following:

$$ \begin{array}{@{}rcl@{}} \Big|\frac{S_{(m)}(s)}{\left( s-p-qsA_{m,m}(s)\right)S_{(m-1)}(s)}-1\Big|^2 &<& 1. \end{array} $$

(101)

Expanding the function that is inside the modulus on the left side in Taylor’s series at point s = 1, and using the fact that S_(m)(1) = 0, we obtain

$$ \begin{array}{@{}rcl@{}} \frac{S_{(m)}(s)}{\left( s-p-qsA_{m,m}(s)\right)S_{(m-1)}(s)}&=&\frac{S_{(m)}^{\prime}(1)}{\left( q(1-A_{m,m}(1))\right)S_{(m-1)}(1)}(s-1)+o(|s-1|). \\ \end{array} $$

(102)

where $S_{(m)}^{\prime }(1)=\frac {d}{ds}S_{(m)}(s)\Big |_{s=1}.$ Using the representation $s=1+\epsilon ({\cos \limits } \phi + i {\sin \limits } \phi ),~~ -{\pi }/2< \phi < {\pi }/2$ and Eq. 102, we can rewrite inequality (101) as

$$ \begin{array}{@{}rcl@{}} 1-2\frac{S_{(m)}^{\prime}(1)}{\left( q(1-A_{m,m}(1))\right)S_{(m-1)}(1)}\epsilon \cos \phi + o(\epsilon) & < & 1. \end{array} $$

(103)

Since $S_{(m)}^{\prime }(1)>0,~S_{(m-1)}(1)>0,~q(1-A_{m,m}(1))>0,~{\cos \limits } \phi >0$ which validates inequality (103). Hence inequality (101) is proved. Now applying Rouch$\acute {e}$’s theorem in Eq. 100 we conclude that S_(m)(s) = 0 (or S(s) = 0) has exactly m roots in the region |s|≤ 1. □