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The convention pii = 0 for all states i is convenient and natural This convention ensures that the sojourn time in a state is unambiguously de ned If there are no absorbing states, it is no restriction to make this convention (the sum of a geometrically distributed number of independent lifetimes with a common exponential distribution is again exponentially distributed) Throughout this chapter the following assumption is made Assumption 411 In any nite time interval the number of jumps is nite with probability 1 De ne now the continuous-time stochastic process {X(t), t 0} by X(t) = the state of the system at time t The process is taken to be right-continuous; that is, at the transition epochs the state of the system is taken as the state just after the transition The process {X(t)} can be shown to be a continuous-time Markov chain It will be intuitively clear that the process has the Markov property by the assumption of exponentially distributed sojourn times in the states Assumption 411 is needed to exclude pathological cases For example, suppose the unbounded state space I = {1, 2, }, take pi,i+1 = 1 and i = i 2 for all i Then transitions occur faster and faster so that the process will ultimately face an explosion of jumps With a nite state space the Assumption 411 is always satis ed Example 411 Inventory control for an in ammable product An in ammable product is stored in a special tank at a lling station Customers asking for the product arrive according to a Poisson process with rate Each customer asks for one unit of the product Any demand that occurs when the tank is out of stock is lost Opportunities to replenish the stock in the tank occur according to a Poisson process with rate The two Poisson processes are assumed to be independent of each other For reasons of security it is only allowed to replenish the stock when the tank is out of stock At those opportunities the stock is replenished with Q units for a given value of Q To work out the long-run average stock in the tank and the long-run fraction of demand that is lost, we need to study the inventory process For any t 0, de ne X(t) = the amount of stock in the tank at time t The stochastic process {X(t), t 0} is a continuous-time Markov chain with state space I = {0, 1, , Q} The sojourn time in each state is exponentially distributed, since both the times between the demand epochs and the times between the replenishment opportunities are exponentially distributed Thus the sojourn time in state i has an exponential distribution with parameter i = , , i = 1, , Q, i = 0.

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The state transitions are governed by a discrete-time Markov chain whose one-step transition probabilities have the simple form pi,i 1 = 1 for i = 1, , Q, p0Q = 1 and the other pij = 0 In nitesimal transition rates Consider the general Markov jump process {X(t)} that was constructed above The sojourn time in any state i has an exponential distribution with mean 1/ i and the state transitions are governed by a Markov chain having one-step transition probabilities pij for i, j I with pii = 0 for all i The Markov process allows for an equivalent representation involving the so-called in nitesimal transition rates To introduce these rates, let us analyse the behaviour of the process in a very small time interval of length t Recall that the exponential (sojourn-time) distribution has a constant failure rate; see Appendix B Suppose that the Markov process {X(t)} is in state i at the current time t The probability that the process will leave state i in the next t time units with t very small equals i t + o( t) by the constant failure rate representation of the exponential distribution If the process leaves state i, it jumps to state j (= i) with probability pij Hence, for any t > 0, P {X(t + t) = j | X(t) = i} = i t pij + o( t), 1 i t + o( t), j = i, j = i,

as t 0 One might argue that in the next t time units state j could be reached from state i by rst jumping from state i to some state k and next jumping in the same time interval from state k to state j However, the probability of two or more state transitions in a very small time interval of length t is of the order ( t)2 and is thus o( t); that is, this probability is negligibly small compared with t as t 0 De ne now qij = i pij , i, j I with j = i

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