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To introduce the uniformization method, consider rst the special case in which the leaving rates i of the states are identical, say i = for all i Then the transition epochs are generated by a Poisson process with rate In this situation an expression for pij (t) is directly obtained by conditioning on the number of Poisson events up to time t and using the n-step transition probabilities of the embedded Markov chain {Xn } However, the leaving rates i are in general not identical Fortunately, there is a simple trick for reducing the case of non-identical leaving rates to the case of identical leaving rates The uniformization method transforms the original continuous-time Markov chain with non-identical leaving rates into an equivalent stochastic process in which the transition epochs are generated by a Poisson process at a uniform rate However, to achieve this, the discrete-time Markov chain describing the state transitions in the transformed process has to allow for self-transitions leaving the state of the process unchanged

De ne now {X n } as the discrete-time Markov chain whose one-step transition probabilities pij are given by pij = ( i / )pij , 1 i / , j = i, j = i,

SMLC Lb Ls (MAP) A MSC Lg (MAP)

.

for any i I Let {N (t), t 0} be a Poisson process with rate such that the process is independent of the discrete-time Markov chain {X n } De ne now the continuous-time stochastic process {X(t), t 0} by X(t) = XN (t) , t 0 (453)

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In other words, the process {X(t)} makes state transitions at epochs generated by a Poisson process with rate and the state transitions are governed by the discretetime Markov chain {Xn } with one-step transition probabilities pij Each time the Markov chain {X n } is in state i, the next transition is the same as in the Markov chain {Xn } with probability i / and is a self-transition with probability 1 i / The transitions out of state i are in fact delayed by a time factor of / i , while the time itself until a state transition from state i is condensed by a factor of i / This heuristically explains why the continuous-time process {X(t)} is probabilistically identical to the original continuous-time Markov chain {X(t)} Another heuristic way to see that the two processes are identical is as follows For any i, j I with j =i P {X(t + t) = j | X(t) = i} = t pij + o( t) = i t pij + o( t) = qij t + o( t) = P {X(t + t) = j | X(t) = i} for t 0

In the next theorem we give a formal proof that the two processes {X(t)} and {X(t)} are probabilistically equivalent Theorem 452 Suppose that the continuous-time Markov chain {X(t)} satis es Assumption 412 Then pij (t) = P {X(t) = j | X(0) = i}, i, j I and t > 0

Proof For any t > 0, de ne the matrix P(t) by P(t) = (pij (t)), i, j I Denote by Q the matrix Q = (qij ), i, j I , where the diagonal elements qii are de ned by qii = i

.

Then Kolmogoroff s forward differential equations can be written as P (t) = P(t)Q for any t > 0 It is left to the reader to verify that the solution of this system of differential equations is given by P(t) = etQ =

n=0

tn n Q , n!

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