13_AS_5B_lec_annotated

As p12 0 0 we have g 12 23 12 13 and hence p12

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Unformatted text preview: del General Markov Model Solution (continued) Solving the above equation we find that P12 (t ) = e −µ23 t µ12 e (µ23 −(µ12 +µ13 ))t + g µ23 − (µ12 + µ13 ) where g is a constant to be determined. As P12 (0) = 0 we have g= −µ12 µ23 − (µ12 + µ13 ) and hence P12 (t ) = −µ12 e −µ23 t − e −(µ12 +µ13 )t µ23 − (µ12 + µ13 ) Finally, we also have P13 (t ) = 1 − P11 (t ) − P12 (t ) 34/60 and notice that Pii (t ) = Pii (t ), and P31 (t ) = P32 (t ) = 0. Actuarial Statistics – Module 5: Parametric methods: Markov Model General Markov Model Computing Transition Probabilities In general cases, numerical methods are required to compute the transition probabilities Consider the forward equation in matrix notation: P (t ) = P (t ) R where P is the transition probability matrix and R is the intensity matrix. Solution is P (t ) = e Rt where ∞ e Rt Rn = n=0 tn n! Use Approximations 35/60 e Rt lim n→∞ I+R t n n Actuarial Statistics – Module 5: Parametric methods:...
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