T1 t2 t3 also 2l 2 1 2 n12 n23 n34 so variance

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Unformatted text preview: ave independently. Over a year we observe that the number of transitions from state i to state j is nij and the total waiting time in state i is Ti . (a) Write down an expression for the likelihood of the transition for the year and derive formulae for maximum likelihood estimates of λ and µ and their standard errors. 53/60 Actuarial Statistics – Module 5: Parametric methods: Markov Model General Markov Model Example (continued) Solution to (a): λn12 +n23 +n34 µn21 +n32 +n43 e −λt1 e −(λ+µ)t2 e −(µ+λ)t3 e −µt4 . Loglikelihood is given by log L = ln λ · (n12 + n23 + n34 ) + ln µ · (n21 + n32 + n43 ) −λ (t1 + t2 + t3 ) − µ (t2 + t3 + t4 ) . Differentiating w.r.t. λ, and setting = 0 then 1 ∂L = (n12 + n23 + n34 ) − (t1 + t2 + t3 ) = 0 ∂λ λ n12 + n23 + n34 ˆ So λ = . t1 + t2 + t3 Also ∂2L ∂λ2 1 = − λ2 (n12 + n23 + n34 ) . So variance can be obtained by ˆ λ2 n12 + n23 + n34 = . n12 + n23 + n24 (t1 + t2 + t3 )2 √ 54/60 Hence standard error: n12 +n23 +n34 . t1 +t2 +t3 √ Actuarial Statistics – Module 5: Parametric methods: Markov Model General Markov Model Example (continued) ∧ ∧ (b) Suppose our estimates were λ = 0.094 and µ = 0.100. It is also given that T1 = 2000, T2 = 3000, T3 = 5000 and T4 = 4000. Is there any statistical evidence that λ < µ? (b) H0 : λ = µ, i .e . λ − µ = 0 vs H1 : λ < µ ˆ Note that λ and µ are asymptotically independent, so ˆ because...
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This document was uploaded on 04/03/2014.

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