# Discrete Observations

In the analysis presented in the preceding three subsections, we assumed that an individual is continuously observed and his or her complete event history during the sample period is provided. However, in many practical situations a researcher may be able to observe the state of an individual only at discrete times. If the observations occur at irregular time intervals, it is probably more

reasonable to assume a continuous-time Markov model rather than a discrete-time model.

We shall derive the likelihood function based on discrete observations in the case of M = 2. The underlying model is the same stationary (or exponential) model we have so far considered. We define

P)k(t) = Prob [rth person is in state к at time t (11.2.40)

given that he or she was in state j at time 0].

Note that this definition differs slightly from the definition (11.1.2) of the same symbol used in the Markov chain model. We shall use the definition

(11.2.1) with the stationarity assumption X‘jk{t) — ttjk for all t.

As before, we shall concentrate on a particular individual and suppress the superscript i. When At is sufficiently small, we have approximately

P12(t + At) = Pn(tUi2At + Pl2m – АгіДГ). (11.2.41)

Dividing (11.2.41) by At and letting At go to 0 yield

^ = 21. (П.2.42)

Performing an analogous operation on Pn, P2i, and P22, we obtain the linear vector differential equation

A solution of (11.2.43) can be shown to be6

P’ = H^H-i, (11.2.44)

where the columns H are characteristic vectors of Л, D is the diagonal matrix consisting of the characteristic roots of Л, and (P1 is the diagonal matrix consisting of exp (djt), dj being the elements of D.

We shall derive D and H. Solving the determinantal equation

yields the two characteristic roots = 0 and d2 = — (A12 + ^21 )• Therefore we have

Let h, be the first column of H (the characteristic vector corresponding to the zero root). Then it should satisfy Ah, = 0, which yields a solution (which is not unique) h, = (A21 , A12)’. Next, the second column h2 of H should satisfy [Л + (Д,2 + A21 )I]h2 = 0, which yields a solution h2 = (— 1, 1)’. Combining the two vectors, we obtain

Finally, inserting (11.2.46) and (11.2.47) into (11.2.44) yields the following expressions for the elements of P’ (putting back the superscript /):

P[, (t) = 1 – у, + уt exp (- S/1) (11.2.48)

Г 1г(0 = У/~ Уі e*p (-Stt)

P‘n (0 “ 1 “ У, ~ (1 “ Уt) exp (-Stt)

P22U) = Уі + (1 “ Уд ехР (~Sit),

where у і = А’гДАІг + 4,) and S, = A‘12 + Ц,. Suppose we observe the rth individual in state jj at time 0 and in state k, at time г,, г = 1, 2,. . . , N. Then the likelihood function is given by

L = f[ptjM – 01-2.49)

1-1

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