On the Markov Three-State Progressive Model

Abstract

In this work we revisit the Markov three-state progressive model. Several characterizations of the Markov condition are discussed, and nonparametric estimators of important targets such as the bivariate distibution of the event times or the transition probabilities are motivated. Three points of special interest are considered: (1) the relative improvements when introducing the Markov condition in the construction of the estimators; (2) bootstrap algorithms to resample under markovianity; and (3) goodness-of-fit testing for the Markov assumption. Simulation studies and some technical derivations are included.

Appendix

In this Appendix we give the technical derivations of three results that have been referred in Sect. 19.2 and Sect. 19.3.

Lemma 1

Assume that $\left( {Z,T} \right)$ is Markov. Then, we have for each t.

$$ \Uppsi (t) = \int\limits_{0}^{t} {\frac{{F_{T} ({\text{d}}y)}}{M(y)}} . $$

Proof

From the bivariate df of $\left( {Z,T} \right)$ we get that the marginal df of T satisfies

$$ F_{T} (t) = F_{Z,T} (t,t) = F_{Z} (t) - \overline{S} (t)\int\limits_{0}^{t} {\frac{{F_{Z} ({\text{d}}y)}}{{\overline{S} (y)}}} . $$

Since $\overline{S}^{'} (t) = - \psi (t)\overline{S} (t)$ we have

$$ F_{T} ({\text{d}}t) = \psi (t)\overline{S} (t)\int\limits_{0}^{t} {\frac{{F_{Z} ({\text{d}}y)}}{{\overline{S} (y)}}{\text{d}}t} . $$

We also have

$$ \begin{aligned} M(y) =& \iint\limits_{{\left\{ {u < y \le v} \right\}}} {F_{Z,T} ({\text{d}}u,{\text{d}}v)} = \iint\limits_{{\left\{ {u < y \le v} \right\}}} {\psi (v)\frac{{\overline{S} (v)}}{{\overline{S} (u)}}F_{Z} ({\text{d}}u){\text{d}}v} \\ =& \int\limits_{y}^{\infty } {\psi (v)\overline{S} (v){\text{d}}v} \int\limits_{0}^{y} {\frac{{F_{Z} ({\text{d}}u)}}{{\overline{S} (u)}}} . \hfill \\ \end{aligned} $$

Hence,

$$ \int\limits_{0}^{t} {\frac{{F_{T} (\hbox{d}y)}}{M(y)}} = \int\limits_{0}^{t} {\frac{{\psi (y)\overline{S} (y)\hbox{d}y}}{{\int_{y}^{\infty } \psi (v)\overline{S} (v)\hbox{d}v}}} = - \log \overline{S} (t), $$

where for the last equality we have used $\overline{S} (t) = \int_{t}^{\infty } {\psi \overline{S} } .$ Since $\Uppsi (t) = - \log \overline{S} (t),$ this completes the proof.

Lemma 2

Assume that $\left( {Z,T} \right)$ is Markov. Then, for each $z < t$ we have $A(t,z) = A(t,0) - A(z,0).$ Conversely, if $\left( {Z,T} \right)$ is not Markov, there exists a pair $\left( {t,z} \right)$ for which $A(t,z) \ne A(t,0) - A(z,0).$

Proof

Write

$$ \begin{aligned} A(t,z) &= \iint\limits_{{\left\{ {z < x < y \le t} \right\}}} {\frac{{F_{Z,T} ({\text{d}}x,{\text{d}}y)}}{{M_{z} (y)}}} = \iint\limits_{{\left\{ {z < x < y \le t} \right\}}} {\frac{\psi (y)}{{M_{z} (y)}} \frac{{\overline{S} (y)}}{{\overline{S} (x)}}F_{Z} ({\text{d}}x){\text{d}}y} \\ &= \int\limits_{z}^{t} \frac{{\psi (y)\overline{S} (y)}}{{M_{z} (y)}}\left[ {\int\limits_{z}^{y} \frac{{F_{Z} ({\text{d}}x)}}{{\overline{S} (x)}}} \right]{\text{d}}y. \hfill \\ \end{aligned} $$

Now,

$$ \begin{aligned} M_{z} (y) &= \iint\limits_{{\left\{ {z < u < y \le v} \right\}}} F_{Z,T} ({\text{d}}u,{\text{d}}v) = \iint\limits_{{\left\{ {z < u < y \le v} \right\}}} \psi (v)\frac{{\overline{S} (v)}} {{\overline{S} (u)}}F_{Z} ({\text{d}}u){\text{d}}v \\ \, &= \int\limits_{y}^{\infty } \psi (v)\overline{S} (v){\text{d}}v\int\limits_{z}^{y} \frac{{F_{Z} ({\text{d}}u)}}{{\overline{S} (u)}}. \hfill \\ \end{aligned} $$

Summarizing, we have

$$ A(t,z) = \int\limits_{z}^{t} {\frac{{\psi (y)\overline{S} (y)}}{{\int_{y}^{\infty } \psi (v)\overline{S} (v){\text{d}}v}}{\text{d}}y} . $$

From this, equation $A(t,z) = A(t,0) - A(z,0)$ is immediately satisfied. The second assertion can be checked by tracing the above steps back.

Lemma 3

$H_{0}^{Q}$ holds if and only if $(U,V)$ conditionally on $U \le V$ is Markov.

Proof

Assume that $(U,V)|U \le V$ is Markov. Then, the conditional df of $(U,V)$ given $U \le V,$ say $F_{U,V|U \le V} ,$ satisfies for $u \le v$

$$ F_{U,V|U \le V} (u,v) = \int\limits_{0}^{u} {\left( {1 - \frac{{\overline{S} (v)}}{{\overline{S} (x)}}} \right)F_{U|U \le V} ({\text{d}}x)} , $$

where $\overline{S} (t) = \exp \left[ { - \int_{0}^{t} \psi } \right]$ stands for the survival function of $V|U = 0,U \le V,$ and $U|U \le V \sim F_{U|U \le V} .$ Then,

$$ \begin{aligned} P(U \le u,V > v|U \le V) =& \iint\limits_{{\left\{ {x \le u,y > v} \right\}}} F_{U,V|U \le V} ({\text{d}}x , {\text{d}}y) \\ \, =& \iint\limits_{{\left\{ {x \le u,y > v} \right\}}} \psi (y)\frac{{\overline{S} (y)}}{{\overline{S} (x)}}F_{U|U \le V} ({\text{d}}x ) {\text{d}}y \\ \, =& \int\limits_{0}^{u} \frac{{F_{U|U \le V} (\hbox{d}x)}}{{\overline{S} (x)}}\int\limits_{v}^{\infty } \psi (y)\overline{S} (y){\text{d}}y, \hfill \\ \end{aligned} $$

and hence $H_{0}^{Q}$ is satisfied. The complementary assertion is proved by tracing the above steps back.