Paper VI visual guide · Visual 24 of 42 · PHASE-TYPE SURVIVAL / TWO SUCCESSIVE STAGES

A zero initial hazard can still lead to certain exit

Traditional Part I: Theorem 14.1. Survival, density, hazard, and moments have distinct meanings.

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Let a transient process move through two successive rate-one stages, then leave the system. With column probabilities, the transient generator and starting vector are

\[ T={\begin{pmatrix}-1&0\\1&-1\end{pmatrix}},\qquad p(0)=e_0. \]

Thus \(p(t)=e^{-t}(1,t)^{\mathsf T}\). Its total remaining mass, exit density, and hazard are

\[ S(t)=(1+t)e^{-t},\qquad f(t)=t e^{-t}, \qquad h(t)=\frac{t}{1+t}. \]

The initial hazard is zero because an exit needs two stages, not one.

Survival, absorbed mass, and hazard are plotted as different curves.
Original drawing. Read the description below; expand the labels and data for selectable text.

Reading the drawing

The solid survival curve decreases, the dashed absorbed mass increases, and the dotted hazard starts at zero. Their meanings and the eventual-exit statement are written beside the plot. The plate keeps probability and rate distinct.

Drawing labels and data 15 labels

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  • 0
  • 1
  • 2
  • 3
  • 4
  • 5
  • 0
  • 0.5
  • 1
  • time \(t\)
  • probability or rate
  • solid: survival \(S(t)\)
  • dashed: absorbed mass
  • dotted: hazard \(h(t)\)

\(h(0)=0\), but eventual exit
has probability one.

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Evidence note from the source edition

Survival falls from one to zero, while absorbed mass \(1-S\) rises to one. The dotted hazard is an instantaneous rate, not a probability, even though it shares the displayed numerical scale. Time is measured in inverse stage-rate units.

Missing transient mass is absorbed mass

The column sums of \(T\) are \(0\) and \(-1\). The exit-rate row is therefore \((0,1)\), and \(f(t)=(0,1)p(t)\). Adding an absorbing state restores a three-state probability vector with total mass one.

Exact integration gives

\[ \int_0^\infty f(t)\,dt=1,\qquad {\mathbb E}[\tau]=2,\qquad {\mathbb E}[\tau^2]=6, \qquad {\operatorname{Var}}(\tau)=2. \]

Equivalently, the mean is \({\mathbf{1}}^{\mathsf T}(-T)^{-1}e_0\). A single exponential waiting time with mean two would have constant hazard \(1/2\) and variance four, so matching the mean alone would not reproduce this distribution.

The hazard formula is used where \(S(t)>0\). The model has that property for every finite \(t\ge0\). No division by an already vanished survival mass is required.

Read this as

Survival tracks mass, density tracks exits per unit time, and hazard conditions on survival. Equal means or equal initial rates do not identify a waiting-time law.

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