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Survival, moments, and competing risks

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E16 Survival, moments, and competing risks

Let \(A\) be a transient state subgenerator. Its off diagonal entries are nonnegative and its column sums are nonpositive. Let \(p_0\) be a transient probability vector. Define

\[ p(t)={\mathrm e}^{tA}p_0, \qquad S(t)={{\boldsymbol{1}}}^Tp(t), \qquad a^T=-{{\boldsymbol{1}}}^TA. \]

The row vector \(a^T\) contains the exit rates.

Theorem E16.1: State dependent killing as survival

Let \(Q\) be a finite Markov generator, let \(V\geq0\), and set

\[ A=Q-D_V. \]

For an initial distribution \(p_0\), define

\[ S_V(t)={{\boldsymbol{1}}}^T{\mathrm e}^{tA}p_0. \]

Then

\[ S_V(t) = {\mathbb E}_{p_0} \left[ \exp\left( -\int_0^tV(X_s){\,\mathrm d} s \right) \right]. \]

Also,

\[ -S_V'(t) = {{\boldsymbol{1}}}^TD_V{\mathrm e}^{tA}p_0. \]

The matrix \(A=Q-D_V\) has nonnegative off diagonal entries and nonpositive column sums, so it is the subgenerator of the killed, possibly nontransient chain. Thus the potential in the finite Feynman Kac formula is a state dependent killing rate in the forward sub-Markov semigroup. Almost-sure absorption follows under the additional spectral stability hypothesis used below.

Proof of Theorem E16.1

Average Theorem E13.2 with terminal function \(f={{\boldsymbol{1}}}\) against the initial distribution. Transposition gives

\[ p_0^T{\mathrm e}^{t(Q^T-D_V)}{{\boldsymbol{1}}} = {{\boldsymbol{1}}}^T{\mathrm e}^{t(Q-D_V)}p_0. \]

Since \({{\boldsymbol{1}}}^TQ=0\), differentiation gives

\[ S_V'(t) = -{{\boldsymbol{1}}}^TD_V{\mathrm e}^{tA}p_0. \]

Theorem E16.2: Survival and hazard identities

The event density and hazard are

\[ f(t)=a^T{\mathrm e}^{tA}p_0=-S'(t), \qquad h(t)=\frac{f(t)}{S(t)} \]

whenever \(S(t)>0\). The survival function is nonincreasing and

\[ S(t) = \exp\left( -\int_0^th(s){\,\mathrm d} s \right). \]

Proof of Theorem E16.2

Because \(A\) is a Metzler matrix, \({\mathrm e}^{tA}\) is nonnegative. The column sum assumption makes \(a^T\) nonnegative. Differentiate \(S(t)\):

\[ S'(t)={{\boldsymbol{1}}}^TA{\mathrm e}^{tA}p_0=-a^T{\mathrm e}^{tA}p_0. \]

The remaining identity follows by integrating \((\log S)'=-h\).

Theorem E16.3: Laplace transform and moments

Assume the spectrum of \(A\) lies in the open left half plane. Then absorption occurs with probability one. For \(s\geq0\),

\[ {\mathbb E}[{\mathrm e}^{-s\tau}] = a^T(sI-A)^{-1}p_0. \]

For every integer \(m\geq1\),

\[ {\mathbb E}[\tau^m] = m!{{\boldsymbol{1}}}^T(-A)^{-m}p_0. \]

Proof of Theorem E16.3

The Laplace transform follows by integrating the density in Theorem E16.2. Stability of \(A\) gives

\[ \int_0^\infty {\mathrm e}^{-st}{\mathrm e}^{tA}{\,\mathrm d} t = (sI-A)^{-1}. \]

At \(s=0\),

\[ a^T(-A)^{-1}p_0 = {{\boldsymbol{1}}}^Tp_0=1. \]

For the moments,

\[ \int_0^\infty t^m{\mathrm e}^{tA}{\,\mathrm d} t = m!(-A)^{-m-1}. \]

Multiplying by \(a^T={{\boldsymbol{1}}}^T(-A)\) gives the result. These are the standard matrix formulas for phase type laws [Neuts1981], [AsmussenNermanOlsson1996].

Proposition E16.4: Residual life as a restarted semigroup

For \(S(t)>0\), define the conditional transient distribution

\[ p_t = \frac{{\mathrm e}^{tA}p_0}{S(t)}. \]

Then

\[ \frac{S(t+s)}{S(t)} = {{\boldsymbol{1}}}^T{\mathrm e}^{sA}p_t. \]

Proof of Proposition E16.4

Use the semigroup law in the numerator \(S(t+s)={{\boldsymbol{1}}}^T{\mathrm e}^{sA}{\mathrm e}^{tA}p_0\).

Theorem E16.5: Competing risk packets

Suppose the exit row splits into nonnegative cause rows

\[ a^T = \sum_{r=1}^q a_r^T. \]

Define

\[ f_r(t)=a_r^T{\mathrm e}^{tA}p_0. \]

Then

\[ f(t)=\sum_{r=1}^qf_r(t). \]

The cause specific hazard is

\[ h_r(t)=\frac{f_r(t)}{S(t)}, \]

and

\[ h(t)=\sum_{r=1}^qh_r(t). \]

If the spectrum of \(A\) lies in the open left half-plane, the eventual probability of cause \(r\) is

\[ \int_0^\infty f_r(t){\,\mathrm d} t = a_r^T(-A)^{-1}p_0. \]

Proof of Theorem E16.5

Every identity follows by linearity and the stable resolvent integral.

Survival and two cause specific event densities from one transient semigroup.
Survival and two cause specific event densities from one transient semigroup.

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Remark E16.6

A finite phase type hazard need not be monotone. The semigroup formulas guarantee nonnegative density and decreasing survival, not monotonicity of the hazard.

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