Parametric survival and reliability

Define $F_{\lambda }(t)=0$ for $t<0$ and $F_{\lambda }(t)=1-e^{-\lambda t}$ for $t\ge 0$; define $F_{k,\eta }(t)=0$ for $t<0$ and $F_{k,\eta }(t)=1-e^{-(t\mathrm{/}\eta {)}^k}$ for $t\ge 0$. Both are continuous at $0$ by [F2], [F4], and [F5]. On $t>0$ their derivatives are the displayed densities and are positive by [F2]. The piecewise densities are Borel measurable: on $(0,\mathrm{\infty })$ they are products of continuous functions, and on $(-\mathrm{\infty },0]$ they vanish; both domains are Borel by [F11]. The mean value theorem in [F3] makes each increasing on $(0,\mathrm{\infty })$. For $t>0$, both $a=\lambda t$ and $a=(t\mathrm{/}\eta {)}^k$ are positive, so strict increase of the exponential gives $e^{-a}<e^0=1$; hence both candidate CDFs are positive there; thus their value $0$ at $0$ is consistent with monotonicity. Their limits are $0$ at $-\mathrm{\infty }$ and $1$ at $+\mathrm{\infty }$, so each is a CDF on $\mathbb{R}$.

Fix $0<a<b$. Each CDF is differentiable on $[a,b]$ with the continuous derivative $f$ from the Statement. By [F6] its Riemann integral is $F(b)-F(a)$; by [F7] this equals its Lebesgue integral with respect to the measure in [F10]. Letting $a=1\mathrm{/}n\downarrow 0$ for integers $n$ large enough that $1\mathrm{/}n<b$, [F8] and continuity at $0$ give ${\int }_{(0,b]}f(u)du=F(b)\mathrm{.}$ This applies separately to the exponential and Weibull formulas.

For either displayed nonnegative density $f$, [F1] defines the measure ${\nu }_f=fdt$ on Lebesgue-measurable sets. By [F12], its restriction to Borel sets is a measure. Step 2.1 gives ${\nu }_f((0,t])=F(t)$ for every $t>0$. Since $f=0$ on $(-\mathrm{\infty },0]$, this part has ${\nu }_f$-measure zero, so ${\nu }_f((-\mathrm{\infty },t])=F(t)$; for $t\le 0$ both sides are $0$. Also ${\nu }_f(\mathbb{R})={\lim }_{n\to \mathrm{\infty }}{\nu }_f((-\mathrm{\infty },n])={\lim }_{n\to \mathrm{\infty }}F(n)=1$ by [F8] and step 1.1. Thus ${\nu }_f$ is a probability law with CDF $F$, and [F9] identifies it with the unique law specified by the survival formula. Consequently $f$ is the stated density.

Dividing the exponential density by its positive survival gives hazard $\lambda$. Dividing the Weibull density by its positive survival gives ${\displaystyle \frac{k}{\eta }}(t\mathrm{/}\eta {)}^{k-1}$ for $t>0$.

Steps 1.1, 2.1, 3.1, and 4.1 establish CDF validity, density-law existence, survival formulas, uniqueness, and the density hazards for both parameter families.

Since $T\ge 0$ and $S_T(0)=1$, $P_T(T=0)=0$. The density law gives $P_T(T\le x)={\int }_{[0,x]}f(u)du$; [F8] makes both endpoint integrals zero, so this equals ${\int }_0^{x}f(u)du$. Therefore $S_T(x)=1-{\int }_0^{x}f(u)du$ for every $x\ge 0$. This integral is at most $1$, so $f\in L^1[0,x]$ for each $x>0$; [F3] then makes $S_T$ absolutely continuous on each compact interval.

The identity is immediate at $t=0$, so fix $0<t<{\tau }_{\ast }$. The survival function is nonincreasing and positive on $[0,t]$, so $S_T(u)\ge S_T(t)>0$ there. For every $a\in \mathbb{R}$, the set $\{u\in [0,t]:S_T(u)>a\}$ is downward closed in $[0,t]$, hence is empty, all of $[0,t]$, or an interval with one endpoint at $0$; each is Lebesgue measurable by [F9]. Since the upper rays generate the Borel sigma-algebra, $S_T$ is Lebesgue measurable on $[0,t]$. The reciprocal map is continuous on $[S_T(t),1]$, so its inverse images of Borel sets are Borel; the preimage identity then makes $1\mathrm{/}{S}_T$ measurable. [F10] gives measurability of the product $f\mathrm{/}{S}_T$. Consequently ${\int }_0^t\frac{f(u)}{S_T(u)}du\le \frac{1}{S_T(t)}{\int }_0^{t}f(u)du<\mathrm{\infty }\mathrm{.}$ Thus ${\mathrm{\Lambda }}_T(x):={\int }_0^{x}f(u)\mathrm{/}{S}_T(u)du$ is absolutely continuous on $[0,t]$ by [F3].

By [F2] and step 1.1, $S_T^{\mathrm{\prime }}=-f$ almost everywhere on $[0,t]$. For $x,y\in [S_T(t),1]$, [F4] gives $\mathrm{\mid }\log x-\log y\mathrm{\mid }\le \frac{\mathrm{\mid }x-y\mathrm{\mid }}{S_T(t)}\mathrm{.}$ This Lipschitz bound and absolute continuity of $S_T$ show directly from the finite disjoint-interval definition that $\log \circ S_T$ is absolutely continuous. Applying [F2] to the integrable function $f\mathrm{/}{S}_T$ gives ${\mathrm{\Lambda }}_T^{\mathrm{\prime }}=f\mathrm{/}{S}_T$ almost everywhere. At almost every point where $S_T^{\mathrm{\prime }}$ exists, [F5] yields $(\log S_T{)}^{\mathrm{\prime }}=S_T^{\mathrm{\prime }}\mathrm{/}{S}_T=-f\mathrm{/}{S}_T=-{\mathrm{\Lambda }}_T^{\mathrm{\prime }}$.

Applying [F6] to $\log \circ S_T$ on $[0,t]$ gives $\log S_T(t)-\log S_T(0)=-{\int }_0^t\frac{f(u)}{S_T(u)}du\mathrm{.}$ Since $S_T(0)=1$ and [F11] gives $\log 1=0$, exponentiating and using [F11] yields the identity for $0<t<{\tau }_{\ast }$; the case $t=0$ was handled in step 1.2.

If ${\tau }_{\ast }<\mathrm{\infty }$, continuity from [F7] makes the zero set $Z:=\{t\ge 0:S_T(t)=0\}$ closed. Since $Z$ is nonempty and bounded below, ${\tau }_{\ast }:=\inf Z$ belongs to $Z$: for every $n\in \mathbb{N}$ choose $z_n\in Z$ with ${\tau }_{\ast }\le z_n<{\tau }_{\ast }+1\mathrm{/}(n+1)$, so $z_n\to {\tau }_{\ast }$ and continuity gives $S_T({\tau }_{\ast })=0$. Nonincreasing survival [F1] then gives $S_T(t)=0$ for $t\ge {\tau }_{\ast }$. Step 3.1 gives ${\mathrm{\Lambda }}_T(t)=-\log S_T(t)$ for $t<{\tau }_{\ast }$. Since $S_T(t)\to 0$ by continuity at ${\tau }_{\ast }$ and [F4], this tends to $+\mathrm{\infty }$ as $t\uparrow {\tau }_{\ast }$. The stated extended convention therefore gives $S_T(t)=e^{-{\mathrm{\Lambda }}_T(t)}$ also for $t\ge {\tau }_{\ast }$.

The identity holds throughout the positive-survival interval and at a finite terminal endpoint under the stated extended convention.

By [F2], $P(T=t)=F_T(t)-F_T(t^{-})$. Also $P(T\ge t)=P(T>t)+P(T=t)$, so [F1] gives $S_T(t^{-})=1-F_T(t^{-}),P(T=t)=S_T(t^{-})-S_T(t)\mathrm{.}$

Divide the identity in step 1.1 by the positive quantity $S_T(t^{-})$. Using the definition of $q_T(t)$ gives $q_T(t)=1-\frac{S_T(t)}{S_T(t^{-})},$ which is equivalent to $S_T(t)=S_T(t^{-})(1-q_T(t))$.

In the finite-support case, $S_T(t_1^{-})=1$, since $T\ge t_1$. For $j>1$, no probability mass lies in $(t_{j-1},t_j)$, so $S_T(t_j^{-})=P(T\ge t_j)=P(T>t_{j-1})=S_T(t_{j-1})$. Applying step 2.1 successively yields $S_T(t_j)=S_T(t_{j-1})(1-q_j)={\prod }_{r=1}^j(1-q_r),$ where the initial factor is $S_T(t_1^{-})=1$.

Step 2.1 proves the jump identity for every atom with positive pre-jump survival, and step 3.1 proves the finite atomic product formula.

For any Borel $B\subseteq [0,\mathrm{\infty })$, the event branch satisfies $P(U\in B,\delta =1)=\int \text{\hspace{0.17em}⁣}\int 1\{t\in B,t\le c\}\lambda e^{-\lambda t}dt{G}_{\psi }(dc)={\int }_B\lambda e^{-\lambda u}{G}_{\psi }([u,\mathrm{\infty }))du\mathrm{.}$ Here we use the event-density version $\lambda e^{-\lambda t}$ for $t\ge 0$, which differs from [F2] only at the Lebesgue-null singleton $\{0\}$ by [F11]. The indicator is product measurable because $B$ is Borel and $\{(t,c):t\le c\}$ is closed. The factor $\lambda e^{-\lambda t}$ is continuous, so [F17] makes the full integrand product measurable. The first equality uses the product law from [A3], [F2], and [F3]; the second integrates the nonnegative indicator in the censor coordinate by Tonelli.

On the censor branch, $U=C$ and $T>C$, so $P(U\in B,\delta =0)={\int }_{B}P(T>u)G_{\psi }(du)={\int }_B{e}^{-\lambda u}{g}_{\psi }(u)\nu (du)\mathrm{.}$ The integrand $1_B(u)e^{-\lambda u}$ is measurable by [F17]. This uses the survival in [F2], independence, and the specified density of $G_{\psi }$; [F12] changes integration against $G_{\psi }=g_{\psi }\nu$ into integration against $g_{\psi }d\nu$.

The formal survival boundary $S_0(t)=1$ cannot be a probability law of a finite-valued event time: for every integer $m$, it would give $P(T\le m)=0$, while $[0,\mathrm{\infty })={\bigcup }_m[0,m]$ and [F7] would then give $P(T\in [0,\mathrm{\infty }))=0$. Thus the boundary value is a likelihood extension unless the event-time model is enlarged to admit $T=\mathrm{\infty }$.

Extend $u\mapsto {\bar{G}}_{\psi }(u)$ to $u<0$ by the constant $G_{\psi }([0,\mathrm{\infty }))=1$. This extension is nonincreasing by inclusion of the tail sets, so [F9] makes it Borel measurable. The event-branch density is measurable by [F17], and the censor-branch density is measurable by [F17] because $g_{\psi }$ is measurable. The event branch is sigma-finite because [F11] gives the finite exhaustion $[0,n]$; the censor branch uses a countable finite-measure exhaustion of $\nu$. Their union across the disjoint branches makes $\rho$ sigma-finite. Steps 1.1 and 1.2 then show that the one-subject observation law is dominated by $\rho$ with the displayed density $h_{\lambda ,\psi }$.

For each $i$, $(T_i,C_i)$ is a measurable map into the product space by [F3]. The map $(t,c)\mapsto (\min (t,c),1\{t\le c\})$ is Borel: minimum is continuous and $\{(t,c):t\le c\}$ is closed. Thus [F10] makes each observed pair measurable under the convention [F1]; taking the finite tuple is measurable since rectangle preimages are finite intersections. The coordinate pullbacks of $h_{\lambda ,\psi }$ are measurable, so repeated use of [F17] makes $D(\mathbf{u},\mathit{\delta }):={\prod }_{i=1}^n{h}_{\lambda ,\psi }(u_i,{\delta }_i)$ measurable. Let $Q$ be its density measure relative to ${\rho }^{\otimes n}$, which exists by [F16]. By step 2.1, each one-subject density integrates to one. Repeated Tonelli integration therefore makes $Q$ a probability measure and gives, for every measurable rectangle $R={\prod }_{i=1}^n{B}_i$, $Q(R)={\prod }_{i=1}^n{\int }_{B_i}{h}_{\lambda ,\psi }d\rho \mathrm{.}$ Independence of the subject pairs and [F3] give the same rectangle probabilities for their joint observation law. Measurable rectangles form a generating $\pi$-system, and the whole product space is a finite-measure exhaustion for both probability measures. The uniqueness clause of [F3], iterated over the $n$ factors, now identifies $Q$ with the joint observation law. Hence $D$ is the asserted density version.

Substituting the two branch formulas from step 2.1 into the product density of step 3.1 gives $L(\lambda ,\psi ;\mathbf{u},\mathit{\delta })={\lambda }^{{\sum }_i{\delta }_i}{e}^{-\lambda {\sum }_i{u}_i}{\prod }_{i:{\delta }_i=1}{\bar{G}}_{\psi }(u_i){\prod }_{i:{\delta }_i=0}{g}_{\psi }(u_i)\mathrm{.}$ By [F5], the experiment and parameter map give the sampling-law context; under the parameter-distinct model [A2], this is the asserted factorization. The density representatives are explicit model inputs, so no arbitrary simultaneous selection of Radon–Nikodym versions is used; the equality is stated separately almost everywhere for each fixed parameter pair.

Fix $\psi$ with $C_{\psi }>0$. For $d>0$ and $E>0$, the log likelihood up to an additive constant is $\mathrm{\ell }(\lambda )=d\log \lambda -E\lambda$. Its derivative is $d\mathrm{/}\lambda -E$, which is positive for $0<\lambda <d\mathrm{/}E$ and negative for $\lambda >d\mathrm{/}E$. By [F13] it is strictly increasing before $d\mathrm{/}E$ and strictly decreasing after it, so its unique global maximum is at $\widehat{\lambda }=d\mathrm{/}E$.

If $d=0$ and $E>0$, then the event factor is $e^{-\lambda E}$, strictly decreasing on $(0,\mathrm{\infty })$ by [F14], and tends to $1$ as $\lambda \downarrow 0$ by [F8]. The supremum is therefore not attained in the ordinary parameter space. If the likelihood alone is extended to include $\lambda =0$, it is attained there.

If $d=0$ and $E=0$, the event factor is identically $1$, so every $\lambda >0$ maximizes it. If $d>0$ and $E=0$, some event has $U_i=T_i=0$; but the exponential law has $P(T_i=0)=0$, so such observations form a probability-zero boundary set.

Steps 1.1, 1.2, 2.1, 3.1, and 4.1 prove the dominated likelihood factorization, and steps 5.1–5.3 establish all stated interior, boundary, and zero-exposure cases.