Nelson–Aalen, Kaplan–Meier, Greenwood

Let $D_{ij}=\mathbf{1}\{U_i=a_j,{\delta }_i=1\}$. By [F1], $Y_i(a_j)=\mathbf{1}\{T_i\ge a_j,C_i\ge a_j\}$ and $D_{ij}=\mathbf{1}\{T_i=a_j,C_i\ge a_j\}$. Independence in [A1] therefore gives $E[Y_i(a_j)]=r_j=P(T\ge a_j)P(C\ge a_j)$ and $E[D_{ij}]=P(T=a_j)P(C\ge a_j)$.

The two indicators are bounded IID sequences by [F2]. For either sequence, the variance of its sample mean is at most $1\mathrm{/}(4n)$, so Chebyshev's inequality gives convergence in probability to its expectation. Since there are finitely many support points, these convergences hold jointly over $j=1,\dots ,m$.

Since $r_j>0$, with probability tending to one $Y_n(a_j)\mathrm{/}n>r_j\mathrm{/}2$. On that event the ratio is the quotient of the two convergent proportions from step 2.1, and its limit is $\frac{P(T=a_j)P(C\ge a_j)}{P(T\ge a_j)P(C\ge a_j)}=P(T=a_j\mid T\ge a_j)=q_j$. The zero-denominator event has probability tending to zero by the same risk-proportion convergence, so the stated zero convention does not change the limit.

Let $k$ be the number of event support points not exceeding $t$. If $k=0$, then $a_1>t\ge 0$, so $a_1>0$ and $T>t$ almost surely; the empty product and sum give the claimed limits. If $k\ge 1$, positivity of $S(t)$ and $P(C\ge t)$ implies $P(T\ge a_j)P(C\ge a_j)>0$ for every $j\le k$, so [F3] applies at each of these finitely many points.

Before the first support point, the product is $1=P(T\ge a_1)$. After the product limit through $a_{j-1}$ converges to $P(T>a_{j-1})$, this equals $P(T\ge a_j)$ because there is no support in $(a_{j-1},a_j)$. By [F1]–[F2], at $a_j$ the product is multiplied by $1-d{N}_n(a_j)\mathrm{/}{Y}_n(a_j)$ and the sum is incremented by $d{N}_n(a_j)\mathrm{/}{Y}_n(a_j)$. [F3] and continuity of finite products and sums give limits $P(T\ge a_j)(1-q_j)=P(T>a_j)=S(a_j)$ and ${\sum }_{\mathrm{\ell }\le j}{q}_{\mathrm{\ell }}$. This also handles $a_1=0$, since $P(T\ge 0)=1$ and the baseline failure increment is included.

Finite induction over $j=1,\dots ,k$ proves both limits at every support point through $t$. On each interval between successive support points, the estimators and their targets are constant, so the same limits hold at the given $t$. The Nelson–Aalen limit is the discrete hazard sum displayed in the statement; the proof does not replace it by a continuous cumulative hazard.

Put $V_i=-U_i$, let $F(x)=P(V_i\le x)$, and let $F_n(x)=n^{-1}{\sum }_{i=1}^n\mathbf{1}\{V_i\le x\}$. Then ${\bar{Y}}_n(u)=F_n(-u)$ and $s_0(u)=F(-u)$. Both CDFs are right-continuous, so their supremum error over $\mathbb{R}$ is the supremum over rational $x$; over $[-t,0]$ it is the supremum over $(\mathbb{Q}\cap [-t,0))\cup \{0\}$ (with $\{0\}$ alone when $t=0$). Thus both suprema are measurable. Fix $0<\epsilon <1$. Choose the finitely many distinct quantiles $x_k=\inf \{x:F(x)\ge k\epsilon \}$ for $1\le k<1\mathrm{/}\epsilon$, together with $x_0=-\mathrm{\infty }$ and $x_{m+1}=+\mathrm{\infty }$, where $m$ is the number of distinct finite cutpoints. Between consecutive distinct quantiles $x_j<x_{j+1}$, the quantile construction gives $F(x_{j+1}-)-F(x_j)\le \epsilon$; any jump at a cutpoint is isolated as its singleton point. At each finite cutpoint, both $F_n(x_j)$ and $F_n(x_j-)$ converge in probability to their expectations by Chebyshev's inequality applied to the IID indicators $\mathbf{1}\{V_i\le x_j\}$ and $\mathbf{1}\{V_i<x_j\}$. For $x\in (x_j,x_{j+1})$, monotonicity gives $F_n(x_j)\le F_n(x)\le F_n(x_{j+1}-)$ and $F(x_j)\le F(x)\le F(x_{j+1}-)$; hence $\mathrm{\mid }{F}_n(x)-F(x)\mathrm{\mid }$ is at most the largest one-sided endpoint error plus $\epsilon$. At singleton cutpoints, direct endpoint convergence applies. Taking the maximum over the finite partition, then $n\to \mathrm{\infty }$ and $\epsilon \downarrow 0$, proves the first convergence.

Let $Z_n={\sup }_{0\le u\le t}\mathrm{\mid }{\bar{Y}}_n(u)-s_0(u)\mathrm{\mid }$. Since $Z_n\le 1$, for every $\eta >0$, $E[Z_n^2]\le {\eta }^2+P(Z_n>\eta )$. Step 1.1 makes the second term tend to zero; then letting $\eta \downarrow 0$ proves the expected-square convergence.

Independence within a subject gives $s_0(u)=P(T_i\ge u,C_i\ge u)=S(u)G(u)$. On $[0,t]$, $s_0(u)\ge c:=S(t)G(t)>0$. Since $S$ is absolutely continuous and $S^{\mathrm{\prime }}(u)=-f(u)=-\lambda (u)S(u)$ almost everywhere, integration of $(\log S{)}^{\mathrm{\prime }}=-\lambda$ gives $\mathrm{\Lambda }(t)=-\log S(t)$. The assumed integral makes ${\sigma }^2(t)$ finite and also gives ${\int }_0^t\lambda (u)du<\mathrm{\infty }$.

For each finite $n$, event times have no ties in $(0,t]$ almost surely: for $i\mathrm{\ne }j$, subject independence and the absolutely continuous law on $[0,t]$ give $P(T_i=T_j\le t)=0$, and there are finitely many pairs. Thus $N_n={\sum }_i{N}_i$ is a simple counting process. Set $Y_n={\sum }_i{Y}_i$, $A_n={\sum }_i{A}_i={\int }_0^{\cdot }{Y}_n(u)\lambda (u)du$, and $M_n={\sum }_i{M}_i=N_n-A_n$. Finite sums preserve the martingale property and square integrability; [F4] gives $E[A_n(t)]={\sum }_{i}E[A_i(t)]<\mathrm{\infty }$. The process $J_n=\mathbf{1}\{Y_n>0\}$ is predictable, and so is ${\eta }_n=Y_n\lambda$ by [F11]; both meet [F4]'s domain, with $A_n={\int }_0^{\cdot }{J}_n(u){\eta }_n(u)du$. Hence the pooled process also satisfies the finite-horizon contract. By [F5], its predictable bracket is $A_n$.

We will identify the completed integrals with pathwise finite-variation integrals for every bounded predictable integrand $H$ used below. The paths of $Y_i$ and ${\bar{Y}}_n$ are left-continuous on $(0,t]$, and $s_0(u)=P(U_i\ge u)$ is left-continuous there by continuity of probabilities along decreasing events. Consequently $s_0^{-1}$, the floored reciprocal $b_n$, its difference $H_n$, and the floored product reciprocal $q_n$ below all have bounded left-continuous paths, with a deterministic bound for each fixed $n$. On deterministic partitions with mesh tending to zero, take the left-step approximants $H^{(m)}(u)=H(t_j)$ on $(t_j,t_{j+1}]$ and set their time-zero value to zero. They are elementary predictable integrands, converge pointwise to $H$ at every $u>0$, and retain the bound. For any subject or pooled contract, [F13] for the finite compensator measure $EdA$ gives $E{\int }_0^t\mathrm{\mid }{H}^{(m)}-H{\mathrm{\mid }}^{2}dA\to 0$, and [F7] gives convergence of their elementary integrals to the completed integral in $L^2(P)$. The paths of $M=N-A$ have finite variation; [F13] applied pathwise to its finite variation measure gives convergence of the same elementary sums to ${\int }_{(0,t]}HdN-{\int }_0^{t}HdA$. Uniqueness of limits in probability identifies these two limits almost surely at the fixed time. The time-zero convention is immaterial because $N_0=A_0=0$ and $A$ is continuous. This proves the pathwise identities needed below without assuming a general semimartingale integration theorem.

Write ${\bar{Y}}_n=Y_n\mathrm{/}n$. By [F8], $E[{\sup }_{u\le t}\mathrm{\mid }{\bar{Y}}_n(u)-s_0(u){\mathrm{\mid }}^2]\to 0$. Let $b_n(u)=1\mathrm{/}\max ({\bar{Y}}_n(u),c\mathrm{/}2)$ and $H_n(u)=b_n(u)-1\mathrm{/}{s}_0(u)$. These are bounded predictable processes: ${\bar{Y}}_n$ is predictable by [F4], and Borel transformations preserve predictability by [F11]. The reciprocal map with floor $c\mathrm{/}2$ is $4\mathrm{/}{c}^2$-Lipschitz, so $E[{\sup }_{u\le t}\mathrm{\mid }{H}_n(u){\mathrm{\mid }}^2]\to 0$.

The energy of $H_n$ for $M_n$ is finite because $H_n$ is bounded and $E[A_n(t)]<\mathrm{\infty }$. Applying [F7] to the pooled contract gives $E\mathrm{\mid }{n}^{-1\mathrm{/}2}{\int }_0^t{H}_{n}d{M}_n{\mathrm{\mid }}^2=n^{-1}E{\int }_0^t{H}_n^{2}d{A}_n=E{\int }_0^t{H}_n^2{\bar{Y}}_n\lambda du\le ({\int }_0^t\lambda du)E[{\sup }_{u\le t}\mathrm{\mid }{H}_n(u){\mathrm{\mid }}^2]\to 0$. Thus this stochastic-integral remainder is $o_P(1)$. On the event $E_n=\{{\sup }_{u\le t}\mathrm{\mid }{\bar{Y}}_n(u)-s_0(u)\mathrm{\mid }<c\mathrm{/}2\}$, whose probability tends to one by [F8], $Y_n(u)>0$ throughout and $b_n=1\mathrm{/}{\bar{Y}}_n$. By the pathwise identification in step 2.1, $d{\bar{N}}_n={\bar{Y}}_n\lambda du+d{\bar{M}}_n$ gives there $\sqrt{n}({\widehat{\mathrm{\Lambda }}}_n-\mathrm{\Lambda })=n^{-1\mathrm{/}2}{\int }_0^t{b}_{n}d{M}_n$. For each bounded predictable step integrand, the finite-sum definition in [F6] makes the integral against $M_n={\sum }_i{M}_i$ equal to the sum of the subject integrals. Approximate $s_0^{-1}$ in the pooled compensator energy by such steps; [F7] gives L² convergence, and $d{A}_i\le d{A}_n$ lets the same approximants converge for each of the finitely many subjects. Since $s_0^{-1}$ is a deterministic Borel time function, it is predictable by [F11]. Passing the finite-sum identity to these L² limits yields $\sqrt{n}({\widehat{\mathrm{\Lambda }}}_n-\mathrm{\Lambda })=n^{-1\mathrm{/}2}{\sum }_{i=1}^n{\int }_0^t{s}_0^{-1}d{M}_i+o_P(1)$.

For one subject, step 2.1 identifies the integral in step 3.1 as ${\varphi }_i^{\mathrm{\Lambda }}(t)=\mathbf{1}\{{\delta }_i=1,U_i\le t\}\mathrm{/}{s}_0(U_i)-{\int }_0^{U_i\wedge t}\lambda (u)\mathrm{/}{s}_0(u)du$, so it is a measurable function of $(T_i,C_i)$. It has mean zero by the martingale property and, by [F7] and Tonelli [F12], variance $E{\int }_0^t{s}_0(u{)}^{-2}d{A}_i(u)={\int }_0^t{s}_0(u{)}^{-2}E[Y_i(u)]\lambda (u)du={\int }_0^t\lambda (u)\mathrm{/}{s}_0(u)du={\sigma }^2(t)$. The integrand is nonnegative and product-measurable because $Y_i$ is predictable and $s_0^{-2}\lambda$ is a deterministic Borel function. Hence these influence variables are IID and square-integrable. The normalized IID sum is bounded in probability by its fixed second moment, so step 3.1 gives a separately established expansion with $o_P(n^{-1\mathrm{/}2})$ remainder. Applying [F9] proves $\sqrt{n}({\widehat{\mathrm{\Lambda }}}_n-\mathrm{\Lambda })\Rightarrow N(0,{\sigma }^2(t))$.

On $E_n$, for all sufficiently large $n$, every event jump has size $x=1\mathrm{/}{Y}_n(u)\le 2\mathrm{/}(nc)\le 1\mathrm{/}2$. For $0\le x\le 1\mathrm{/}2$, the power series for $-\log (1-x)-x$ gives $\mathrm{\mid }\log (1-x)+x\mathrm{\mid }\le x^2$. There are at most $n$ event jumps, so $\mathrm{\mid }\log {\widehat{S}}_n(t)+{\widehat{\mathrm{\Lambda }}}_n(t)\mathrm{\mid }\le 4\mathrm{/}(n{c}^2)$. Consequently $\sqrt{n}\{{\widehat{S}}_n-e^{-{\widehat{\mathrm{\Lambda }}}_n}\}{\to }_{P}0$. Step 4.1 also gives ${\widehat{\mathrm{\Lambda }}}_n-\mathrm{\Lambda }=O_P(n^{-1\mathrm{/}2})$; Taylor's formula for $e^{-x}$ on $[0,\mathrm{\infty })$ then yields $\sqrt{n}({\widehat{S}}_n-S)=-S(t)n^{-1\mathrm{/}2}{\sum }_i{\varphi }_i^{\mathrm{\Lambda }}(t)+o_P(1)$. Taking $g(x)=e^{-x}$ and derivative $g^{\mathrm{\prime }}(\mathrm{\Lambda })=-S(t)$ in [F10] transfers the terminal normal limit, giving $N(0,S(t{)}^2{\sigma }^2(t))$.

For sufficiently large $n$, define the bounded predictable $q_n(u)=[\max ({\bar{Y}}_n(u),c\mathrm{/}2)\max ({\bar{Y}}_n(u)-1\mathrm{/}n,c\mathrm{/}4){]}^{-1}$. On $E_n$, $\mathrm{\Delta }{N}_n(u)=1$ at event times and $q_n(u)=1\mathrm{/}[{\bar{Y}}_n(u)({\bar{Y}}_n(u)-1\mathrm{/}n)]$, so $n{\sum }_{u\le t}\mathrm{\Delta }{N}_n(u)\mathrm{/}[Y_n(u)(Y_n(u)-\mathrm{\Delta }{N}_n(u))]={\int }_0^t{q}_{n}d{\bar{N}}_n$. The uniform convergence in [F8] implies $q_n{\bar{Y}}_n\to 1\mathrm{/}{s}_0$ uniformly in probability, while $q_n$ is uniformly bounded. Using step 2.1, decompose $\int q_{n}d{\bar{N}}_n=\int q_n{\bar{Y}}_n\lambda du+n^{-1}\int q_{n}d{M}_n$. The first term tends to ${\sigma }^2(t)$ by boundedness and $\int \lambda <\mathrm{\infty }$; by [F3]–[F4] the second has second moment at most $n^{-1}(8\mathrm{/}{c}^2{)}^{2}E[A_1(t)]$, which tends to zero. Therefore the scaled Greenwood sum tends to ${\sigma }^2(t)$, and ${\widehat{S}}_n(t{)}^2{\to }_{P}S(t{)}^2$ by step 5.1. Since $P(E_n)\to 1$ and Greenwood's formula is defined on $E_n$ for large $n$, the zero extension agrees with the formula on an event whose probability tends to one. It follows that $n{\widehat{V}}_G^{\circ }(t){\to }_{P}S(t{)}^2{\sigma }^2(t)$, which equals $\mathrm{Var}({\varphi }_i^S(t))$ by step 4.1. This is a first-order variance estimate, not a finite-sample identity.