Consistency of the weighted logrank variance estimate

Since $d_j=1$, the variance formula [F2] reduces to $V_j=Y_{1,n}(t_j)Y_{0,n}(t_j)\mathrm{/}{Y}_n(t_j{)}^2=p_n(t_j)(1-p_n(t_j))$ when $Y_n(t_j)>1$. If $Y_n(t_j)=1$, both groups' product is zero and [F1] assigns variance zero. Thus $\frac{{\widehat{V}}_{n,w_n}}{n}=\frac{1}{n}{\int }_0^{\tau }{w}_n(t{)}^2{p}_n(t)(1-p_n(t))d{N}_n(t)\mathrm{.}$

Uniform convergence of $w_n$ to $w$ and of $p_n$ to $\bar{z}$, together with their uniform bounds, gives ${\sup }_{t\le \tau }\mathrm{\mid }{w}_n(t{)}^2{p}_n(t)(1-p_n(t))-w(t{)}^2\bar{z}(t)(1-\bar{z}(t))\mathrm{\mid }{\to }_{P}0$. Because $N_n(\tau )\le n$, replacing the integrand in step 1.1 by $h(t):=w(t{)}^2\bar{z}(t)(1-\bar{z}(t))$ changes the normalized integral by at most this supremum, hence by $o_P(1)$.

The aggregate process $N_n$ is simple because no two subjects jump simultaneously, and $A_n(\tau )\le n\mathrm{\Lambda }(\tau )$. The deterministic $h$ is bounded and predictable, so [F4]-[F6] give $\mathbb{E}\left[{\{\frac{1}{n}{\int }_0^{\tau }h(t)d{M}_n(t)\}}^2\right]=\frac{1}{n^2}\mathbb{E}{\int }_0^{\tau }h(t{)}^{2}d{A}_n(t)\le \frac{\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}^2\mathrm{\Lambda }(\tau )}{n}⟶0.$ Hence the martingale part of the normalized variance sum tends to zero in probability.

The compensator part of that sum is $\frac{1}{n}{\int }_0^{\tau }h(t)d{A}_n(t)={\int }_0^{\tau }h(t)\frac{Y_n(t)}{n}\lambda (t)dt\mathrm{.}$ Uniform convergence of $Y_n\mathrm{/}n$ to $y$ and ${\int }_0^{\tau }\lambda (t)dt<\mathrm{\infty }$ make this converge in probability to ${\int }_0^{\tau }h(t)y(t)\lambda (t)dt$. Since $\bar{z}=y_1\mathrm{/}y$, the limit is ${\int }_0^{\tau }w(t{)}^2{y}_0(t)y_1(t)y(t{)}^{-1}\lambda (t)dt={\sigma }_w^2$. Combining this with steps 2.1 and 3.1 proves the assertion.