Cox score martingale representation and random-risk-mean replacement

Differentiating the finite product in Cox proportional-hazards model and partial likelihood gives $U_n({\beta }_0)={\sum }_i{\int }_0^{\tau }\{Z_i(t)-{\bar{Z}}_n(t,{\beta }_0)\}d{N}_i(t)\mathrm{.}$ By [F4], this pathwise integral can be decomposed as the L² stochastic integral and its compensator integral. Writing $d{N}_i=d{M}_i+Y_i{e}^{{\beta }_0^T{Z}_i}{\lambda }_{0}dt$, the compensator contribution is zero because ${\sum }_i{Y}_i(t)e^{{\beta }_0^T{Z}_i(t)}\{Z_i(t)-{\bar{Z}}_n(t,{\beta }_0)\}=0$ whenever the risk denominator is positive; outside the risk set every summand is zero.

Set $\mathcal{M}={\sum }_i{M}_i$ and $H_n(t)={\bar{Z}}_n(t,{\beta }_0)-\bar{z}(t)$. The process $H_n$ is predictable by [F1]–[F3] and the deterministic measurability of $\bar{z}$. For bounded predictable step approximations, the finite-sum definition [L2] is linear in the integrator; the isometry [L4] and $A_{\cdot }={\sum }_i{A}_i$ pass that identity to the L² limits. Thus $n^{-1\mathrm{/}2}{U}_n({\beta }_0)=n^{-1\mathrm{/}2}{\sum }_i{\xi }_i-n^{-1\mathrm{/}2}{\int }_0^{\tau }{H}_n(t)d\mathcal{M}(t)$. No simultaneous failures make $N_{\cdot }={\sum }_i{N}_i$ a simple counting process, and the sum of the subject contracts is a square-integrable martingale contract with compensator $A_{\cdot }={\sum }_i{A}_i$.

The predictable aggregate integrand $H_n$ is bounded by $2K$. Its compensator energy satisfies $n^{-1}E{\int }_0^{\tau }\mathrm{\parallel }{H}_n(t){\mathrm{\parallel }}_2^{2}d{A}_{\cdot }(t)=E{\int }_0^{\tau }\mathrm{\parallel }{H}_n(t){\mathrm{\parallel }}_2^2{S}_n^{(0)}(t,{\beta }_0){\lambda }_0(t)dt\le e^{K\mathrm{\parallel }{\beta }_0{\mathrm{\parallel }}_2}E{\int }_0^{\tau }\mathrm{\parallel }{H}_n(t){\mathrm{\parallel }}_2^2{\lambda }_0(t)dt⟶0$ by [F2]'s integrated squared-error conclusion, without a time supremum. Thus [L4] shows the aggregate correction in step 2.1 tends to zero in L² after division by $\sqrt{n}$, and therefore is $o_P(1)$.

By [F4], ${\xi }_i$ equals the event sum minus the compensator Lebesgue integral with integrand $Z_i-\bar{z}$. Both are measurable functions of the $i$th subject record and the deterministic function $\bar{z}$, so the ${\xi }_i$ are IID. The integral isometry gives mean zero. Applying its scalar form to each coordinate and polarizing the identity gives the cross-coordinate isometry, hence $E[{\xi }_i{\xi }_i^T]={\int }_0^{\tau }E[Y_i(t)e^{{\beta }_0^T{Z}_i(t)}\{Z_i(t)-\bar{z}(t){\}}^{\otimes 2}]{\lambda }_0(t)dt=I$. The equality follows by expanding the outer product and using $\bar{z}=s^{(1)}\mathrm{/}{s}^{(0)}$. Independence and centering give $E\mathrm{\parallel }{n}^{-1\mathrm{/}2}{\sum }_i{\xi }_i{\mathrm{\parallel }}_2^2=\mathrm{tr}(I)$. For $M>0$, nonnegativity gives $M^{2}P(\mathrm{\parallel }{n}^{-1\mathrm{/}2}{\sum }_i{\xi }_i{\mathrm{\parallel }}_2>M)\le E\mathrm{\parallel }{n}^{-1\mathrm{/}2}{\sum }_i{\xi }_i{\mathrm{\parallel }}_2^2=\mathrm{tr}(I)$, so the normalized sum is $O_P(1)$ by [L5]. Combining this with step 3.1 proves the representation and the stated moment properties.