Efficient influence function in proportional hazards model

Question

I was hoping someone could help me with this problem in the cox proportional hazards model.

I am given the following setup.

T is a non-negative random variable with continous distribution and hazard function $\lambda_T(t)$. T has density $f_T(t) = \lambda_T(t) S(t)$ and $S(t) = P(T>t)$. Also $F(t) = P(T \leq t)$ is the distribution function.

If I have $n$ observations of $T$. Note no censoring is assumed. Can anyone tell me how I arrive at an $\textbf{efficient influence function}$ for $S(t_0)$ where $t_0$ is a fixed time point.

note $\sqrt{n} ( \hat{S(t_0)} - S(t_0) ) = \sum_{i=1}^{n} \phi(T_i)$ where we have $\phi(T_i)$ is the influence function. This leads to an efficient estimator of $\hat{S}$

Answer 1

Write $\mathbb{P}_n$ as the empirical expectation. The estimator $\hat\psi = \mathbb{P}_n I(T>t_0)$ satisfies that $$\sqrt{n}(\hat\psi - S(t_0)) = \sqrt{n}\mathbb{P}_n [I(T>t_0)-S(t_0)] + 0,$$ showing that $\hat\psi$ is asymptotically linear with influence function $I(T>t_0)-S(t_0)$. Since the model is nonparametric, this must be the efficient influence function.

Another, direct, way to find the efficient influence function is through the canonical gradient. Let $\epsilon$ parametrize a one dimensional parametric submodel. Then \begin{align*} \frac{\mathrm{d}}{\mathrm{d}\epsilon} \mathbb{E}_\epsilon[T>t_0] \mid_{\epsilon = 0} &= \frac{\mathrm{d}}{\mathrm{d}\epsilon} \int_{t_0}^\infty f_\epsilon(s)\,\mathrm{d}s \mid_{\epsilon = 0} \\ &= \int_{t_0}^\infty f_\epsilon(s) R_\epsilon(s) \,\mathrm{d}s \mid_{\epsilon = 0} \\ &= \mathbb{E}_\epsilon [I(T>t_0) R_\epsilon(T)] \mid_{\epsilon = 0} \\ &= \mathbb{E}_\epsilon [\left\{ I(T>t_0) - S_\epsilon(t_0) \right\} R_\epsilon(T)] \mid_{\epsilon = 0} \\ &= \mathbb{E}[\left\{ I(T>t_0) - S(t_0) \right\} R_0(T)], \end{align*} where $R_\epsilon$ is the score. Since $I(T>t_0) - S(t_0)$ is mean zero, it must be the canonical gradient and the efficient influence function.

Answer 2

I'm not much of an expert on influence functions; I'll start with the working definition provided in this answer by Michael Chernick: "The influence function for a parameter...essentially measures the difference between the parameter estimate when the data point is included compared with when it is left out."

In your case you want to know how removing particular event times from the observation set (maybe more precisely, making small changes in observed event times $T_i$) affect an estimate of survival at a particular time, $\hat S(t_0)$. In your situation with a non-parametric survival function estimate,* that might be the Kaplan-Meier estimate, or the survival function derived from the Nelson-Aalen estimate of cumulative hazard. So ask yourself the following questions:

If $T_i > t_0$, is $\hat S(t_0)$ affected if you omit observation $i$ or make a (small) change in its observed time?

If $T_i = t_0$ (an event perhaps of 0 probability in principle, but maybe of some practical interest), what happens to $\hat S(t_0)$ if you omit observation $i$ or make a (small) change in its observed time?

If $T_i < t_0$, what happens to $\hat S(t_0)$ if you omit observation $i$ or make a (small) change in its observation time?

The Wikipedia entry shows the derivation of the Kaplan-Meier estimate based on maximum likelihood, which might help put the above into a more formal argument.

Although you ask in the context of no censoring, also consider what happens to $\hat S(t_0)$ if there are small changes in censoring times that aren't close to $t_0$.

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*Although the question was originally posed in terms of a Cox regression, discussion in comments clarified that the question is about a non-parametric estimate of a single survival curve. A "semi-parametric" Cox regression makes no parametric assumptions about the baseline hazard, with parametric modeling of covariate effects on hazard. If the "influence function" is defined in terms of small changes in observed event times with unaltered covariate values, this type of argument can be extended to Cox models. In Cox models, however, the "influence" of interest is generally in how each of $n$ individual cases, with associated covariate values, affects estimates of each of the $p$ regression coefficients.

Answer 3

I'll use the Gateaux derivative approach to derive the efficient influence function of the survival function $S(t)$, assuming there is no censoring.

By definition, the estimand $\Phi(P) = S(t_0) = 1 - F(t_0) = 1 - \int_0^{t_0} f(t) dt$.

The efficient influence function (EIF) is \begin{align} \phi(\tilde{t}) &= \frac{d \Phi(P_s)}{d s} \vert_{s=0} \\ &= \frac{d}{d s} [1 - \int_0^{t_0} \{s \mathbb{1}_{\tilde{t}}(t) + (1 - s) f(t)\} dt] \\ &= -\int_0^{t_0} \mathbb{1}_{\tilde{t}}(t) dt + \int_0^{t_0} f(t) dt \\ &= [1 - \int_0^{t_0} \mathbb{1}_{\tilde{t}}(t) dt] - [1 - \int_0^{t_0} f(t) dt] \\ &= [1 - \mathbb{1}(\tilde{t} \le t_0)] - S(t_0) \\ &= \mathbb{1}(\tilde{t} > t_0) - S(t_0) \end{align}

WLOG (change the notation a bit), the EIF is $\phi(t) = \mathbb{1}(t > t_0) - S(t_0)$. Further, we can show that \begin{align} \mathrm{Var}[\phi(t)] &= \mathrm{Var}[\mathbb{1}(t > t_0) - S(t_0)] \\ &= \mathrm{E}[\{\mathbb{1}(t > t_0) - S(t_0)\}^2] - \{\mathrm{E}[\mathbb{1}(t > t_0) - S(t_0)]\}^2 \\ &= \mathrm{E}[\{\mathbb{1}(t > t_0) - S(t_0)\}^2] - 0 \\ &= \mathrm{E}[\mathbb{1}(t > t_0)^2] - 2\mathrm{E}[\mathbb{1}(t > t_0)]S(t_0) + S(t_0)^2 \\ &= S(t_0) - 2 S(t_0)^2 + S(t_0)^2 \\ &= S(t_0) [1 - S(t_0)] \end{align}

The resulting EIF is exactly what you should expect without censoring. Since it's famous that the EIF for CDF is $\mathbb{1}(t \le t_0) - F(t_0)$, which suggests that the empirical CDF estimator is efficient.

Reference for the Gateaux derivative approach: Hines, Oliver, et al. "Demystifying statistical learning based on efficient influence functions." The American Statistician 76.3 (2022): 292-304.