# Efficient influence function in proportional hazards model

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 $$\hat{S}$$

• is there a particular text book you are working from in regards to this? Apr 14, 2021 at 18:46
• what does $O_p$ here represent? Apr 14, 2021 at 18:51
• $O_p$ is a term converging to zero. Apr 14, 2021 at 18:54
• No not a particular book actually. Apr 14, 2021 at 18:55
• As this is a Cox regression model, are you asking about estimating the influence of removing one event time-point on the baseline survival function? Put another way, would the individual with the event at that time point still be included at prior times (effectively censored at what was really the event time)?
– EdM
Apr 14, 2021 at 21:54

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.

• How did you get the first line? Is this just an appeal to asymptotic properties of any empirical estimator that takes the form of the empirical expectation? May 10 at 14:53
• @Casey It's just rewriting the expression, since $\sqrt{n} \mathbb{P}_n S(t_0) = \sqrt{n} S(t_0) \mathbb{P}_n 1 = \sqrt{n} S(t_0)$.
– Ben
May 10 at 15:03
• Oh that's clever, is that usually how these arguments go? Or is there some remainder that gets pushed into the O() term? I'm trying to use this great example to gain more intuition behind the general "algorithm" for finding influence curves May 10 at 15:15

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 $$t_0$$.

*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.