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I've seen it claimed (e.g. in these notes ) that "Schoenfeld residuals are, in principle, independent of time."

Can this be right? Consider the following situation:

  • You are using a Cox model to predict the hazard ratio of death, with sex as a categorical covariate.

  • The hazard for men is higher than the hazard for women.

  • Therefore, the proportion of men in the sample decreases with time.

  • Therefore, the expected value of the "male" dummy variable (call it $X$) decreases with time.

  • The Schoenfeld residuals are distributed like $X - E(X)$. But this isn't independent of time! For instance, the Schoenfeld residual is greater than 0 only if $X = 1$, but $p(X = 1)$ is not independent of time.

I understand that the expected value (over data $X$) of the Schoenfeld residual is independent of time (indeed, it is always 0 if the PH assumption holds). But the above case seems like a counterexample to the stronger assertion that the entire distribution is independent of time. Am I missing something, or is the claim incorrect?

Ben Kuhn
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1 Answers1

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I'll start out by defining the Schoenfeld residuals. Let $D$ be the set of individuals that fail before censoring, and let $R_i$ be the set of individuals at risk at the time of the failure of individual $i$, $i\in D$. The Schoenfeld residual for the $i$'th invidual, $i \in D$, and the $j$'th covariate is defined as

$$ \hat{r}_{ij} = X_{ij} - \sum_{k\in R_i} X_{kj}\exp(\hat{\beta}'X_k)/\sum_{k\in R_i}\exp(\hat{\beta}'X_k) $$

Let's take a look at the quote "Schoenfeld residuals are, in principle, independent of time". Could the author mean, that the Schoenfeld residuals have the same distribution at all times $t$? Therneau and Grambsch (Modeling Survival Data, Springer-Verlag, 2000, p. 128) state that a plot of the cumulative Schoenfeld residuals can be compared to a Brownian bridge. It's a bit vague what exactly that means. Wei (Testing goodness-of-fit for propotional hazards model with censored observations, J. Am. Statist. Assoc. 79, 649-52, 1994) states that if we only consider a dichotomous covariate then the standardized score process converges in probability to a Brownian bridge. This is useful as the increments of the score process are the Schoenfeld residuals. When ${W_t}$ is a Brownian motion, $$ B_t = W_t - tW_1, \ t\in [0,1] $$

is a Brownian bridge. An increment of the Brownian bridge from time $s$ to time $t$ is a Gaussian random variable and using the above representation one can derive the mean and variance, none of which depend on the specific time points, only on $t - s$. Thus, asymptotically the residuals do have the same distribution in some sense.

Finally, the author might have meant something else, something that Schoenfeld himself mentions in the original article (Schoenfeld, Partial residuals for the proportional hazards regression model, Biometrika 69, 239-41), if I understand him correctly. He states that "The present paper defines residuals which do not depend on time so that the $i$'th residual can be plotted against $t_i$ to test the proportional hazards assumption". However, when he defines the residuals, he's conditioning on the risk sets. He also finds an asymptotic distribution of the residuals, but again conditional on the risk sets. This means that the residuals actually have discrete distributions, as each residuals only can take on as many values as there are individuals in its risk set. In this case, it makes sense to talk of them as being independent of time, simply because we've conditioned on the risk sets, thus there is nothing temporal left, only a distribution of "who-dies-at-time-i".

swmo
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  • I'm sorry for forgetting about this answer! It seems like there are a number of looser interpretations of "independent of time" that could work here ("the *cumulative* Schoenfeld residuals are *asymptotically* independent of time" or "the Schoenfeld residuals are independent of time *conditional on the risk set*). Too bad the author of the notes wasn't more precise... – Ben Kuhn Apr 16 '15 at 20:47