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I'm wondering about the Wald test when applied to regression coefficients in the Cox PH model.

In linear regression, you have to estimate $\sigma^2$ separately from the mean, which means the standard error for the coefficients is based on an estimate, leading to using t-scores to test the coefficients rather than Z-scores (i.e. the standard error for $\hat{\beta}$ is $\sqrt{s^2(X^TX)^{-1}_{jj}}$ instead of $\sqrt{\sigma^2(X^TX)^{-1}_{jj}}$).

In the Cox case, Z-scores are shown in the R output, because Wald tests are done on the coefficients. The Wald test assumes the coefficient is normally distributed: $$\frac{\hat{\beta}}{se(\hat{\beta})}\sim N(0,1)$$

but why doesn't it follow a t-distribution, since the standard error is estimated? I realize the standard errors are not computed the same way as in linear regression (although I'm not 100% clear on that), but it just seems like a t-distribution should be used. I must be missing a property of the Wald test. On Wikipedia it says "The square root of the single-restriction Wald statistic can be understood as a (pseudo) t-ratio that is, however, not actually t-distributed except for the special case of linear regression with normally distributed errors. In general, it follows an asymptotic z distribution", but I don't really understand what that means. Any help is appreciated!

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Asymptotic theory is an important basis for Cox models and other types of models fit by maximum (partial) likelihood. The tests are based on the behavior of statistics as the sample size becomes increasingly large. In that limit of very large sample size, the normal distribution of coefficient estimates holds. At finite sample sizes there's no assurance that a t distribution would hold, however, unlike the situation with sampling from a normal distribution. So the tests are based on the asymptotic normality.

With small sample sizes, likelihood ratio tests are typically more reliable than Wald tests, but they require refitting the model over a range of coefficient values. A way to proceed is outlined in this answer. I'm not sure whether there is any built-in way to do this for Cox models in R, but I recall that SAS can do this directly.

This page discusses related matters in the context of logistic regression, which fits models similarly.

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  • $\begingroup$ If you found this answer helpful, then please consider upvoting and/or accepting it. $\endgroup$ Apr 25, 2022 at 13:57
  • $\begingroup$ @EdM thank you for the answer, but couldn't you say that the t-test in linear regression is also based on large sample size? The confidence interval becomes narrower at a larger n. I guess that the difference is that the finite sample t-distribution is known, but the finite sample Wald statistic distribution is not? Yet both can be used for a small sample hypothesis test. So is it just that when you use likelihood to estimate, you need a different method for the CI (e.g. Wald)? $\endgroup$
    – fmtcs
    Apr 25, 2022 at 16:08
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    $\begingroup$ @fmtcs unlike the Cox Wald test, the t-test in linear regression is based on the exact theoretical distribution of the mean value for a sample of specified size taken from a normal distribution, when both the mean and the standard deviation are determined from the data. See the Wikipedia page for a brief introduction, or the derivation in an introductory mathematical statistics text. Confidence intervals get smaller with increasing $n$ because the t-test is based on the standard error of the mean, which decreases with $\sqrt n$. $\endgroup$
    – EdM
    Apr 25, 2022 at 16:50

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