# Wald test for Cox PH coefficients: Z distributed

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!

• @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 $\sqrt n$.