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It seems to me that they are basically calculate the same thing. Since we already have t-test, why do we need a squared version (wald test)? Does wald test have its own advantage? For example in Cox model, why can't we just use t test?

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There are similarities between a Wald test and a t-test, as this page describes.* They both are used to determine whether a coefficient value is significantly different from 0 (or from any value specified in a null hypothesis). They typically arise in different contexts that show why we need both.

Consider the situation for testing whether a coefficient is significantly different from 0 in two scenarios: one for the mean value of a set of numbers examined with a t-test, and another for a coefficient in a Cox regression examined with a Wald test. In both cases you use the ratio of the observed difference from 0 to a measure of the error in that estimate. The error term, however, is determined in different ways.

For testing whether a mean value of a set of numbers is different from 0, the t-test takes into account the fact that you are estimating both the mean value and the variance from a particular number of observations. It assumes that the underlying population is normally distributed and uses the properties of normal distributions to calculate the error estimate. This gives an exact answer, under that assumption, for the probability that the observed difference of the mean value from 0 could have arisen by chance if the true population mean were 0.

The error estimate for coefficients determined by maximizing likelihoods (partial likelihoods with Cox regressions) are determined in a different way. This answer describes how the matrix of second derivatives of the log-likelihood, calculated at the maximum likelihood, can be transformed into an estimate of the variance-covariance matrix for all coefficients in a multiple regression setting.

For the case of a single coefficient that matrix is a single value, the variance of the coefficient estimate. The Wald test for a single coefficient assumes that the coefficient estimate is normally distributed. That's different from the assumptions for the t-test; in this case you aren't sampling a specified number of times from a normal distribution but rather using your estimate of the variance directly. Note that with small sample sizes that assumption of normally distributed coefficient estimates might not hold.

This page shows how the t-test and the Wald test become equivalent as sample size increases.


*The form of the Wald statistic that you have in mind, for testing against a chi-square distribution, is the square of that presented on the linked page.

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  • $\begingroup$ Thanks for the detailed answer. As you explained that I got coefficient estimate in a different way in Cox model. I still don't understand why I can't use t-statistics even though I "using I estimate of the variance directly". Which part of the t-statistics does this "different way getting variance" violated so that I can't use t-test? $\endgroup$ – unicorn May 31 at 2:34
  • $\begingroup$ @unicorn the theory behind the Wald test for maximum likelihood estimates is a model in which the coefficient estimate is normally distributed, not t-distributed. You must compare any observed statistic (here, the ratio of coefficient estimate to standard error of its estimate) against the distribution supported by the underlying theory. If the coefficient estimate was determined by maximum likelihood: a test based on a normal distribution. If the estimate comes from sampling a normal distribution a certain number of times: a t-distribution with corresponding degrees of freedom. $\endgroup$ – EdM May 31 at 16:37
  • $\begingroup$ If the difference is underlying assumption of normal distribution vs student's t-distribution, why can't I use z-test instead of Wald test? I did some further research. As far as I know, the reason we are using Wald test here is because I need to test more than one variable in multiple regression Cox model, right? $\endgroup$ – unicorn Jun 1 at 11:15
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    $\begingroup$ @unicorn For testing a single coefficient, the Z-test and the Wald test are equivalent; the square of a standard normally distributed variable is distributed as chi-square with 1 degree of freedom, as used in the corresponding Wald test. A thread linked in the answer thus essentially calls the Z-test a Wald test in that context. As you note the Wald test has the advantage of extending easily to a test of any linear combination of several coefficients simultaneously once you have an estimate of the variance-covariance matrix for the coefficient estimates. $\endgroup$ – EdM Jun 1 at 13:04
  • $\begingroup$ Got it. Thanks for your patient explanation. $\endgroup$ – unicorn Jun 2 at 3:57

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