1
$\begingroup$

The linked sites (link1, link2) demonstrate that the likelihood ratio tests and the corresponding one- and two-sample t-tests are equivalent. However, based on my understanding, the null distribution of the likelihood ratio test (i.e., chi-square distribution) is an approximation, whereas the t-distribution in the t-tests is not. Despite this difference, why are they equivalent?

$\endgroup$

1 Answer 1

1
$\begingroup$

You need to distinguish two different things:

  1. With a likelihood ratio statistic, $\Lambda$, the chi-squared distribution is an asymptotic approximation to the distribution of $-2\log(\Lambda)$ under quite broad conditions.

  2. Under the conditions where some t-test is exact, the corresponding likelihood ratio statistic might be shown to be monotonic in $|t|$ or $t^2$ (whichever is convenient), and therefore the two statistics should lead to equivalent tests, if the correct null distribution for that statistic is used; it won't be distributed as chi-squared in small samples.

Note in particular, that the square of a $t$-distributed random variable has an $F$ distribution, and that asymptotically as the denominator d.f. goes to infinity, the $F$ distribution goes to a scaled chi-squared. You may be able to derive an equivalent statistic to $-2\log(\Lambda)$ that has a small-sample $F$ distribution and in the limit as $n$ goes to infinity that $F$ will be proportional to a chi-squared with the same d.f. as the numerator d.f. of the F statistic (and indeed the same d.f. as the $t$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.