The two tests below seem to be the same to me. But when I tried an example and constructed the test statistics I couldn't see how they are the same.

$H_0: \lambda=1~~~~H_a: \lambda\not=1$

$H_0: \ln(\lambda)=0~~~~H_a: \ln(\lambda)\not=0$

Use exponential distribution with form $\lambda$=mean

Then the MLE of the mean is sample average $\bar X$, with asymptotic variance $\lambda^2$

Consider the transformation $\ln(\lambda)$, then we get the MLE is $\ln\bar X$ with asymptotic variance 1.

So the test statistics should be:

$\sqrt{n}\frac{\bar X-1}{\bar X}\to Z$

$\sqrt{n}\ln(\bar X)\to Z$

So did I make an error (I asked somebody else and they said it looked fine) or do these tests just converge to the same thing while looking different the whole time?

Edit: I should be using Wald test here instead I think.

  • $\begingroup$ "... variance $\bar{X}^2/n$" ?? $\endgroup$ Jul 28, 2016 at 20:07
  • $\begingroup$ @Dilip Ya that was a type sorry. $\endgroup$
    – VCG
    Jul 28, 2016 at 20:12

1 Answer 1


From Wooldridge's cross section book, there is a problem where a non-linear transformation changes the hypothesis outcome, so the two are not equivalent.


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