I'm currently reading All of Statistics by Larry Wasserman. The question is about the Delta Method, page 160 in the pdf (I have a hard copy, there it's page 133). The authors defines $\hat{\theta}$ or $\hat{\theta}_n$ as notation for the point estimator of $\theta$. In the Delta Method theorem he states (I quote):

If $\tau = g(\theta)$" where $g$ is differential and $g(\theta)\neq0 > $ then $$\frac{\hat{\tau}_n-\tau}{\hat{\operatorname{se}}(\hat{\tau})} \to > N(0,1) $$ in distribution, where $\hat{\tau}_n = g(\hat{\theta}_n)$ and $\hat{\operatorname{se}}(\hat{\tau}_n)=|g'(\hat{\theta})|\hat{\operatorname{se}}(\hat{\theta}_n)$.

Does the author really mean $|g'(\hat{\theta})| = |g'(\hat{\theta}_n)|$ etc? Why is he even mixing these up in the same expression? Is there any reason to do so?

  • 1
    $\begingroup$ Did you have the curiosity to check what is the delta method entry on Wikipedia? We could have then maybe escaped this question. $\endgroup$
    – Xi'an
    Oct 2, 2015 at 20:04
  • 2
    $\begingroup$ @Xi'an I suspect the source of confusion is the use of the estimate instead of the population parameter which is the method of Wikipedia. $\endgroup$
    – JohnK
    Oct 2, 2015 at 20:06
  • $\begingroup$ @JohnK yes its not about the theorem, just the authors notation. The book was suggested to me. $\endgroup$
    – math
    Oct 2, 2015 at 20:09

2 Answers 2


I don't think it is a typo. It seems to me that the author uses the notation $\widehat{\theta}_n$ and $\widehat{\theta}$ interchangeably. The former just emphasizes the asymptotics. Recall that the mle is consistent and thus if $g(\theta)$ is continuous, by the CMT the standard error is estimated consistently.


I think that it's a typo. Also, strictly speaking, the standard error is in asymtotic sense and exact standard error is not equal to that value.

  • $\begingroup$ thanks for you anwser. Where do you see the "exact standard error" in his theorem? I just see approximated ones. $\endgroup$
    – math
    Oct 2, 2015 at 19:48
  • $\begingroup$ I am sypaying that the approximated one is asymptotic result. $\endgroup$
    – user13148
    Oct 2, 2015 at 19:49

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