In All of Statistics, chapter 11 (pg. 183), Larry Wasserman states in his description of the Wald Test:

We are testing the null hypothesis $ \hat{\theta} = \theta_0 $ versus the alternative hypothesis $ \theta \neq \theta_0 $.

He then says we should assume that $ \hat{\theta} $ is asymptotically normal, i.e. converges in distribution to a standard normal: $$ \frac{\sqrt{n}(\hat{\theta} - \theta_0)}{\hat{\text{se}}} \rightsquigarrow N(0, 1). $$

(Wasserman uses $ \rightsquigarrow $ to denote converging in distribution.)

My impression is that this is a reasonable assumption because of the Central Limit Theorem.

But then, when proving the following theorem:

Asymptotically the Wald test has size $ \alpha $, that is, $$ \mathbb{P}_{\theta_0}\left(\lvert Z \rvert > z_{\alpha/2}\right) \rightarrow \alpha $$ as $ n \rightarrow \infty $,

He says:

Under $ \theta = \theta_0 $, $ (\hat{\theta} - \theta_0)/\hat{se} \rightsquigarrow N(0, 1). $

I don't see how $ (\hat{\theta} - \theta_0)/\hat{se} \rightsquigarrow N(0, 1) $ follows from the fact that $$ \frac{\sqrt{n}(\hat{\theta} - \theta_0)}{\hat{\text{se}}} \rightsquigarrow N(0, 1). $$

Doesn't this imply that $$ \lim_{n \rightarrow \infty} P\left(\frac{\sqrt{n}(\hat{\theta} - \theta_0)}{\hat{\text{se}}} \leq z\right) = \lim_{n \rightarrow \infty} P\left((\hat{\theta} - \theta_0)/\hat{se} \leq z\right), $$ which is not true?

Is this a mistake or am I missing something?

  • 1
    $\begingroup$ I cannot find these passages in Chapter 11 of the book. I see the results in Chapter 10, and I don't see the factor of $\sqrt n$ anywhere in the book in defining asymptotic normality (which is good, because if you are normalizing by the standard error then the $\sqrt n$ shouldn't be there). So if you want help you will need to be explicit about the edition of the book you are looking at and where precisely all of these things are defined. $\endgroup$
    – guy
    Commented Apr 3, 2020 at 22:45
  • $\begingroup$ So I was using a PDF version that was based on the course notes, but I just downloaded the e-textbook version and noticed that you're right. In section 10.1, the $ \sqrt{n} $ term has been removed and was presumably a mistake. $\endgroup$
    – an1lam
    Commented Apr 3, 2020 at 22:55
  • $\begingroup$ @guy: Do you want to leave a short answer just noting that this is wrong because the sample standard error includes the $ \sqrt{n} $ term and I'll mark it as correct so that it's clear this is resolved? $\endgroup$
    – an1lam
    Commented Apr 3, 2020 at 22:59

1 Answer 1


It’s an error, the sample standard error shouldn’t have square root of $ n $.


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