# How can these two expressions both converge in distribution to N(0, 1)?

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?

• 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.
– guy
Apr 3, 2020 at 22:45
• 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. Apr 3, 2020 at 22:55
• @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? Apr 3, 2020 at 22:59

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