# Why approximate delta-method Variance isn't multiplied by $\frac{1}{n}$?

I'm reading Casella-Berger chapter 10, where they introduce asymptotic evaluations. I don't seem quite to understand how the factor $$\sqrt{n}$$ works when we are using asymptotic evaluations in order to estimate the approximate variance of estimators.

Following their example (on p. 473, here's a screenshot), given that, according to the asymptotic efficiency of MLEs (where $$\hat{\theta}$$ is a MLE from an iid sample, $$\tau(.)$$ is continuous and $$I(\theta)$$ is the information number for a single observation):

$$\sqrt{n} \left[ \tau(\hat{\theta}) - \tau(\theta) \right] \rightarrow_d N \left(0, \frac{1}{I(\theta)}\right)$$

and that, according to the Delta method, if $$\sqrt{n} \left[ \hat{\theta} - \theta \right] \rightarrow_d N\left(0, \sigma^2\right)$$, then:

$$\sqrt{n} \left[ h(\hat{\theta}) - h(\theta) \right]\rightarrow_d N(0, \sigma^2[h^{'}(\theta)]^2)$$

Why is it that (according to Casella-Berger, p. 473), the approximate variance of the estimator is given by:

$$Var(h(\hat{\theta})|\theta) \approx \frac{[h^{'}(\theta)]^2}{I_n(\theta)}$$

$$Var(h(\hat{\theta})|\theta) \approx \frac{[h^{'}(\theta)]^2}{I_n(\theta)} \cdot \frac{1}{n}$$
I can't understand why, in this example and others (e.g. at p.243, example 5.5.25), the approximate variance of the estimator doesn't get multiplied by the $$\frac{1}{n}$$ term. Can anyone help me understand? Thanks in advance.
• You are confusing the unit-level and sample-level information. The first expression is you write $\sqrt n\{\tau(\widehat \theta) - \tau(\theta)\} \to N(0,I_n(\theta)^{-1})$ presumes that $I_n$ is the unit-level Fisher information, whereas your screenshot clearly has $I_n(\theta) = E \frac{\partial^2}{\partial \theta^2} L(\theta \mid \mathbf X)$ as the sample-level Fisher information, which by linearity of expectation is $n$ times the unit-level information in this case.
Briefly, $$h(\cdot)$$ is a change of variable, and not a statistic. In the assumption of the $$\delta$$-method, we already have that $$\hat{\theta}$$ is scaled by $$\sqrt{n}$$ so that the limiting distribution has a finite, non-zero variance. As a corollary we can say $$\text{Var}(\theta) \approx I_n(\theta)^{-1} / n$$ for large $$n$$, and a similar statement can be made about the change of variable, $$h(\hat{\theta})$$.