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:

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

instead of:

$Var(h(\hat{\theta})|\theta) \approx \frac{[h^{'}(\theta)]^2}{I_n(\theta)} \cdot \frac{1}{n}$

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.

  • 2
    $\begingroup$ 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. $\endgroup$
    – guy
    Apr 29, 2021 at 15:59
  • $\begingroup$ Yep, that totally eluded me. Thank you so much, now it's clear. $\endgroup$ Apr 29, 2021 at 16:28

1 Answer 1


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, $h(\hat{\theta})$.


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