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To compute the (approximate) limiting (asymptotic) distribution of a function of a statistic with known (asymptotically normal) variance, the delta method can be invoked:

$\sqrt{n}[g(\hat{\theta}) - g(\theta)] \rightarrow_{d} N(0, \sigma^2[g'(\theta)]^2)$

In calculating a (1-$\alpha$)100% confidence interval for $g(\theta)$, is it necessary to divide Var[$g(\theta)$] = $\sigma^2[g'(\theta)]^2$ by $\sqrt{n}$?

I see some sources that do this, along with others that do not, which leads to some confusion.

I would say yes, since for example $\bar{X} \sim N(\mu, \frac{\sigma^2}{n})$.

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If $$ \sqrt{n}[g(\hat{\theta}) - g(\theta)] \rightarrow_{d} N(0, \sigma^2[g'(\theta)]^2) $$ (note the typo you made in the variance term) then for a "large" $n$ it is approximately true that $$ \sqrt{n}[g(\hat{\theta}) - g(\theta)] \sim N(0, \sigma^2[g'(\theta)]^2) $$ which is true if and only if $$ g(\hat{\theta})\sim N\left(g(\theta), \frac{\sigma^2[g'(\theta)]^2}{n}\right). $$

Here $\frac{\sigma^2[g'(\theta)]^2}{n}$ is the (approximate) variance of $g(\hat{\theta})$, which means its (approximate) standard deviation is $\frac{\sigma|g'(\theta)|}{\sqrt{n}}$.

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    $\begingroup$ Oops, thanks for the heads up regarding the typo. I have corrected this in the post. $\endgroup$ – compbiostats Sep 9 at 14:15

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