# Delta Method Confidence Interval: Dividing by $\sqrt{n}$

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 $$\bar{X} \sim N(\mu, \frac{\sigma^2}{n})$$.

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 $$\frac{\sigma|g'(\theta)|}{\sqrt{n}}$$.