Confidence interval for MLE estimator?

Question

I have an MLE estimator which is asymptotically normally distributed with mean $\beta$ and variance $\beta^2/n$. How do I get an approximate confidence interval for this estimator?

I know usually two ways to do it: if we know the variance, we construct a confidence interval using normal quantiles, and if we don't know it, we plug in a sample standard deviation and use t-quantiles. But now the variance is not known but we also don't have a sample standard deviation at hand, so what to do?

What I could do is plug in the MLE estimator in my variance, and then I can construct a normal confidence interval using the standard $z$-quantiles, but how is this assumption warranted?

Answer 1

You know that $\beta_{MLE} \overset{p}{\to} \beta$, and thus you can use Slutsky's theorem.

\begin{align*} \sqrt{n}(\beta_{MLE} - \beta) &\overset{d}{\to} N(0, \beta^2)\\ \dfrac{\sqrt{n}(\beta_{MLE} - \beta)}{\beta_{MLE}} &\overset{d}{\to} N(0, 1)\\ \sqrt{n}\left(1 - \dfrac{\beta}{\beta_{MLE}}\right) &\overset{d}{\to} N(0, 1) \end{align*}

Using this you can make asymptotic confidence intervals.

Thanks to Glen_b, you could also do the following without Slutsky's Theorem. This is an alternative solution.

\begin{align*} \sqrt{n}(\beta_{MLE} - \beta) &\overset{d}{\to} N(0, \beta^2)\\ \dfrac{\sqrt{n}(\beta_{MLE} - \beta)}{\beta} &\overset{d}{\to} N(0, 1)\\ \sqrt{n}\left(\dfrac{\beta_{MLE}}{\beta} - 1\right) &\overset{d}{\to} N(0, 1) \end{align*}

This can then similarly be used to make asymptotic confidence intervals.