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?

  • 1
    $\begingroup$ Its warranted due to Slutsky's Theorem. I have expanded this comment in the answer below. $\endgroup$ – Greenparker Apr 12 '16 at 15:10

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.

  • 1
    $\begingroup$ Out of curiosity, do you see a reason you could not replace $\beta_{MLE}$ on the denominator in the second line with $\beta$ and then simplifying in similar fashion to back out an interval for $\beta$? It's not any simpler, but it would seem to save the need of invoking Slutsky (which could be an advantage for people that hadn't heard of it) $\endgroup$ – Glen_b -Reinstate Monica Apr 13 '16 at 10:04
  • $\begingroup$ @Glen_b I think you are right, and I have edited the answer to include that alternative solution. Thanks! $\endgroup$ – Greenparker Apr 13 '16 at 12:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.