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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?

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

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    $\begingroup$ 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

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