# Why does asymptotic efficiency require normality?

Greene (Econometric Analysis), states the defintiion of Asymptotic Efficiency.

My question is, why does this definition contain a reference to normal distribution? We already know by the central limit theorems that under certain conditions, the estimator will be normally distributed in the limit.

So I surmise that this additional requirement is added to rule out the estimators that are not asymptotically normally distributed.

However, if an estimator is not asymptotically normally distributed, will it then not also necessarily have an infinite variance (or undefined variance)? In which case it cannot possibly have the lowest possible asymptotic covariance.

Where am I going wrong? Why do we need to add the requirement of normality?

Consider for example estimating the upper bound of the uniform distribution $U[0,\theta]$. It is well known that the sample maximum $y_{(n)}$ is the MLE, for a random sample $y$.
$y_{(n)}$ is not unbiased so let us bias-correct it. Its cdf is \begin{eqnarray*} F_{y_{(n)}}(x)&=&\Pr\{Y_1\leqslant x,\ldots,Y_n\leqslant x\}\\ &=&\Pr\{Y_1\leqslant x\}^n\\ &=&\begin{cases} 0&\qquad\text{for}\quad x<0\\ \left(\frac{x}{\theta}\right)^n&\qquad\text{for}\quad 0\leqslant x\leqslant\theta\\ 1&\qquad\text{for}\quad x>\theta \end{cases} \end{eqnarray*} Thus, its density is $$f_{y_{(n)}}(x)= \begin{cases} \frac{n}{\theta}\left(\frac{x}{\theta}\right)^{n-1}&\qquad\text{for}\quad 0\leqslant x\leqslant\theta\\ 0&\qquad\text{else} \end{cases}$$
Hence, \begin{eqnarray*} E[Y_{(n)}]&=&\int_0^\theta x\frac{n}{\theta}\left(\frac{x}{\theta}\right)^{n-1}dx\\ &=&\int_0^\theta n\left(\frac{x}{\theta}\right)^{n}dx\\ &=&\frac{n}{n+1}\theta \end{eqnarray*} Thus, an unbiased estimator is given by $$\frac{n+1}{n}y_{(n)},$$ as $$E\left[\frac{n+1}{n}Y_{(n)}\right]=\frac{n+1}{n}\frac{n}{n+1}\theta=\theta$$
Its variance is \begin{eqnarray*} Var\left(\frac{n+1}{n}Y_{(n)}\right)&=&\left(\frac{n+1}{n}\right)^2Var(y_{(n)})\\ &=&\left(\frac{n+1}{n}\right)^2\left[E(Y^2)-\left(\frac{n}{n+1}\theta\right)^2\right]\\ &=&\left(\frac{n+1}{n}\right)^2\left[\int_0^\theta n\left(\frac{x^{n+1}}{\theta^{n}}\right)dx-\left(\frac{n}{n+1}\theta\right)^2\right]\\ &=&\left(\frac{n+1}{n}\right)^2\left[\frac{n}{n+2}\frac{\theta^{n+2}}{\theta^n}-\left(\frac{n}{n+1}\theta\right)^2\right]\\ &=&\left(\frac{n+1}{n}\right)^2\left[\frac{n}{n+2}\theta^2-\left(\frac{n}{n+1}\theta\right)^2\right]\\ &=&\left(\frac{n+1}{n}\right)^2\left[\frac{n(n+1)^2\theta^2-n^2(n+2)\theta^2}{(n+2)(n+1)^2}\right]\\ &=&\frac{1}{n(n+2)}\theta^2 \end{eqnarray*} Another unbiased estimator is given by twice the sample mean. It has variance $$Var(2\bar X)=\theta^2/n$$ So we see that the variance of the MLE converges to zero at a faster rate, $O(n^{-2})$ than that of twice the sample mean, $O(n^{-1})$, so that, so to speak , the MLE is infinitely more efficient asympotically.