The Question:
Given independent random variables $X_1,\dots,X_n \sim N(\mu,\mu^2)$ where $\mu>0$ is unknown, find the MLE $\hat \mu$ of $\mu$.
My Attempt:
Yes, I know this is a really standard question, but I got a really strange answer so I am not sure. We go through the usual drill:
$$L(\mu)=(2\pi \mu^2)^{-n/2}\exp\Bigl(\frac{-1}{2\mu^2} \sum_{i=1}^n(x_i-\mu)^2 \Bigr)$$
$$\ell (\mu)=-\frac n2 \ln(2\pi\mu^2)-\frac{1}{2\mu^2}\sum_{i=1}^n(x_i-\mu)^2 $$
\begin{align} \ \frac{d\ell}{d\mu} & = \Bigl(-\frac n\mu\Bigl)+\Bigl(\frac{1}{\mu^2}\sum_{i=1}^n(x_i-\mu)\Bigl)+\Bigl(\frac{1}{\mu^3}\sum_{i=1}^n(x_i-\mu)^2\Bigl) \\ \ & = \Bigl(-\frac n\mu \Bigl) -\Bigl(\frac n\mu +\frac{1}{\mu^2}\sum_{i=1}^nx_i \Bigl)+ \Bigl(\frac{1}{\mu^3}\sum_{i=1}^nx_i^2-\frac{2}{\mu^2}\sum_{i=1}^nx_i+\frac n\mu \Bigl) \\ \ & = -\frac n\mu - \frac{1}{\mu^2}\sum_{i=1}^nx_i+\frac{1}{\mu^3}\sum_{i=1}^nx_i^2 \end{align}
Thus
\begin{align} \ & \frac{d\ell}{d\mu}=0 \\ \ \implies & n\mu^2+\mu \sum_{i=1}^nx_i-\sum_{i=1}^nx_i^2=0 \\ \ \implies & \mu = \frac{-\sum_{i=1}^nx_i + \sqrt{\bigl(\sum_{i=1}^nx_i \bigr)^2+4n\sum_{i=1}^nx_i^2}}{2n} \end{align}
... and I feel that I have done something wrong, since there is no way I can use this disgusting thing to do the next part of the question.
Any help would be much appreciated!
OK, so seems like there is nothing wrong with that disgusting expression for $\hat \mu$. The next part of the question is to show that the asymptotic variance is $\mu^2/3n$ which actually does not require the expression for $\hat \mu$:
\begin{align} \ \frac{d^2 \ell}{d \ell ^2} & = \frac{d}{d \ell}\Bigl(-\frac n\mu-\frac{1}{\mu^2}\sum_{i=1}^nx_i+\frac{1}{\mu^3}\sum_{i=1}^nx_i^2\Bigl) \\ \ & = \frac{n}{\mu^2}+\frac{2}{\mu^3}\sum_{i=1}^nx_i-\frac{3}{\mu^4}\sum_{i=1}^nx_i^2 \\ \end{align}
Hence, the asymptotic variance is the reciprocal of the Fisher Information, which is:
\begin{align} \ I(\mu)^{-1} & =\biggl(\Bbb E\Bigl[-\frac{d^2 \ell}{d \ell ^2}\Bigr]\biggr)^{-1} \\ \ & =\biggl(-\frac{n}{\mu^2}-\frac{2}{\mu^3}\sum_{i=1}^n\Bbb E(x_i)+\frac{3}{\mu^4}\sum_{i=1}^n\Bbb E(x_i^2) \biggr)^{-1} \\ \ & = \biggl(-\frac{n}{\mu^2}-\frac{2}{\mu^2}(n\mu)+\frac{3}{\mu^4}(2\mu^2)\biggr)^{-1} \\ \ & = \frac{\mu^2}{3n} \end{align}
noting that $\Bbb E(x_i^2)=Var(x_i)+[\Bbb E(x_i)]^2=\mu^2+\mu^2=2\mu^2$