Find the MLE of $N(\theta,\theta)$

Question

Suppose $X_1,\ldots,X_n$ are iid $N(\theta,\theta)$, with $\theta\in(0,\infty)$. Find the MLE of $\theta$. I got $\frac{\partial logL(x|\theta)}{\partial \theta}=-\frac{n}{2}\frac{1}{\theta}+\frac{\sum_{i=1}^nx_{i}^2}{2}\frac{1}{\theta^2}-\frac{n}{2}$. Then, I let $\frac{\partial logL(x|\theta)}{\partial \theta}=0$. Then, I have $-\frac{n}{2}\theta^2-\frac{n}{2}\theta+\frac{1}{2}\sum_{i=1}^nx_{i}^2=0$. Then, I find the solution is weird. Am I wrong?

Answer 1

I see there is some mistake in your calculations..

Given $x_1,\ldots,x_n$ the log-likelihood is proportional to $$ \ell\colon \mathbf{R}^+\to \mathbf{R}\colon \theta\mapsto -n\ln \theta-\sum_{i=1}^n\left(\frac{x_i}{\theta}-1\right)^2. $$ Since we have to maximize $-\ell$, it is enough to see that $$ \ell^\prime(\theta) \propto \frac{n}{\theta}-\sum_{i=1}^n\frac{2x_i}{\theta^2}\left(\frac{x_i}{\theta}-1\right). $$ Then, if I am not wrong, the solution is $$ \hat{\theta}=-\overline{x}+\sqrt{\overline{x}^2+2\overline{x^2}}. $$