Find the MLE of $N(\theta,\theta)$ 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?
 A: 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
$$
\hat{\theta}=-\overline{x}+\sqrt{\overline{x}^2+2\overline{x^2}}.
$$
A: Derivation
I think, you are (were) on the right way. Let me continue your computations:
Suppose $X_1,\ldots,X_n$ are iid $N(\theta,\theta)$, with $\theta\in(0,\infty)$. Find the MLE of $\theta$.
$$
  \frac{\partial \log L(x|\theta)}{\partial \theta}=-\frac{n}{2}\frac{1} 
  {\theta}+\frac{\sum_{i=1}^nx_{i}^2}{2}\frac{1}{\theta^2}-\frac{n}{2}
$$
$$
\frac{\partial \log L(x|\theta)}{\partial \theta}=0
$$
$$
  -\frac{n}{2}\theta^2-\frac{n}{2}\theta+\frac{1}{2}\sum_{i=1}^nx_{i}^2=0
$$
And multiply by 2
$$
  -n\theta^2-n\theta+\sum_{i=1}^nx_{i}^2=0
$$
Now solve for $\theta$:
$$
\theta_{1,2} = \frac{n \pm \sqrt{n^2 + 4n\sum_{i=1}^nx_{i}^2}}{-2n}
             = \frac{-1 \mp \sqrt{1 + 4\frac{1}{n}\sum_{i=1}^nx_{i}^2}}{2}
$$
Since $1 < \sqrt{1 + 4\frac{1}{n}\sum_{i=1}^nx_{i}^2}$ we have only one solution for $\theta > 0$
$$
\bar{\theta}^{MLE}_n = \frac{\sqrt{1 + 4\frac{1}{n}\sum_{i=1}^nx_{i}^2}-1}{2}
       = \frac{1}{2}\left(\sqrt{1 + 4 \overline{x^2}} - 1\right)
$$
Proof of consistency
Let's proof a.s. convergance to true $\theta$ using LLN and second moment of the normal distribution.
Accourding to LLN (https://en.wikipedia.org/wiki/Law_of_large_numbers)
$$
  \overline{x^2} =
  \frac{1}{n}\sum_{i=1}^nx_{i}^2
  \underset{n\rightarrow\infty}{\overset{a.s.}{\longrightarrow}}
  \mathbb{E}[x_i^2]
$$
Second moment of $N(\mu, \sigma^2)$ is $\mathbb{E}[x_i^2] =\mu^2 + \sigma^2 = \theta^2 + \theta$
$$
  \bar{\theta}^{MLE}_n =
  \frac{\sqrt{1 + 4\overline{x^2}}-1}{2} =
  \frac{\sqrt{1 + 4\theta^2 + 4\theta}-1}{2} =
  \frac{\sqrt{(1 + 2\theta)^2}-1}{2} =
  \theta
$$
Simulation with R
You can verify this with the following simple R-script (n=100000, $\theta$ = 0.25) :
theta <- 0.25
n <- 100000
z <- rnorm(n, mean = theta, sd = sqrt(theta))
theta_MLE = (1 + sqrt(1 + 4*sum(z^2)/n))/2 - 1

# Output
theta_MLE
theta - theta_MLE

Output:
[1] 0.2515441
[1] -0.001544133

