Understanding MLE of $N(\theta, \theta)$ The maximum likelihood estimator of $N(\theta, \theta)$ is
$$
\bar{\theta}^{MLE}_n = \frac{1}{2}(\sqrt{1 + 4 \overline{x^2}} - 1)
$$
for $\overline{x^2}=\frac{1}{n}\sum_{i=1}^nx_{i}^2$
(See https://math.stackexchange.com/a/3478232/735298 for derivation)
Now playing with this estimator I tried some simple $x_i$s, like, 


*

*$x_1=x_2=...=x_n=1$, that gives me


$$
\bar{\theta}^{MLE}_n = \frac{1}{2}(\sqrt{1 + 4\cdot 1} - 1) = \frac{1}{2}(\sqrt{5} - 1)=0.618...
$$
I expected here to get 1, since intuitively the mean 1 with any variance would be the most probable one.
Also for $x_1=x_2=...=x_n=0$ we get expected $\bar{\theta}^{MLE}_n=0$.
My questions:


*

*What's wrong with my intuition described above?

*What is the explanation of the $\bar{\theta}^{MLE}_n = 0.618$ for all $x_i=1$?

 A: Note that parameter $\theta$ is not only the mean of the distribution but also the variance. If all samples are $1$, the sample mean is $1$ and the sample variance is $0$. How should estimator for $\theta$ react to such a difference? It gives out something in between.
For zero values: there is complete agreement - the sample mean is $0$, the sample variance is $0$. It would be strange if the estimate did not give zero.
A: 
What's wrong with my intuition described above? What is the explanation of the $\bar{\theta}^{MLE}_n = 0.618$ for all $x_i=1$?

The problem with the data samples  $$x_1=x_2=...=x_n=1$$ is that "are they really generated by $\mathcal{N}(\theta,\theta)$ ?" I mean yes, the mean is $1$. But is their variance $1$ ? Especially when $n$ grows larger and larger !!? No. Roughly speaking, the sampling distribution model has a hard time believing the data.
In contrast, the data samples $$x_1=x_2=...=x_n=0$$ fit the model $\mathcal{N}(\theta,\theta)$ perfectly for $\theta = 0$, and thus you could explain the data using the sampling distribution. That said, you could use any software such as Python or MATLAB to generate pseudorandom numbers sampled from a normal distribution with equal mean and variance. For example, on MATLAB, the following generates $10^4$ samples from $\mathcal{N}(1,1)$:
x = 1 + randn(10000,1)
0.5*(sqrt(1 + 4*mean(x.^2)) - 1)

which gives an answer $\sim 1$
