Unbiased estimator of $ 1 + \mu^{2}$ from a Normal population 
Question: If $ x_{1}, x_{2}, x_{3},...x_{n}$ is a random sample from a $Normal$ $population$ $N(\mu,1)$ then  what is the unbiased estimator of $ 1 + \mu^{2}$ ?

I began finding the mean and variance of $X$
$$E(X) = \mu \  , \ Var(X) = 1$$
And the expectation of $X^{2}$  is
$$  E(X^{2}) = 1 + \mu^{2} $$
Hence is the unbiased estimator of $ 1 + \mu^{2}$ =  $\frac{1}{n}\sum_{i=1}^{n} x_{i}^{2}$ ?
Is my method and solution correct?
 A: Any function of the data is called an estimator. There is no such thing as "THE" estimator of a quantity. Various estimators can have different properties.
You have shown (correctly) that your estimator $\tfrac{1}{n}\sum x_i^2$ is unbiased for $1+\mu^2$. You could consider other estimators and they may have different properties (e.g. smaller or larger variance).
A: You can use the properties of a non-central chi squared distribution to construct a non-biased estimator from the sum $$S_1 = \sum_{i=1}^n x_i^2$$
This sum $S_1$ has the mean $n+n\mu^2$. So $S/n$ will have the mean $1+\mu^2$, and is indeed an unbiased estimator.

A more efficient estimator (an estimator with lower variance) is to be made with $S_2 = \left( \sum x_i \right)^2$. If we scale this like $\frac{1}{n} S_2$ it will be distributed as non central chi squared distribution with 1 degree of freedom and mean $1+n\mu^2$.
This distribution for $\frac{1}{n} S_2$  has a lower variance than the case with $S_1$. You can verify this easily based on the lower mean and lower degrees of freedom (and the variance is a function of both of those).
So, the estimator $\frac{1}{n^2} S_2 + 1 - \frac{1}{n}$ will be a more efficient estimator.

The last solution is an estimator. And like bdeonovic stresses in his answer it is not the estimator.
However, there is a type of estimator which is unique and that is the minimum variance unbiased estimator (MVUE).
The estimator $\frac{1}{n^2} S_2 + 1 - \frac{1}{n}$ is the MVUE.
This is because it is a function based on the complete and sufficient statistic $\sum x_i$. See also the Lehmann Sheffé theorem.
