In [1, p. 60 exercise 3.4] we are asked to prove that the estimate
\hat{\mu}_x = \frac{1}{N}\sum\limits^{N-1}_{n=0}x[n] (1)

is an unbiased estimator, given \left\{x[0],x[1],...,x[N]\right\} are independent and identically distributed according to a N(\mu_x,\sigma^{2}_x) distribution. Furthermore we are asked to also find the variance of the estimator.
Solution: The mean of the estimator \hat{\mu}_x is given by
=\frac{1}{N} \cdot N \cdot \mu_x

Thus the estimator \hat{\mu}_x is unbiased. The variance of the estimator about the mean \mu_x is given by:
=\frac{1}{N^2}\sum\limits^{N-1}_{n=0}\sum\limits^{N-1}_{k=0}E\{x[n]x[k]\} -\mu^2_x  (2)

Considering the fact that the samples are independent we can set E\{x[n]x[k]\}=E\{x[n]\}E\{x[k]\}=\mu^2_x for n\neq k and obtain
\sum\limits^{N-1}_{n=0}\sum\limits^{N-1}_{k=0}E\{x[n]x[k]\}= \sum\limits^{N-1}_{n=0}E\{x^2[n]\}
 +{\sum\limits^{N-1}_{n=0, n \neq k}\sum\limits^{N-1}_{k=0}}E\{x[n]\}E\{x[k]\}
=N \cdot E\{x^2[n]\} +N(N-1)\mu^2_x (3)

Thus it remains to obain the value of E\{x^2[n]\} in order to determine the variance of the estimator. We can determine E\{x^2[n]\} by the variance \sigma^2_x of x[n]:
=E\{x^2[n]\}-\mu^2_x \Rightarrow
E\{x^2[n]\}=\sigma^2_x+\mu^2_x (4)

And thus the variance of the estimator (2), considering (3) and (4), is given by:
E\left\{\left(\hat{\mu}_x-\mu_x\right)^2\right\}=\frac{1}{N} (\sigma^2_x+\mu^2_x+(N-1)\mu^2_x) -\mu^2_x
=\frac{1}{N} \sigma^2_x. (5)

We have proved thus that the estimator is unbiased and have obtained the variance of the estimator \hat{\mu_x}. QED.

[1] Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”, Prentice Hall, ISBN: 0-13-598582-X.