From the definition of the variance:
$$\sigma^2 = \frac{1}{N-1}\sum_{i=0}^{N-1}(x_i-\mu)^2 \tag{1}$$
and mean:
$$\mu = \frac{1}{N}\sum^{N-1}_{i=0}x_i \tag{2}$$
how is it possible to derive the variance for running statistics, which is:
$$\sigma^2 = \frac{1}{N-1}\left[\sum^{N-1}_{i=0}x_i^2-\frac{1}{N}\left(\sum_{i=0}^{N-1}x_i \right)^2\right] \tag{3}$$
why the term $2x_i\mu$ is not considered?
Starting from $(1)$, I get:
$$\sigma^2 = \frac{1}{N-1}\left[\sum^{N-1}_{i=0}x_i^2-\sum_{i=0}^{N-1}2x_i \mu\right]+\frac{\mu^2}{N-1} $$
What you should have for the unbiased sample estimator of the variance, using $\bar x=\frac1N\sum\limits_{i=0}^{N-1}x_i$, is $$s^2 = \frac{1}{N-1}\sum\limits_{i=0}^{N-1}(x_i-\bar x)^2 \\ =\frac{1}{N-1}\sum\limits_{i=0}^{N-1}\left(x_i^2 -2x_i\bar x+\bar x^2 \right) \\ =\frac{1}{N-1}\left[\sum\limits_{i=0}^{N-1}x_i^2 -2 \sum\limits_{i=0}^{N-1}x_i\bar x+N\bar x^2 \right]\\ =\frac{1}{N-1}\left[\sum\limits_{i=0}^{N-1}x_i^2 -2\sum\limits_{i=0}^{N-1}x_i \frac1N\sum\limits_{i=0}^{N-1}x_i+N\frac1N\sum\limits_{i=0}^{N-1}x_i\frac1N\sum\limits_{i=0}^{N-1}x_i \right]\\ =\frac{1}{N-1}\left[\sum\limits_{i=0}^{N-1}x_i^2 -\frac1N\left(\sum\limits_{i=0}^{N-1}x_i\right)^2 \right]$$