Is it possible to compute the variance without computing the mean first? I have a list of values of a random variable $x \in \mathbb R$. Is it possible to find the varience $\overline{(x - \overline x)^2}$ without computing the mean $\overline x$ first? That is to process the list only once.

The sample variance with mean is calculated as:
$$ v_{X}=\frac{1}{n-1}\sum_{i=1}^{n}(x_{i}-\overline{x})^{2} $$
And the sample variance without mean as:
$$ v_{X}=\frac{1}{n-1}\left [ \sum_{i=1}^{n}x_{i}^{2}-\frac{1}{n}\left ( \sum_{i=1}^{n}x_{i} \right ) ^{2}\right ] $$
