I was looking at the Welford online algorithm to update the variance. If we denote by $M_n = \sum_{i=1}^n (x_i-\overline{x}_n)^2$ where $\overline{x}_n = (\sum_{i=1}^n x_i)/n$ is the sample mean, I want to verify that $M_n = M_{n-1}+(x_n-\overline{x}_{n-1})(x_n-\overline{x}_n)$ which is the claim on wikipedia. But I get \begin{align} \sum_{i=1}^{n-1}(x_i-\overline{x}_{n-1})^2+(x_n-\overline{x}_{n-1})(x_n-\overline{x}_n)\\ =\sum_{i=1}^n(x_i-\overline{x}_{n-1})^2-(x_n-\overline{x}_{n-1})^2+(x_n-\overline{x}_{n-1})(x_n-\overline{x}_n)\\ = \sum_{i=1}^n(x_i-\overline{x}_{n-1})^2-(x_n-\overline{x}_{n-1})(\overline{x}_{n-1}-\overline{x}_n)\\ = \sum_{i=1}^n(x_i-\overline{x}_{n-1})^2-n(\overline{x}_n-\overline{x}_{n-1})^2 \end{align}
where the last step is because \begin{align} \overline{x}_n-\overline{x}_{n-1} = \frac{(n-1)\overline{x}_{n-1}+x_n}{n}- \frac{n\overline{x}_{n-1}}{n}=\frac{x_n-\overline{x}_{n-1}}{n} \end{align}
Then
\begin{align} M_{n-1}+(x_n-\overline{x}_{n-1})(x_n-\overline{x}_n)=\sum_{i=1}^n[(x_i-\overline{x}_{n-1})^2-(\overline{x}-\overline{x}_{n-1})^2]\\ =\sum_{i=1}^n (x_i-\overline{x}_{n-1}+\overline{x}_n-\overline{x}_{n-1})(x_i-\overline{x}_{n-1}-\overline{x}_n+\overline{x}_{n-1})\\ =\sum_{i=1}^n (x_i-\overline{x}_{n-1}+\overline{x}_n-\overline{x}_{n-1})(x_i-\overline{x}_n) \end{align}
which is not the right conclusion.