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I need to know the dimostration of Running Variance's formula: $$ \sigma_n^2 =\frac{(n-1)\sigma_{n-1}^{2}+(x_n-\overline{x}_{n-1})\cdot(x_n-\overline{x}_{n})}{n} $$

Michael Hardy
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Carlo Cavicchia
  • 35

1 Answers1

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With the definition $Q_n= n\sigma_n^2\,$ we have to show $$Q_n- Q_{n-1} = (x_n-\bar{x}_{n-1})\cdot(x_n-\bar{x}_n).$$ Firstly derive a formula for $Q_n:$ $$Q_n = \sum_{i=1}^n(x_i-\bar{x}_n)^2=\sum_{i=1}^n(x_i^2-2 x_i\bar{x}_n +\bar{x}_n^2)\\ = \sum_{i=1}^n x_i^2- 2\bar{x}_n \sum_{i=1}^nx_i + \sum_{i=1}^n\bar{x}_n^2\\ = \sum_{i=1}^n x_i^2- 2\bar{x}_n (n\bar{x}_n) + n\bar{x}_n^2\\ = \sum_{i=1}^n x_i^2- n\bar{x}_n^2$$ then start computing the difference $$Q_n- Q_{n-1} = \sum_{i=1}^n x_i^2- n\bar{x}_n^2 - \sum_{i=1}^{n-1} x_i^2+(n-1)\bar{x}_{n-1}^2\\ = x_n^2 - n\bar{x}_n^2 + (n-1)\bar{x}_{n-1}^2\\ = x_n^2 - \bar{x}_{n-1}^2 + n(\bar{x}_{n-1}^2-\bar{x}_{n}^2)\\ =x_n^2 - \bar{x}_{n-1}^2 + n(\bar{x}_{n-1}-\bar{x}_{n})(\bar{x}_{n-1}+\bar{x}_{n}).$$ Now use $n(\bar{x}_{n-1}-\bar{x}_{n})=\bar{x}_{n-1}-x_n$ which follows from $$n\bar{x}_n = \sum_{i=1}^n x_i = \sum_{i=1}^{n-1} x_i + x_n = (n-1)\bar{x}_{n-1} + x_n$$ and get for the difference: $$Q_n- Q_{n-1} = x_n^2 - \bar{x}_{n-1}^2 + (\bar{x}_{n-1}-x_n)(\bar{x}_{n-1}+\bar{x}_{n})\\ =x_n^2 - \bar{x}_{n-1}^2 + \bar{x}_{n-1}^2 + \bar{x}_{n-1}\bar{x}_{n}-x_n\bar{x}_{n-1}-x_n\bar{x}_{n}\\ =x_n^2 + \bar{x}_{n-1}\bar{x}_{n}-x_n\bar{x}_{n-1}-x_n\bar{x}_{n}\\ =(x_n-\bar{x}_{n-1})(x_n-\bar{x}_n) $$

gammatester
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