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How can I compute the standard deviation in an incremental way (using the new value and the last computed mean and/or std deviation) ?

for the non incremental way, I just do something like: http://upload.wikimedia.org/wikipedia/en/math/c/c/7/cc79944cf31b14406acd0a8ec8498688.png

mean = Mean(list)
for i = 0 to list.size
   stdev = stdev + (list[i] - mean)^2
stdev = sqrRoot( stdev / list.size )
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2 Answers 2

up vote 14 down vote accepted

I think the easiest way to do this is with an orthogonality trick. I'll show how to incrementally compute the variance instead of the standard deviation. Let $X_1, X_2, ...$ be an iid sequence of random variables with $\bar X = n^{-1} \sum_{j = 1} ^ n X_j$ and $s^2_n$ defined similarly as the $n$'th sample variance (I use a denominator of $n-1$ instead of $n$ in your picture to keep things unbiased for the variance, but you can use the same argument just adding $1$ to all the terms with $n$ in them). First write $$ s^2_n = \frac{\sum_{j = 1} ^ n (X_j - \bar X_n)^2}{n - 1} = \frac{\sum_{j = 1} ^ n (X_j - \bar X_{n - 1} + \bar X_{n - 1} + \bar X_n)^2}{n - 1}. $$ Expand this to get $$ s^2_n = \frac{(n - 2)s^2_{n - 1} + (n - 1) (\bar X_{n - 1} - \bar X_n)^2 + 2 \sum_{j = 1} ^ {n - 1} (X_j - \bar X_{n - 1})(\bar X_{n - 1} - \bar X_n) + (X_n - \bar X_{n} ^ 2)}{n - 1} $$ and it is easy to show that the summation term above is equal to $0$ which gives $$ s^2_n = \frac{(n - 2)s^2_{n - 1} + (n - 1)(\bar X_{n - 1} - \bar X_n)^2 + (X_n - \bar X_{n})^2}{n - 1}. $$

EDIT: I assumed you already have an incremental expression for the sample mean. It is much easier to get that: $\bar X_n = n^{-1}[X_n + (n-1)\bar X_{n-1}]$.

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7  
Nice answer. Note that your formula can be simplified to $$s_n^2={n-2\over n-1}s^2_{n-1}+{1\over n}(X_n-\bar X_{n-1})^2.$$ –  Byron Schmuland Jan 27 '12 at 18:49
    
Thanks Guy for the great explaination (2). it is exactly what i was looking for. but can you please elaborate how the summation term equals zero as i seem to be missing a trick... –  user55490 Jan 7 '13 at 18:09

The standard deviation is a function of the totals $T_{\alpha}=\sum_{i=1}^{N}x_{i}^{\alpha}$ for $\alpha=0,1,2$, each of which can be calculated incrementally in an obvious way. In particular, $E[X]=T_{1}/T_{0}$ and $E[X^2]=T_{2}/T_{0}$, and the standard deviation is $$ \sigma = \sqrt{\text{Var}[X]} = \sqrt{E[X^2]-E[X]^2} = \frac{1}{T_0}\sqrt{T_{0}T_{2}-T_{1}^2}. $$ By maintaining totals of higher powers ($T_{\alpha}$ for $\alpha \ge 3$), you can derive similar "incremental" expressions for the skewness, kurtosis, and so on.

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I think the @user wanted a formula that lets you incrementally calculate the standard deviation as you increase the data set in a way that is computationally efficient, not this. –  guy Jan 27 '12 at 17:52

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