I'd like to calculate a standard deviation for a very large (but known) number of sample values, with the highest accuracy possible. The number of samples is larger than can be efficiently stored in memory.

The basic variance formula is:

$\sigma^2 = \frac{1}{N}\sum (x - \mu)^2$

... but this formulation depends on knowing the value of $\mu$ already.

$\mu$ can be calculated cumulatively -- that is, you can calculate the mean without storing every sample value. You just have to store their sum.

But to calculate the variance, is it necessary to store every sample value? Given a stream of samples, can I accumulate a calculation of the variance, without a need for memory of each sample? Put another way, is there a formulation of the variance which doesn't depend on foreknowledge of the exact value of $\mu$ before the whole sample set has been seen?

  • $\begingroup$ This very same issue was discussed on dsp.SE, and the answers were very similar, with numerically unstable methods proposed in one answer, and more stable methods described by others. $\endgroup$ Commented Mar 4, 2012 at 17:04
  • $\begingroup$ math.stackexchange.com/questions/198336/… $\endgroup$
    – sanmai
    Commented Apr 12, 2023 at 4:51

2 Answers 2


I'm a little late to the party, but it appears that this method is pretty unstable, but that there is a method that allows for streaming computation of the variance without sacrificing numerical stability.

Cook describes a method from Knuth, the punchline of which is to initialize $m_1 = x_1$, and $v_1 = 0$, where $m_k$ is the mean of the first $k$ values. From there,

$$ \begin{align*} m_k & = m_{k-1} + \frac{x_k - m_{k-1}}k \\ v_k & = v_{k-1} + (x_k - m_{k-1})(x_k - m_k) \end{align*} $$

The mean at this point is simply extracted as $m_k$, and the variance is $\sigma^2 = \frac{v_k}{k-1}$. It's easy to verify that it works for the mean, but I'm still working on grokking the variance.

  • 4
    $\begingroup$ +1, excellent! I didn't feel a simple upvote would be sufficient to express my appreciation of this answer, so there's an extra 50 rep bounty coming your way in 24 hours. $\endgroup$ Commented Mar 4, 2012 at 20:12
  • 1
    $\begingroup$ Isn't the final variance calculation should be Vk / k (not k-1)? $\endgroup$
    – Eyal Roth
    Commented Oct 10, 2013 at 15:49
  • $\begingroup$ @errr: it has to do with using biased vs unbiased estimator $\endgroup$
    – alexey
    Commented Mar 4, 2016 at 2:03
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    $\begingroup$ $v_k = \sum_i^k (x_i-m_k)^2$, so $v_k = v_{k-1} + x_k^2 - \frac{(k-1)m_k^2 - km_{k-1}^2}{k(k-1)}$ $\endgroup$
    – GuLearn
    Commented May 5, 2017 at 2:02
  • $\begingroup$ To test whether it's stable or not refer to the Examples in en.wikipedia.org/wiki/Algorithms_for_calculating_variance. The goes through finding the variance for (4, 7, 13, 16) then for (1e8 + 4, 1e8 + 7, 1e8 + 13, 1e8 + 16) and then for (1e9 + 4, 1e9 + 7, 1e9 + 13, 1e9 + 16). In theory they should all return the same variance but because of the imprecisions in the double arithmetic the result deviate from the expected variance. $\endgroup$
    – thomas.han
    Commented Jun 30, 2018 at 16:58

You can keep two running counters - one for $\sum_i x_i$ and another for $\sum_i x_i^2$. Since variance can be written as $$ \sigma^2 = \frac{1}{N} \left[ \sum_i x_i^2 - \frac{(\sum_i x_i)^2}{N} \right] $$

you can compute the variance of the data that you have seen thus far with just these two counters. Note that the $N$ here is not the total length of all your samples but only the number of samples you have observed in the past.

  • 1
    $\begingroup$ Wait -- shouldn't the N be inside the the expected values, such that the right side is divided by $N^2$? $\endgroup$
    – user6677
    Commented Feb 5, 2011 at 22:05
  • $\begingroup$ e.g. $$ \sigma^2 = \frac{(\sum_i x_i^2)}{N} - (\frac{\sum_i x_i}{N})^2 $$ $\endgroup$
    – user6677
    Commented Feb 5, 2011 at 22:27
  • $\begingroup$ @user6677: You are indeed right. Thanks for the correction. $\endgroup$
    – Dinesh
    Commented Feb 5, 2011 at 22:30
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    $\begingroup$ Note the sum of squares especially is at risk of overflow. $\endgroup$
    – dfrankow
    Commented Dec 29, 2014 at 18:06
  • 6
    $\begingroup$ This is numerically unstable. If your mean is large compared to your stdev, your estimate will be way off. Knuth's algo is better (see other answer). $\endgroup$
    – Leopd
    Commented Feb 18, 2015 at 1:40

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