Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:

mean = Mean(list)
for i = 0 to list.size
   stdev = stdev + (list[i] - mean)^2
stdev = sqrRoot( stdev / list.size )
share|cite|improve this question
up vote 25 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: $\bar X_n = n^{-1}[X_n + (n-1)\bar X_{n-1}]$.

share|cite|improve this answer
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
@user55490: $$ \sum_{j = 1} ^ {n - 1} (X_j - \bar X_{n - 1}) = (sum_{j = 1} ^ {n - 1} (X_j) - (n - 1) \bar X_{n - 1}) = (n - 1) (\bar X_{n - 1} - \bar X_{n - 1})$$ – Jean Hominal Oct 26 '15 at 14:24

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.

share|cite|improve this answer
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
This is actually computationally efficient, because you just need to count the number of samples, the sum of the samples, and the sum of the squares of each sample. Once you need to "read" the standard deviation (or average,) you calculate the square root, which costs two multuplies, one subtraction, one square root, and one divide. On most modern CPUs, this is highly efficient (and very cache friendly.) – Jon Watte Aug 1 '14 at 3:39
Furthermore, this methods allows to compute easily a running standard deviation as well, which is what i was looking for. in addition to adding the new value to the sum of values and squares, remove the oldest value from them as well. Sorry for the late comment, but I found this page after googling this issue and I hope it will let future people find this more easily. – Quentin Feb 11 at 16:31

Another way of saying the above is that you need to keep a count, an incremental sum of values and an incremental sum of squares.
Let $N$ be the count of values seen so far, $S = \sum_{1,N} x_i$ and $Q = \sum_{1,N} x_i^2$ (where both $S$ and $Q$ are maintained incrementally). Then any stage,
the mean is $\frac{S}{N}$
and the variance is $\frac{Q}{N} - \left( \frac{S}{N}\right)^2$
An advantage of this method is that it is less prone to rounding errors after a long series of calculations.

share|cite|improve this answer

What you refer to as an incremental computation is very close to the computer scientist's notion of an online algorithm. There is in fact a well-known online algorithm for computing the variance and thus square root of a sequence of data, documented here.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.