Algorithms for calculating variance play a major role in statistical computing. A key problem in the design of good algorithms for this problem is that formulas for the variance may involve sums of squares, which can lead to numerical instability as well as to arithmetic overflow when dealing with large values.
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The formula for calculating the variance of an entire population of size n is:
The formula for calculating an unbiased estimate of the population variance from a finite sample of n observations is:
Therefore a naive algorithm to calculate the estimated variance is given by the following:
def naive_variance(data): n = 0 Sum = 0 Sum_sqr = 0 for x in data: n = n + 1 Sum = Sum + x Sum_sqr = Sum_sqr + x*x mean = Sum/n variance = (Sum_sqr  Sum*mean)/(n  1) return variance
This algorithm can easily be adapted to compute the variance of a finite population: simply divide by n instead of n − 1 on the last line.
Because sum_sqr
and sum * mean
can be
very similar numbers, the precision of
the result can be much less than the inherent precision of the floatingpoint arithmetic used to perform
the computation. This is particularly bad if the standard deviation
is small relative to the mean.
An alternate approach, using a different formula for the variance,
is given by the following pseudocode:
def two_pass_variance(data): n = 0 sum1 = 0 for x in data: n = n + 1 sum1 = sum1 + x mean = sum1/n sum2 = 0 for x in data: sum2 = sum2 + (x  mean)**2 variance = sum2/(n  1) return variance
This algorithm is often more numerically reliable than the naïve algorithm for large sets of data, although it can be worse if much of the data is very close to but not precisely equal to the mean and some are quite far away from it.
The results of both of these simple algorithms (I and II) can depend inordinately on the ordering of the data and can give poor results for very large data sets due to repeated roundoff error in the accumulation of the sums. Techniques such as compensated summation can be used to combat this error to a degree.
The compensatedsummation version of the algorithm above reads:
def compensated_variance(data): n = 0 sum1 = 0 for x in data: n = n + 1 sum1 = sum1 + x mean = sum1/n sum2 = 0 sumc = 0 for x in data: sum2 = sum2 + (x  mean)**2 sumc = sumc + (x  mean) variance = (sum2  sumc**2/n)/(n  1) return variance
It is often useful to be able to compute the variance in a single pass, inspecting each value x_{i} only once; for example, when the data are being collected without enough storage to keep all the values, or when costs of memory access dominate those of computation. For such an online algorithm, a recurrence relation is required between quantities from which the required statistics can be calculated in a numerically stable fashion.
The following formulas can be used to update the mean and (estimated) variance of the sequence, for an additional element x_{new}. Here, m denotes the estimate of the population mean(using the sample mean), s^{2}_{n1} the estimate of the sample variance, s^{2}_{n} the estimate of the population variance, and n the number of elements in the sequence before the addition.
It turns out that a more suitable quantity for updating is the sum of squares of differences from the (current) mean, , here denoted M_{2}:
A numerically stable algorithm is given below. It also computes the mean. This algorithm is due to Knuth,^{[1]} who cites Welford.^{[2]}
def online_variance(data): n = 0 mean = 0 M2 = 0 for x in data: n = n + 1 delta = x  mean mean = mean + delta/n M2 = M2 + delta*(x  mean) # This expression uses the new value of mean variance_n = M2/n variance = M2/(n  1) return variance
This algorithm is much less prone to loss of precision due to massive cancellation, but might not be as efficient because of the division operation inside the loop. For a particularly robust twopass algorithm for computing the variance, first compute and subtract an estimate of the mean, and then use this algorithm on the residuals.
A slightly more convenient form allows one to calculate the standard deviation without having to explicitly calculate the new mean. If n is the number of elements in the sequence after the addition of the new element, then one has
When the observations are weighted, West (1979)^{[3]} suggests this incremental algorithm:
def weighted_incremental_variance(dataWeightPairs): n = 0 mean = 0 S = 0 sumweight = 0 for x, weight in dataWeightPairs: # Alternately "for x in zip(data, weight):" n = n + 1 temp = weight + sumweight Q = x  mean R = Q * weight / temp S = S + sumweight * Q * R mean = mean + R sumweight = temp Variance = S * n / ((n1) * sumweight) # if sample is the population, omit n/(n1) return Variance
Chan et al.^{[4]} note that the above online algorithm III is a special case of an algorithm that works for any partition of the sample X into sets X^{A}, X^{B}:
This may be useful when, for example, multiple processing units may be assigned to discrete parts of the input.
Terriberry^{[5]} extends Chan's formulae to calculating the third and fourth central moments, needed for example when estimating skewness and kurtosis:
Here the M_{k} are again the sums of powers of differences from the mean , giving
For the incremental case (i.e., B = {x}), this simplifies to:
By preserving the value δ / n, only one division operation is needed and the higherorder statistics can thus be calculated for little incremental cost.
Pébay^{[6]} further extends these results to arbitraryorder central moments, for the incremental and the pairwise cases. One can also find there similar formulas for covariance.
Assume that all floating point operations use the standard IEEE 754 doubleprecision arithmetic. Consider the sample (4, 7, 13, 16) from an infinite population. Based on this sample, the estimated population mean is 10, and the unbiased estimate of population variance is 30. Both Algorithm I and Algorithm II compute these values correctly. Next consider the sample (10^{8} + 4, 10^{8} + 7, 10^{8} + 13, 10^{8} + 16), which gives rise to the same estimated variance as the first sample. Algorithm II computes this variance estimate correctly, but Algorithm I returns 29.333333333333332 instead of 30. While this loss of precision may be tolerable and viewed as a minor flaw of Algorithm I, it is easy to find data that reveal a major flaw in the naive algorithm: Take the sample to be (10^{9} + 4, 10^{9} + 7, 10^{9} + 13, 10^{9} + 16). Again the estimated population variance of 30 is computed correctly by Algorithm II, but the naive algorithm now computes it as −170.66666666666666. This is a serious problem with Algorithm I and is due to catastrophic cancellation in the subtraction of two similar numbers at the final stage of the algorithm.
