Calculate variance from a stream of sample values
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?
statistics algorithms
asked Feb 5 '11 at 21:18
user6677
173115
add a comment |
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?
statistics algorithms
asked Feb 5 '11 at 21:18
user6677
173115
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.
– Dilip Sarwate
Mar 4 '12 at 17:04
add a comment |
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?
statistics algorithms
asked Feb 5 '11 at 21:18
user6677
173115
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?
statistics algorithms
statistics algorithms
asked Feb 5 '11 at 21:18
user6677
173115
asked Feb 5 '11 at 21:18
user6677
173115
asked Feb 5 '11 at 21:18
user6677
173115
asked Feb 5 '11 at 21:18
user6677
173115
asked Feb 5 '11 at 21:18
user6677
173115
173115
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.
– Dilip Sarwate
Mar 4 '12 at 17:04
add a comment |
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.
– Dilip Sarwate
Mar 4 '12 at 17:04
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.
– Dilip Sarwate
Mar 4 '12 at 17:04
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.
– Dilip Sarwate
Mar 4 '12 at 17:04
add a comment |
2 Answers
2
active
oldest
votes
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.
answered Feb 5 '11 at 21:35
Dinesh
2,1691923
1
Wait -- shouldn't the N be inside the the expected values, such that the right side is divided by $N^2$?
– user6677
Feb 5 '11 at 22:05
e.g. $$ sigma^2 = frac{(sum_i x_i^2)}{N} - (frac{sum_i x_i}{N})^2 $$
– user6677
Feb 5 '11 at 22:27
@user6677: You are indeed right. Thanks for the correction.
– Dinesh
Feb 5 '11 at 22:30
5
Note the sum of squares especially is at risk of overflow.
– dfrankow
Dec 29 '14 at 18:06
2
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).
– Leopd
Feb 18 '15 at 1:40
|
show 2 more comments
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.
edited Mar 4 '12 at 20:03
Ilmari Karonen
19.3k25182
answered Mar 4 '12 at 15:59
Dan Lecocq
44143
2
+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.
– Ilmari Karonen
Mar 4 '12 at 20:12
1
Isn't the final variance calculation should be Vk / k (not k-1)?
– Eyal Roth
Oct 10 '13 at 15:49
@errr: it has to do with using biased vs unbiased estimator
– alexey
Mar 4 '16 at 2:03
1
$v_k = sum_i^k (x_i-m_k)^2$, so $v_k = v_{k-1} + x_k^2 - frac{(k-1)m_k^2 - km_{k-1}^2}{k(k-1)}$
– GuLearn
May 5 '17 at 2:02
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.
– thomas.han
Jun 30 at 16:58
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Feb 5 '11 at 21:35
Dinesh
2,1691923
1
Wait -- shouldn't the N be inside the the expected values, such that the right side is divided by $N^2$?
– user6677
Feb 5 '11 at 22:05
e.g. $$ sigma^2 = frac{(sum_i x_i^2)}{N} - (frac{sum_i x_i}{N})^2 $$
– user6677
Feb 5 '11 at 22:27
@user6677: You are indeed right. Thanks for the correction.
– Dinesh
Feb 5 '11 at 22:30
5
Note the sum of squares especially is at risk of overflow.
– dfrankow
Dec 29 '14 at 18:06
2
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).
– Leopd
Feb 18 '15 at 1:40
|
show 2 more comments
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.
answered Feb 5 '11 at 21:35
Dinesh
2,1691923
1
Wait -- shouldn't the N be inside the the expected values, such that the right side is divided by $N^2$?
– user6677
Feb 5 '11 at 22:05
e.g. $$ sigma^2 = frac{(sum_i x_i^2)}{N} - (frac{sum_i x_i}{N})^2 $$
– user6677
Feb 5 '11 at 22:27
@user6677: You are indeed right. Thanks for the correction.
– Dinesh
Feb 5 '11 at 22:30
5
Note the sum of squares especially is at risk of overflow.
– dfrankow
Dec 29 '14 at 18:06
2
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).
– Leopd
Feb 18 '15 at 1:40
|
show 2 more comments
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.
answered Feb 5 '11 at 21:35
Dinesh
2,1691923
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.
answered Feb 5 '11 at 21:35
Dinesh
2,1691923
edited Feb 5 '11 at 22:31
answered Feb 5 '11 at 21:35
Dinesh
2,1691923
answered Feb 5 '11 at 21:35
Dinesh
2,1691923
answered Feb 5 '11 at 21:35
Dinesh
2,1691923
2,1691923
1
Wait -- shouldn't the N be inside the the expected values, such that the right side is divided by $N^2$?
– user6677
Feb 5 '11 at 22:05
e.g. $$ sigma^2 = frac{(sum_i x_i^2)}{N} - (frac{sum_i x_i}{N})^2 $$
– user6677
Feb 5 '11 at 22:27
@user6677: You are indeed right. Thanks for the correction.
– Dinesh
Feb 5 '11 at 22:30
5
Note the sum of squares especially is at risk of overflow.
– dfrankow
Dec 29 '14 at 18:06
2
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).
– Leopd
Feb 18 '15 at 1:40
|
show 2 more comments
1
Wait -- shouldn't the N be inside the the expected values, such that the right side is divided by $N^2$?
– user6677
Feb 5 '11 at 22:05
e.g. $$ sigma^2 = frac{(sum_i x_i^2)}{N} - (frac{sum_i x_i}{N})^2 $$
– user6677
Feb 5 '11 at 22:27
@user6677: You are indeed right. Thanks for the correction.
– Dinesh
Feb 5 '11 at 22:30
5
Note the sum of squares especially is at risk of overflow.
– dfrankow
Dec 29 '14 at 18:06
2
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).
– Leopd
Feb 18 '15 at 1:40
1
1
Wait -- shouldn't the N be inside the the expected values, such that the right side is divided by $N^2$?
– user6677
Feb 5 '11 at 22:05
Wait -- shouldn't the N be inside the the expected values, such that the right side is divided by $N^2$?
– user6677
Feb 5 '11 at 22:05
e.g. $$ sigma^2 = frac{(sum_i x_i^2)}{N} - (frac{sum_i x_i}{N})^2 $$
– user6677
Feb 5 '11 at 22:27
e.g. $$ sigma^2 = frac{(sum_i x_i^2)}{N} - (frac{sum_i x_i}{N})^2 $$
– user6677
Feb 5 '11 at 22:27
@user6677: You are indeed right. Thanks for the correction.
– Dinesh
Feb 5 '11 at 22:30
@user6677: You are indeed right. Thanks for the correction.
– Dinesh
Feb 5 '11 at 22:30
5
5
Note the sum of squares especially is at risk of overflow.
– dfrankow
Dec 29 '14 at 18:06
Note the sum of squares especially is at risk of overflow.
– dfrankow
Dec 29 '14 at 18:06
2
2
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).
– Leopd
Feb 18 '15 at 1:40
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).
– Leopd
Feb 18 '15 at 1:40
|
show 2 more comments
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.
edited Mar 4 '12 at 20:03
Ilmari Karonen
19.3k25182
answered Mar 4 '12 at 15:59
Dan Lecocq
44143
2
+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.
– Ilmari Karonen
Mar 4 '12 at 20:12
1
Isn't the final variance calculation should be Vk / k (not k-1)?
– Eyal Roth
Oct 10 '13 at 15:49
@errr: it has to do with using biased vs unbiased estimator
– alexey
Mar 4 '16 at 2:03
1
$v_k = sum_i^k (x_i-m_k)^2$, so $v_k = v_{k-1} + x_k^2 - frac{(k-1)m_k^2 - km_{k-1}^2}{k(k-1)}$
– GuLearn
May 5 '17 at 2:02
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.
– thomas.han
Jun 30 at 16:58
add a comment |
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.
edited Mar 4 '12 at 20:03
Ilmari Karonen
19.3k25182
answered Mar 4 '12 at 15:59
Dan Lecocq
44143
2
+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.
– Ilmari Karonen
Mar 4 '12 at 20:12
1
Isn't the final variance calculation should be Vk / k (not k-1)?
– Eyal Roth
Oct 10 '13 at 15:49
@errr: it has to do with using biased vs unbiased estimator
– alexey
Mar 4 '16 at 2:03
1
$v_k = sum_i^k (x_i-m_k)^2$, so $v_k = v_{k-1} + x_k^2 - frac{(k-1)m_k^2 - km_{k-1}^2}{k(k-1)}$
– GuLearn
May 5 '17 at 2:02
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.
– thomas.han
Jun 30 at 16:58
add a comment |
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.
edited Mar 4 '12 at 20:03
Ilmari Karonen
19.3k25182
answered Mar 4 '12 at 15:59
Dan Lecocq
44143
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.
edited Mar 4 '12 at 20:03
Ilmari Karonen
19.3k25182
answered Mar 4 '12 at 15:59
Dan Lecocq
44143
edited Mar 4 '12 at 20:03
Ilmari Karonen
19.3k25182
edited Mar 4 '12 at 20:03
Ilmari Karonen
19.3k25182
edited Mar 4 '12 at 20:03
Ilmari Karonen
19.3k25182
19.3k25182
answered Mar 4 '12 at 15:59
Dan Lecocq
44143
answered Mar 4 '12 at 15:59
Dan Lecocq
44143
answered Mar 4 '12 at 15:59
Dan Lecocq
44143
44143
2
+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.
– Ilmari Karonen
Mar 4 '12 at 20:12
1
Isn't the final variance calculation should be Vk / k (not k-1)?
– Eyal Roth
Oct 10 '13 at 15:49
@errr: it has to do with using biased vs unbiased estimator
– alexey
Mar 4 '16 at 2:03
1
$v_k = sum_i^k (x_i-m_k)^2$, so $v_k = v_{k-1} + x_k^2 - frac{(k-1)m_k^2 - km_{k-1}^2}{k(k-1)}$
– GuLearn
May 5 '17 at 2:02
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.
– thomas.han
Jun 30 at 16:58
add a comment |
2
+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.
– Ilmari Karonen
Mar 4 '12 at 20:12
1
Isn't the final variance calculation should be Vk / k (not k-1)?
– Eyal Roth
Oct 10 '13 at 15:49
@errr: it has to do with using biased vs unbiased estimator
– alexey
Mar 4 '16 at 2:03
1
$v_k = sum_i^k (x_i-m_k)^2$, so $v_k = v_{k-1} + x_k^2 - frac{(k-1)m_k^2 - km_{k-1}^2}{k(k-1)}$
– GuLearn
May 5 '17 at 2:02
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.
– thomas.han
Jun 30 at 16:58
2
2
+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.
– Ilmari Karonen
Mar 4 '12 at 20:12
+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.
– Ilmari Karonen
Mar 4 '12 at 20:12
1
1
Isn't the final variance calculation should be Vk / k (not k-1)?
– Eyal Roth
Oct 10 '13 at 15:49
Isn't the final variance calculation should be Vk / k (not k-1)?
– Eyal Roth
Oct 10 '13 at 15:49
@errr: it has to do with using biased vs unbiased estimator
– alexey
Mar 4 '16 at 2:03
@errr: it has to do with using biased vs unbiased estimator
– alexey
Mar 4 '16 at 2:03
1
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$v_k = sum_i^k (x_i-m_k)^2$, so $v_k = v_{k-1} + x_k^2 - frac{(k-1)m_k^2 - km_{k-1}^2}{k(k-1)}$
– GuLearn
May 5 '17 at 2:02
$v_k = sum_i^k (x_i-m_k)^2$, so $v_k = v_{k-1} + x_k^2 - frac{(k-1)m_k^2 - km_{k-1}^2}{k(k-1)}$
– GuLearn
May 5 '17 at 2:02
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.
– thomas.han
Jun 30 at 16:58
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.
– thomas.han
Jun 30 at 16:58
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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.
– Dilip Sarwate
Mar 4 '12 at 17:04