Calculate variance without calculating the mean Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How would you explain covariance to someone who understands only the mean?When does a distribution not have a mean or a variance?How to calculate the variance for mean of means?Using the median for calculating VarianceHigher variance in sample mean or sample median?Calculate variance from mean and variance of binsHow do you use the CLT without knowing variance?Calculate the variance from variancesHow to calculate the variance for the mean of means of binomial only i.d. variables?Calculating new Mean and Variance after observation removalcalculating mean and variance of non-stationary time dependent samplesCalculating the variance of sample, knowing the mean of population
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Calculate variance without calculating the mean
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How would you explain covariance to someone who understands only the mean?When does a distribution not have a mean or a variance?How to calculate the variance for mean of means?Using the median for calculating VarianceHigher variance in sample mean or sample median?Calculate variance from mean and variance of binsHow do you use the CLT without knowing variance?Calculate the variance from variancesHow to calculate the variance for the mean of means of binomial only i.d. variables?Calculating new Mean and Variance after observation removalcalculating mean and variance of non-stationary time dependent samplesCalculating the variance of sample, knowing the mean of population
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Can we calculate the variance without using the mean as the 'base' point?
mathematical-statistics variance mean median scale-estimator
$endgroup$
add a comment |
$begingroup$
Can we calculate the variance without using the mean as the 'base' point?
mathematical-statistics variance mean median scale-estimator
$endgroup$
3
$begingroup$
Given $mathbbE(X^2)<infty$, the variance is given by $sigma^2 = mathbbE((X-mathbbE(X))^2)$ by definition. The formular simplifies to $sigma^2 =mathbbE(X^2) - mathbbE(X)^2$. I.e., for the variance you need $mathbbE(X)$. Of course you could define your own dispersion measure using some other statistic...or use one from the answers.
$endgroup$
– BloXX
Mar 26 at 7:56
5
$begingroup$
Short answer: Lots of other ways to summarize variability (dispersion, spread, scale) but none of the others would be the variance. (In fact, the variance can be defined without reference to the mean.)
$endgroup$
– Nick Cox
Mar 26 at 8:28
3
$begingroup$
Yes: given data $X,$ compute the covariance of $(X,X)$ as described at stats.stackexchange.com/a/18200/919. This method never computes the mean.
$endgroup$
– whuber♦
Mar 26 at 13:15
add a comment |
$begingroup$
Can we calculate the variance without using the mean as the 'base' point?
mathematical-statistics variance mean median scale-estimator
$endgroup$
Can we calculate the variance without using the mean as the 'base' point?
mathematical-statistics variance mean median scale-estimator
mathematical-statistics variance mean median scale-estimator
edited Apr 4 at 7:02
Ferdi
3,85952355
3,85952355
asked Mar 26 at 7:46
WillWill
411
411
3
$begingroup$
Given $mathbbE(X^2)<infty$, the variance is given by $sigma^2 = mathbbE((X-mathbbE(X))^2)$ by definition. The formular simplifies to $sigma^2 =mathbbE(X^2) - mathbbE(X)^2$. I.e., for the variance you need $mathbbE(X)$. Of course you could define your own dispersion measure using some other statistic...or use one from the answers.
$endgroup$
– BloXX
Mar 26 at 7:56
5
$begingroup$
Short answer: Lots of other ways to summarize variability (dispersion, spread, scale) but none of the others would be the variance. (In fact, the variance can be defined without reference to the mean.)
$endgroup$
– Nick Cox
Mar 26 at 8:28
3
$begingroup$
Yes: given data $X,$ compute the covariance of $(X,X)$ as described at stats.stackexchange.com/a/18200/919. This method never computes the mean.
$endgroup$
– whuber♦
Mar 26 at 13:15
add a comment |
3
$begingroup$
Given $mathbbE(X^2)<infty$, the variance is given by $sigma^2 = mathbbE((X-mathbbE(X))^2)$ by definition. The formular simplifies to $sigma^2 =mathbbE(X^2) - mathbbE(X)^2$. I.e., for the variance you need $mathbbE(X)$. Of course you could define your own dispersion measure using some other statistic...or use one from the answers.
$endgroup$
– BloXX
Mar 26 at 7:56
5
$begingroup$
Short answer: Lots of other ways to summarize variability (dispersion, spread, scale) but none of the others would be the variance. (In fact, the variance can be defined without reference to the mean.)
$endgroup$
– Nick Cox
Mar 26 at 8:28
3
$begingroup$
Yes: given data $X,$ compute the covariance of $(X,X)$ as described at stats.stackexchange.com/a/18200/919. This method never computes the mean.
$endgroup$
– whuber♦
Mar 26 at 13:15
3
3
$begingroup$
Given $mathbbE(X^2)<infty$, the variance is given by $sigma^2 = mathbbE((X-mathbbE(X))^2)$ by definition. The formular simplifies to $sigma^2 =mathbbE(X^2) - mathbbE(X)^2$. I.e., for the variance you need $mathbbE(X)$. Of course you could define your own dispersion measure using some other statistic...or use one from the answers.
$endgroup$
– BloXX
Mar 26 at 7:56
$begingroup$
Given $mathbbE(X^2)<infty$, the variance is given by $sigma^2 = mathbbE((X-mathbbE(X))^2)$ by definition. The formular simplifies to $sigma^2 =mathbbE(X^2) - mathbbE(X)^2$. I.e., for the variance you need $mathbbE(X)$. Of course you could define your own dispersion measure using some other statistic...or use one from the answers.
$endgroup$
– BloXX
Mar 26 at 7:56
5
5
$begingroup$
Short answer: Lots of other ways to summarize variability (dispersion, spread, scale) but none of the others would be the variance. (In fact, the variance can be defined without reference to the mean.)
$endgroup$
– Nick Cox
Mar 26 at 8:28
$begingroup$
Short answer: Lots of other ways to summarize variability (dispersion, spread, scale) but none of the others would be the variance. (In fact, the variance can be defined without reference to the mean.)
$endgroup$
– Nick Cox
Mar 26 at 8:28
3
3
$begingroup$
Yes: given data $X,$ compute the covariance of $(X,X)$ as described at stats.stackexchange.com/a/18200/919. This method never computes the mean.
$endgroup$
– whuber♦
Mar 26 at 13:15
$begingroup$
Yes: given data $X,$ compute the covariance of $(X,X)$ as described at stats.stackexchange.com/a/18200/919. This method never computes the mean.
$endgroup$
– whuber♦
Mar 26 at 13:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The median absolute deviation is defined as
$$textMAD(X) = textmedian |X-textmedian(X)|$$
and is considered an alternative to the standard deviation. But this is not the variance. In particular, it always exists, whether or not $X$ allows for moments. For instance, the MAD of a standard Cauchy is equal to one since
$$underbrace<1)_text0 is the median=arctan(1)/pi-arctan(-1)/pi=frac12$$
$endgroup$
7
$begingroup$
Newcomers to this idea should watch out also for mean absolute deviation from the mean (mean deviation, often) and median absolute deviation from the mean. I don't recall mean absolute deviation from the median, but am open to examples. The abbreviation MAD, unfortunately, has been applied variously, so trust people's code first, then their algebraic or verbal definition, but use of an abbreviation MAD only not at all. In symmetric distributions, and some others, MAD as defined here is half the interquartile range. (Punning on MAD I resist as a little too obvious.)
$endgroup$
– Nick Cox
Mar 26 at 8:23
3
$begingroup$
Also, note that software implementations of the median absolute deviation function can scale the MAD value by a constant factor from the form presented in this answer, so that its value coincides with the standard deviation for a normal distribution.
$endgroup$
– EdM
Mar 26 at 8:30
$begingroup$
@EdM Excellent point. Personally I dislike that practice unless people use some different term. It's no longer the MAD!
$endgroup$
– Nick Cox
Mar 26 at 8:35
1
$begingroup$
@NickCox: the appeal of centring on the median is that the quantity always exists, whether or not the distribution enjoys a mean. This is the definition found in Wikipedia.
$endgroup$
– Xi'an
Mar 26 at 9:20
$begingroup$
MAD is Mutually Assured Destruction
$endgroup$
– kjetil b halvorsen
Mar 26 at 10:05
|
show 2 more comments
$begingroup$
There is already a solution for this question on Math.stackexchange:
I summarize the answers:
- You can use that the variance is $overlinex^2 - overline x^2$, which takes only one pass (computing the mean and the mean of the squares simultaneously), but can be more prone to roundoff error if the variance is small compared with the mean.
- How about sum of squared pairwise differences ? Indeed, you can check by direct computation that
$$
2v_X = frac1n(n-1)sum_1 le i < j le n(x_i - x_j)^2.
$$
- The sample variance without mean is calculated as:
$$ v_X=frac1n-1left [ sum_i=1^nx_i^2-frac1nleft ( sum_i=1^nx_i right ) ^2right ] $$
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The median absolute deviation is defined as
$$textMAD(X) = textmedian |X-textmedian(X)|$$
and is considered an alternative to the standard deviation. But this is not the variance. In particular, it always exists, whether or not $X$ allows for moments. For instance, the MAD of a standard Cauchy is equal to one since
$$underbrace<1)_text0 is the median=arctan(1)/pi-arctan(-1)/pi=frac12$$
$endgroup$
7
$begingroup$
Newcomers to this idea should watch out also for mean absolute deviation from the mean (mean deviation, often) and median absolute deviation from the mean. I don't recall mean absolute deviation from the median, but am open to examples. The abbreviation MAD, unfortunately, has been applied variously, so trust people's code first, then their algebraic or verbal definition, but use of an abbreviation MAD only not at all. In symmetric distributions, and some others, MAD as defined here is half the interquartile range. (Punning on MAD I resist as a little too obvious.)
$endgroup$
– Nick Cox
Mar 26 at 8:23
3
$begingroup$
Also, note that software implementations of the median absolute deviation function can scale the MAD value by a constant factor from the form presented in this answer, so that its value coincides with the standard deviation for a normal distribution.
$endgroup$
– EdM
Mar 26 at 8:30
$begingroup$
@EdM Excellent point. Personally I dislike that practice unless people use some different term. It's no longer the MAD!
$endgroup$
– Nick Cox
Mar 26 at 8:35
1
$begingroup$
@NickCox: the appeal of centring on the median is that the quantity always exists, whether or not the distribution enjoys a mean. This is the definition found in Wikipedia.
$endgroup$
– Xi'an
Mar 26 at 9:20
$begingroup$
MAD is Mutually Assured Destruction
$endgroup$
– kjetil b halvorsen
Mar 26 at 10:05
|
show 2 more comments
$begingroup$
The median absolute deviation is defined as
$$textMAD(X) = textmedian |X-textmedian(X)|$$
and is considered an alternative to the standard deviation. But this is not the variance. In particular, it always exists, whether or not $X$ allows for moments. For instance, the MAD of a standard Cauchy is equal to one since
$$underbrace<1)_text0 is the median=arctan(1)/pi-arctan(-1)/pi=frac12$$
$endgroup$
7
$begingroup$
Newcomers to this idea should watch out also for mean absolute deviation from the mean (mean deviation, often) and median absolute deviation from the mean. I don't recall mean absolute deviation from the median, but am open to examples. The abbreviation MAD, unfortunately, has been applied variously, so trust people's code first, then their algebraic or verbal definition, but use of an abbreviation MAD only not at all. In symmetric distributions, and some others, MAD as defined here is half the interquartile range. (Punning on MAD I resist as a little too obvious.)
$endgroup$
– Nick Cox
Mar 26 at 8:23
3
$begingroup$
Also, note that software implementations of the median absolute deviation function can scale the MAD value by a constant factor from the form presented in this answer, so that its value coincides with the standard deviation for a normal distribution.
$endgroup$
– EdM
Mar 26 at 8:30
$begingroup$
@EdM Excellent point. Personally I dislike that practice unless people use some different term. It's no longer the MAD!
$endgroup$
– Nick Cox
Mar 26 at 8:35
1
$begingroup$
@NickCox: the appeal of centring on the median is that the quantity always exists, whether or not the distribution enjoys a mean. This is the definition found in Wikipedia.
$endgroup$
– Xi'an
Mar 26 at 9:20
$begingroup$
MAD is Mutually Assured Destruction
$endgroup$
– kjetil b halvorsen
Mar 26 at 10:05
|
show 2 more comments
$begingroup$
The median absolute deviation is defined as
$$textMAD(X) = textmedian |X-textmedian(X)|$$
and is considered an alternative to the standard deviation. But this is not the variance. In particular, it always exists, whether or not $X$ allows for moments. For instance, the MAD of a standard Cauchy is equal to one since
$$underbrace<1)_text0 is the median=arctan(1)/pi-arctan(-1)/pi=frac12$$
$endgroup$
The median absolute deviation is defined as
$$textMAD(X) = textmedian |X-textmedian(X)|$$
and is considered an alternative to the standard deviation. But this is not the variance. In particular, it always exists, whether or not $X$ allows for moments. For instance, the MAD of a standard Cauchy is equal to one since
$$underbrace<1)_text0 is the median=arctan(1)/pi-arctan(-1)/pi=frac12$$
edited Mar 26 at 12:13
ukemi
1053
1053
answered Mar 26 at 7:59
Xi'anXi'an
59.5k897368
59.5k897368
7
$begingroup$
Newcomers to this idea should watch out also for mean absolute deviation from the mean (mean deviation, often) and median absolute deviation from the mean. I don't recall mean absolute deviation from the median, but am open to examples. The abbreviation MAD, unfortunately, has been applied variously, so trust people's code first, then their algebraic or verbal definition, but use of an abbreviation MAD only not at all. In symmetric distributions, and some others, MAD as defined here is half the interquartile range. (Punning on MAD I resist as a little too obvious.)
$endgroup$
– Nick Cox
Mar 26 at 8:23
3
$begingroup$
Also, note that software implementations of the median absolute deviation function can scale the MAD value by a constant factor from the form presented in this answer, so that its value coincides with the standard deviation for a normal distribution.
$endgroup$
– EdM
Mar 26 at 8:30
$begingroup$
@EdM Excellent point. Personally I dislike that practice unless people use some different term. It's no longer the MAD!
$endgroup$
– Nick Cox
Mar 26 at 8:35
1
$begingroup$
@NickCox: the appeal of centring on the median is that the quantity always exists, whether or not the distribution enjoys a mean. This is the definition found in Wikipedia.
$endgroup$
– Xi'an
Mar 26 at 9:20
$begingroup$
MAD is Mutually Assured Destruction
$endgroup$
– kjetil b halvorsen
Mar 26 at 10:05
|
show 2 more comments
7
$begingroup$
Newcomers to this idea should watch out also for mean absolute deviation from the mean (mean deviation, often) and median absolute deviation from the mean. I don't recall mean absolute deviation from the median, but am open to examples. The abbreviation MAD, unfortunately, has been applied variously, so trust people's code first, then their algebraic or verbal definition, but use of an abbreviation MAD only not at all. In symmetric distributions, and some others, MAD as defined here is half the interquartile range. (Punning on MAD I resist as a little too obvious.)
$endgroup$
– Nick Cox
Mar 26 at 8:23
3
$begingroup$
Also, note that software implementations of the median absolute deviation function can scale the MAD value by a constant factor from the form presented in this answer, so that its value coincides with the standard deviation for a normal distribution.
$endgroup$
– EdM
Mar 26 at 8:30
$begingroup$
@EdM Excellent point. Personally I dislike that practice unless people use some different term. It's no longer the MAD!
$endgroup$
– Nick Cox
Mar 26 at 8:35
1
$begingroup$
@NickCox: the appeal of centring on the median is that the quantity always exists, whether or not the distribution enjoys a mean. This is the definition found in Wikipedia.
$endgroup$
– Xi'an
Mar 26 at 9:20
$begingroup$
MAD is Mutually Assured Destruction
$endgroup$
– kjetil b halvorsen
Mar 26 at 10:05
7
7
$begingroup$
Newcomers to this idea should watch out also for mean absolute deviation from the mean (mean deviation, often) and median absolute deviation from the mean. I don't recall mean absolute deviation from the median, but am open to examples. The abbreviation MAD, unfortunately, has been applied variously, so trust people's code first, then their algebraic or verbal definition, but use of an abbreviation MAD only not at all. In symmetric distributions, and some others, MAD as defined here is half the interquartile range. (Punning on MAD I resist as a little too obvious.)
$endgroup$
– Nick Cox
Mar 26 at 8:23
$begingroup$
Newcomers to this idea should watch out also for mean absolute deviation from the mean (mean deviation, often) and median absolute deviation from the mean. I don't recall mean absolute deviation from the median, but am open to examples. The abbreviation MAD, unfortunately, has been applied variously, so trust people's code first, then their algebraic or verbal definition, but use of an abbreviation MAD only not at all. In symmetric distributions, and some others, MAD as defined here is half the interquartile range. (Punning on MAD I resist as a little too obvious.)
$endgroup$
– Nick Cox
Mar 26 at 8:23
3
3
$begingroup$
Also, note that software implementations of the median absolute deviation function can scale the MAD value by a constant factor from the form presented in this answer, so that its value coincides with the standard deviation for a normal distribution.
$endgroup$
– EdM
Mar 26 at 8:30
$begingroup$
Also, note that software implementations of the median absolute deviation function can scale the MAD value by a constant factor from the form presented in this answer, so that its value coincides with the standard deviation for a normal distribution.
$endgroup$
– EdM
Mar 26 at 8:30
$begingroup$
@EdM Excellent point. Personally I dislike that practice unless people use some different term. It's no longer the MAD!
$endgroup$
– Nick Cox
Mar 26 at 8:35
$begingroup$
@EdM Excellent point. Personally I dislike that practice unless people use some different term. It's no longer the MAD!
$endgroup$
– Nick Cox
Mar 26 at 8:35
1
1
$begingroup$
@NickCox: the appeal of centring on the median is that the quantity always exists, whether or not the distribution enjoys a mean. This is the definition found in Wikipedia.
$endgroup$
– Xi'an
Mar 26 at 9:20
$begingroup$
@NickCox: the appeal of centring on the median is that the quantity always exists, whether or not the distribution enjoys a mean. This is the definition found in Wikipedia.
$endgroup$
– Xi'an
Mar 26 at 9:20
$begingroup$
MAD is Mutually Assured Destruction
$endgroup$
– kjetil b halvorsen
Mar 26 at 10:05
$begingroup$
MAD is Mutually Assured Destruction
$endgroup$
– kjetil b halvorsen
Mar 26 at 10:05
|
show 2 more comments
$begingroup$
There is already a solution for this question on Math.stackexchange:
I summarize the answers:
- You can use that the variance is $overlinex^2 - overline x^2$, which takes only one pass (computing the mean and the mean of the squares simultaneously), but can be more prone to roundoff error if the variance is small compared with the mean.
- How about sum of squared pairwise differences ? Indeed, you can check by direct computation that
$$
2v_X = frac1n(n-1)sum_1 le i < j le n(x_i - x_j)^2.
$$
- The sample variance without mean is calculated as:
$$ v_X=frac1n-1left [ sum_i=1^nx_i^2-frac1nleft ( sum_i=1^nx_i right ) ^2right ] $$
$endgroup$
add a comment |
$begingroup$
There is already a solution for this question on Math.stackexchange:
I summarize the answers:
- You can use that the variance is $overlinex^2 - overline x^2$, which takes only one pass (computing the mean and the mean of the squares simultaneously), but can be more prone to roundoff error if the variance is small compared with the mean.
- How about sum of squared pairwise differences ? Indeed, you can check by direct computation that
$$
2v_X = frac1n(n-1)sum_1 le i < j le n(x_i - x_j)^2.
$$
- The sample variance without mean is calculated as:
$$ v_X=frac1n-1left [ sum_i=1^nx_i^2-frac1nleft ( sum_i=1^nx_i right ) ^2right ] $$
$endgroup$
add a comment |
$begingroup$
There is already a solution for this question on Math.stackexchange:
I summarize the answers:
- You can use that the variance is $overlinex^2 - overline x^2$, which takes only one pass (computing the mean and the mean of the squares simultaneously), but can be more prone to roundoff error if the variance is small compared with the mean.
- How about sum of squared pairwise differences ? Indeed, you can check by direct computation that
$$
2v_X = frac1n(n-1)sum_1 le i < j le n(x_i - x_j)^2.
$$
- The sample variance without mean is calculated as:
$$ v_X=frac1n-1left [ sum_i=1^nx_i^2-frac1nleft ( sum_i=1^nx_i right ) ^2right ] $$
$endgroup$
There is already a solution for this question on Math.stackexchange:
I summarize the answers:
- You can use that the variance is $overlinex^2 - overline x^2$, which takes only one pass (computing the mean and the mean of the squares simultaneously), but can be more prone to roundoff error if the variance is small compared with the mean.
- How about sum of squared pairwise differences ? Indeed, you can check by direct computation that
$$
2v_X = frac1n(n-1)sum_1 le i < j le n(x_i - x_j)^2.
$$
- The sample variance without mean is calculated as:
$$ v_X=frac1n-1left [ sum_i=1^nx_i^2-frac1nleft ( sum_i=1^nx_i right ) ^2right ] $$
edited Mar 26 at 17:08
answered Mar 26 at 16:48
FerdiFerdi
3,85952355
3,85952355
add a comment |
add a comment |
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Given $mathbbE(X^2)<infty$, the variance is given by $sigma^2 = mathbbE((X-mathbbE(X))^2)$ by definition. The formular simplifies to $sigma^2 =mathbbE(X^2) - mathbbE(X)^2$. I.e., for the variance you need $mathbbE(X)$. Of course you could define your own dispersion measure using some other statistic...or use one from the answers.
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– BloXX
Mar 26 at 7:56
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Short answer: Lots of other ways to summarize variability (dispersion, spread, scale) but none of the others would be the variance. (In fact, the variance can be defined without reference to the mean.)
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– Nick Cox
Mar 26 at 8:28
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Yes: given data $X,$ compute the covariance of $(X,X)$ as described at stats.stackexchange.com/a/18200/919. This method never computes the mean.
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– whuber♦
Mar 26 at 13:15