Could you calculate the variance of data using the median or something other than the mean?












7












$begingroup$


Could we calculate the variance without using the mean as the 'base' point?






mathematical-statistics variance mean median scale-estimator







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Given $mathbb{E}(X^2)<infty$, the variance is given by $sigma^2 = mathbb{E}((X-mathbb{E}(X))^2)$ by definition. The formular simplifies to $sigma^2 =mathbb{E}(X^2) - mathbb{E}(X)^2$. I.e., for the variance you need $mathbb{E}(X)$. Of course you could define your own dispersion measure using some other statistic...or use one from the answers.
    $endgroup$
    – BloXX
    2 days ago








  • 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
    2 days ago






  • 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♦
    2 days ago
















7












$begingroup$


Could we calculate the variance without using the mean as the 'base' point?






mathematical-statistics variance mean median scale-estimator







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Given $mathbb{E}(X^2)<infty$, the variance is given by $sigma^2 = mathbb{E}((X-mathbb{E}(X))^2)$ by definition. The formular simplifies to $sigma^2 =mathbb{E}(X^2) - mathbb{E}(X)^2$. I.e., for the variance you need $mathbb{E}(X)$. Of course you could define your own dispersion measure using some other statistic...or use one from the answers.
    $endgroup$
    – BloXX
    2 days ago








  • 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
    2 days ago






  • 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♦
    2 days ago














7












7








7


1



$begingroup$


Could we calculate the variance without using the mean as the 'base' point?






mathematical-statistics variance mean median scale-estimator







share|cite|improve this question











$endgroup$




Could 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Ferdi

3,86742355




3,86742355










asked 2 days ago









WillWill

411




411








  • 3




    $begingroup$
    Given $mathbb{E}(X^2)<infty$, the variance is given by $sigma^2 = mathbb{E}((X-mathbb{E}(X))^2)$ by definition. The formular simplifies to $sigma^2 =mathbb{E}(X^2) - mathbb{E}(X)^2$. I.e., for the variance you need $mathbb{E}(X)$. Of course you could define your own dispersion measure using some other statistic...or use one from the answers.
    $endgroup$
    – BloXX
    2 days ago








  • 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
    2 days ago






  • 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♦
    2 days ago














  • 3




    $begingroup$
    Given $mathbb{E}(X^2)<infty$, the variance is given by $sigma^2 = mathbb{E}((X-mathbb{E}(X))^2)$ by definition. The formular simplifies to $sigma^2 =mathbb{E}(X^2) - mathbb{E}(X)^2$. I.e., for the variance you need $mathbb{E}(X)$. Of course you could define your own dispersion measure using some other statistic...or use one from the answers.
    $endgroup$
    – BloXX
    2 days ago








  • 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
    2 days ago






  • 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♦
    2 days ago








3




3




$begingroup$
Given $mathbb{E}(X^2)<infty$, the variance is given by $sigma^2 = mathbb{E}((X-mathbb{E}(X))^2)$ by definition. The formular simplifies to $sigma^2 =mathbb{E}(X^2) - mathbb{E}(X)^2$. I.e., for the variance you need $mathbb{E}(X)$. Of course you could define your own dispersion measure using some other statistic...or use one from the answers.
$endgroup$
– BloXX
2 days ago






$begingroup$
Given $mathbb{E}(X^2)<infty$, the variance is given by $sigma^2 = mathbb{E}((X-mathbb{E}(X))^2)$ by definition. The formular simplifies to $sigma^2 =mathbb{E}(X^2) - mathbb{E}(X)^2$. I.e., for the variance you need $mathbb{E}(X)$. Of course you could define your own dispersion measure using some other statistic...or use one from the answers.
$endgroup$
– BloXX
2 days ago






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
2 days ago




$begingroup$
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
2 days ago




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♦
2 days ago




$begingroup$
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♦
2 days ago










2 Answers
2






active

oldest

votes


















11












$begingroup$

The median absolute deviation is defined as
$$text{MAD}(X) = text{median} |X-text{median}(X)|$$
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
$$underbrace{Bbb P(|X-0|<1)}_text{0 is the median}=arctan(1)/pi-arctan(-1)/pi=frac{1}{2}$$






share|cite|improve this answer











$endgroup$









  • 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
    2 days ago








  • 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
    2 days ago










  • $begingroup$
    @EdM Excellent point. Personally I dislike that practice unless people use some different term. It's no longer the MAD!
    $endgroup$
    – Nick Cox
    2 days ago






  • 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
    2 days ago










  • $begingroup$
    MAD is Mutually Assured Destruction
    $endgroup$
    – kjetil b halvorsen
    2 days ago



















3












$begingroup$

There is already a solution for this question on Math.stackexchange:



I summarize the answers:




  1. You can use that the variance is $overline{x^2} - overline {x}^2$, 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.




  1. How about sum of squared pairwise differences ? Indeed, you can check by direct computation that


$$
2v_X = frac{1}{n(n-1)}sum_{1 le i < j le n}(x_i - x_j)^2.
$$





  1. The sample variance without mean is calculated as:
    $$ v_{X}=frac{1}{n-1}left [ sum_{i=1}^{n}x_{i}^{2}-frac{1}{n}left ( sum_{i=1}^{n}x_{i} right ) ^{2}right ] $$






share|cite|improve this answer











$endgroup$














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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    11












    $begingroup$

    The median absolute deviation is defined as
    $$text{MAD}(X) = text{median} |X-text{median}(X)|$$
    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
    $$underbrace{Bbb P(|X-0|<1)}_text{0 is the median}=arctan(1)/pi-arctan(-1)/pi=frac{1}{2}$$






    share|cite|improve this answer











    $endgroup$









    • 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
      2 days ago








    • 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
      2 days ago










    • $begingroup$
      @EdM Excellent point. Personally I dislike that practice unless people use some different term. It's no longer the MAD!
      $endgroup$
      – Nick Cox
      2 days ago






    • 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
      2 days ago










    • $begingroup$
      MAD is Mutually Assured Destruction
      $endgroup$
      – kjetil b halvorsen
      2 days ago
















    11












    $begingroup$

    The median absolute deviation is defined as
    $$text{MAD}(X) = text{median} |X-text{median}(X)|$$
    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
    $$underbrace{Bbb P(|X-0|<1)}_text{0 is the median}=arctan(1)/pi-arctan(-1)/pi=frac{1}{2}$$






    share|cite|improve this answer











    $endgroup$









    • 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
      2 days ago








    • 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
      2 days ago










    • $begingroup$
      @EdM Excellent point. Personally I dislike that practice unless people use some different term. It's no longer the MAD!
      $endgroup$
      – Nick Cox
      2 days ago






    • 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
      2 days ago










    • $begingroup$
      MAD is Mutually Assured Destruction
      $endgroup$
      – kjetil b halvorsen
      2 days ago














    11












    11








    11





    $begingroup$

    The median absolute deviation is defined as
    $$text{MAD}(X) = text{median} |X-text{median}(X)|$$
    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
    $$underbrace{Bbb P(|X-0|<1)}_text{0 is the median}=arctan(1)/pi-arctan(-1)/pi=frac{1}{2}$$






    share|cite|improve this answer











    $endgroup$



    The median absolute deviation is defined as
    $$text{MAD}(X) = text{median} |X-text{median}(X)|$$
    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
    $$underbrace{Bbb P(|X-0|<1)}_text{0 is the median}=arctan(1)/pi-arctan(-1)/pi=frac{1}{2}$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 2 days ago









    ukemi

    1053




    1053










    answered 2 days ago









    Xi'anXi'an

    58.9k897364




    58.9k897364








    • 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
      2 days ago








    • 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
      2 days ago










    • $begingroup$
      @EdM Excellent point. Personally I dislike that practice unless people use some different term. It's no longer the MAD!
      $endgroup$
      – Nick Cox
      2 days ago






    • 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
      2 days ago










    • $begingroup$
      MAD is Mutually Assured Destruction
      $endgroup$
      – kjetil b halvorsen
      2 days ago














    • 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
      2 days ago








    • 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
      2 days ago










    • $begingroup$
      @EdM Excellent point. Personally I dislike that practice unless people use some different term. It's no longer the MAD!
      $endgroup$
      – Nick Cox
      2 days ago






    • 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
      2 days ago










    • $begingroup$
      MAD is Mutually Assured Destruction
      $endgroup$
      – kjetil b halvorsen
      2 days ago








    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
    2 days ago






    $begingroup$
    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
    2 days ago






    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
    2 days ago




    $begingroup$
    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
    2 days ago












    $begingroup$
    @EdM Excellent point. Personally I dislike that practice unless people use some different term. It's no longer the MAD!
    $endgroup$
    – Nick Cox
    2 days ago




    $begingroup$
    @EdM Excellent point. Personally I dislike that practice unless people use some different term. It's no longer the MAD!
    $endgroup$
    – Nick Cox
    2 days ago




    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
    2 days ago




    $begingroup$
    @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
    2 days ago












    $begingroup$
    MAD is Mutually Assured Destruction
    $endgroup$
    – kjetil b halvorsen
    2 days ago




    $begingroup$
    MAD is Mutually Assured Destruction
    $endgroup$
    – kjetil b halvorsen
    2 days ago













    3












    $begingroup$

    There is already a solution for this question on Math.stackexchange:



    I summarize the answers:




    1. You can use that the variance is $overline{x^2} - overline {x}^2$, 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.




    1. How about sum of squared pairwise differences ? Indeed, you can check by direct computation that


    $$
    2v_X = frac{1}{n(n-1)}sum_{1 le i < j le n}(x_i - x_j)^2.
    $$





    1. The sample variance without mean is calculated as:
      $$ v_{X}=frac{1}{n-1}left [ sum_{i=1}^{n}x_{i}^{2}-frac{1}{n}left ( sum_{i=1}^{n}x_{i} right ) ^{2}right ] $$






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      There is already a solution for this question on Math.stackexchange:



      I summarize the answers:




      1. You can use that the variance is $overline{x^2} - overline {x}^2$, 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.




      1. How about sum of squared pairwise differences ? Indeed, you can check by direct computation that


      $$
      2v_X = frac{1}{n(n-1)}sum_{1 le i < j le n}(x_i - x_j)^2.
      $$





      1. The sample variance without mean is calculated as:
        $$ v_{X}=frac{1}{n-1}left [ sum_{i=1}^{n}x_{i}^{2}-frac{1}{n}left ( sum_{i=1}^{n}x_{i} right ) ^{2}right ] $$






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        There is already a solution for this question on Math.stackexchange:



        I summarize the answers:




        1. You can use that the variance is $overline{x^2} - overline {x}^2$, 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.




        1. How about sum of squared pairwise differences ? Indeed, you can check by direct computation that


        $$
        2v_X = frac{1}{n(n-1)}sum_{1 le i < j le n}(x_i - x_j)^2.
        $$





        1. The sample variance without mean is calculated as:
          $$ v_{X}=frac{1}{n-1}left [ sum_{i=1}^{n}x_{i}^{2}-frac{1}{n}left ( sum_{i=1}^{n}x_{i} right ) ^{2}right ] $$






        share|cite|improve this answer











        $endgroup$



        There is already a solution for this question on Math.stackexchange:



        I summarize the answers:




        1. You can use that the variance is $overline{x^2} - overline {x}^2$, 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.




        1. How about sum of squared pairwise differences ? Indeed, you can check by direct computation that


        $$
        2v_X = frac{1}{n(n-1)}sum_{1 le i < j le n}(x_i - x_j)^2.
        $$





        1. The sample variance without mean is calculated as:
          $$ v_{X}=frac{1}{n-1}left [ sum_{i=1}^{n}x_{i}^{2}-frac{1}{n}left ( sum_{i=1}^{n}x_{i} right ) ^{2}right ] $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago

























        answered 2 days ago









        FerdiFerdi

        3,86742355




        3,86742355






























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