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

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?Is the mean smaller than the medianUsing the median for calculating VarianceIs the sample mean a better point estimate of the population median than the sample median?Higher variance in sample mean or sample median?Calculate variance from mean and variance of binsusing Mean or MedianHow to calculate the variance for the mean of means of binomial only i.d. variables?Mean and variance of the median estimatorCould you derive the mean and variance in this theorem?

7

1

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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

edited yesterday

Ferdi

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asked yesterday

WillWill

411

$endgroup$

• 
  
  
  

  
  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
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  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♦
  yesterday
  

  
  
  
  

add a comment | 

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

edited yesterday

Ferdi

3,86742355

asked yesterday

WillWill

411

$endgroup$

• 
  
  
  

  
  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
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  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♦
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

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

edited yesterday

Ferdi

3,86742355

asked yesterday

WillWill

411

$endgroup$

share|cite|improve this question

edited yesterday

Ferdi

3,86742355

asked yesterday

WillWill

411

share|cite|improve this question

edited yesterday

Ferdi

3,86742355

asked yesterday

WillWill

411

edited yesterday

Ferdi

3,86742355

edited yesterday

Ferdi

3,86742355

asked yesterday

WillWill

411

asked yesterday

WillWill

411

2 Answers
2

active

oldest

votes

11

$begingroup$

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
$$underbraceX-0_text0 is the median=arctan(1)/pi-arctan(-1)/pi=frac12$$

share|cite|improve this answer

edited yesterday

ukemi

1053

answered yesterday

Xi'anXi'an

58.9k897364

$endgroup$

• 
  
  
  

  
  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
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  MAD is Mutually Assured Destruction
  $endgroup$
  – kjetil b halvorsen
  yesterday
  

  
  
  
  

 | 
show 2 more comments

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 $overlinex^2 - overline x^2$, 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.

---

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

$$
2v_X = frac1n(n-1)sum_1 le i < j le n(x_i - x_j)^2.
$$

---

3. The sample variance without mean is calculated as:
   $$ v_X=frac1n-1left [ sum_i=1^nx_i^2-frac1nleft ( sum_i=1^nx_i right ) ^2right ] $$
   

share|cite|improve this answer

edited yesterday

answered yesterday

FerdiFerdi

3,86742355

$endgroup$

add a comment | 

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11

$begingroup$

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
$$underbraceX-0_text0 is the median=arctan(1)/pi-arctan(-1)/pi=frac12$$

share|cite|improve this answer

edited yesterday

ukemi

1053

answered yesterday

Xi'anXi'an

58.9k897364

$endgroup$

• 
  
  
  

  
  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
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  MAD is Mutually Assured Destruction
  $endgroup$
  – kjetil b halvorsen
  yesterday
  

  
  
  
  

 | 
show 2 more comments

11

$begingroup$

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
$$underbraceX-0_text0 is the median=arctan(1)/pi-arctan(-1)/pi=frac12$$

share|cite|improve this answer

edited yesterday

ukemi

1053

answered yesterday

Xi'anXi'an

58.9k897364

$endgroup$

• 
  
  
  

  
  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
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  MAD is Mutually Assured Destruction
  $endgroup$
  – kjetil b halvorsen
  yesterday
  

  
  
  
  

 | 
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
$$underbraceX-0_text0 is the median=arctan(1)/pi-arctan(-1)/pi=frac12$$

share|cite|improve this answer

edited yesterday

ukemi

1053

answered yesterday

Xi'anXi'an

58.9k897364

$endgroup$

share|cite|improve this answer

edited yesterday

ukemi

1053

answered yesterday

Xi'anXi'an

58.9k897364

edited yesterday

ukemi

1053

edited yesterday

ukemi

1053

answered yesterday

Xi'anXi'an

58.9k897364

answered yesterday

Xi'anXi'an

58.9k897364

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 $overlinex^2 - overline x^2$, 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.

---

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

$$
2v_X = frac1n(n-1)sum_1 le i < j le n(x_i - x_j)^2.
$$

---

3. The sample variance without mean is calculated as:
   $$ v_X=frac1n-1left [ sum_i=1^nx_i^2-frac1nleft ( sum_i=1^nx_i right ) ^2right ] $$
   

share|cite|improve this answer

edited yesterday

answered yesterday

FerdiFerdi

3,86742355

$endgroup$

add a comment | 

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 $overlinex^2 - overline x^2$, 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.

---

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

$$
2v_X = frac1n(n-1)sum_1 le i < j le n(x_i - x_j)^2.
$$

---

3. The sample variance without mean is calculated as:
   $$ v_X=frac1n-1left [ sum_i=1^nx_i^2-frac1nleft ( sum_i=1^nx_i right ) ^2right ] $$
   

share|cite|improve this answer

edited yesterday

answered yesterday

FerdiFerdi

3,86742355

$endgroup$

add a comment | 

$begingroup$

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

I summarize the answers:

1. 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.

---

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

$$
2v_X = frac1n(n-1)sum_1 le i < j le n(x_i - x_j)^2.
$$

---

3. The sample variance without mean is calculated as:
   $$ v_X=frac1n-1left [ sum_i=1^nx_i^2-frac1nleft ( sum_i=1^nx_i right ) ^2right ] $$
   

share|cite|improve this answer

edited yesterday

answered yesterday

FerdiFerdi

3,86742355

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

FerdiFerdi

3,86742355

answered yesterday

FerdiFerdi

3,86742355

answered yesterday

FerdiFerdi

3,86742355