Is it possible to compute the variance without computing the mean first?
$begingroup$
I have a list of values of a random variable $x in mathbb R$. Is it possible to find the varience $overline{(x - overline x)^2}$ without computing the mean $overline x$ first? That is to process the list only once.
probability statistics
asked Jan 24 '12 at 16:39
YrogirgYrogirg
1,7361545
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add a comment |
$begingroup$
I have a list of values of a random variable $x in mathbb R$. Is it possible to find the varience $overline{(x - overline x)^2}$ without computing the mean $overline x$ first? That is to process the list only once.
probability statistics
asked Jan 24 '12 at 16:39
YrogirgYrogirg
1,7361545
$endgroup$
add a comment |
$begingroup$
I have a list of values of a random variable $x in mathbb R$. Is it possible to find the varience $overline{(x - overline x)^2}$ without computing the mean $overline x$ first? That is to process the list only once.
probability statistics
asked Jan 24 '12 at 16:39
YrogirgYrogirg
1,7361545
$endgroup$
I have a list of values of a random variable $x in mathbb R$. Is it possible to find the varience $overline{(x - overline x)^2}$ without computing the mean $overline x$ first? That is to process the list only once.
probability statistics
probability statistics
asked Jan 24 '12 at 16:39
YrogirgYrogirg
1,7361545
asked Jan 24 '12 at 16:39
YrogirgYrogirg
1,7361545
edited Jan 24 '12 at 16:42
user21436
asked Jan 24 '12 at 16:39
YrogirgYrogirg
1,7361545
asked Jan 24 '12 at 16:39
YrogirgYrogirg
1,7361545
asked Jan 24 '12 at 16:39
YrogirgYrogirg
1,7361545
1,7361545
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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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.
answered Jan 24 '12 at 16:53
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
oh, how could I forget this.
$endgroup$
– Yrogirg
Jan 24 '12 at 17:24
$begingroup$
And should have stated from wikipedia en.wikipedia.org/wiki/Algorithms_for_calculating_variance At first I thought it would be impossible and there would be no such algorithm.
$endgroup$
– Yrogirg
Jan 24 '12 at 18:48
add a comment |
$begingroup$
How about sum of squared pairwise differences ? Indeed, you can check by direct computation that
$$
2v_X = frac{1}{n(n-1)}sum_{1 le i < j le n}(x_i - x_j)^2.
$$
Disclaimer: This is not meant to be efficient, just another way to represent the variance, without first computing the mean.
answered Dec 14 '18 at 7:10
dohmatobdohmatob
3,672629
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add a comment |
$begingroup$
The sample variance with mean is calculated as:
$$ v_{X}=frac{1}{n-1}sum_{i=1}^{n}(x_{i}-overline{x})^{2} $$
And the sample variance without mean as:
$$ v_{X}=frac{1}{n-1}left [ sum_{i=1}^{n}x_{i}^{2}-frac{1}{n}left ( sum_{i=1}^{n}x_{i} right ) ^{2}right ] $$
answered Sep 3 '17 at 16:34
FreemanFreeman
1184
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Jan 24 '12 at 16:53
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
oh, how could I forget this.
$endgroup$
– Yrogirg
Jan 24 '12 at 17:24
$begingroup$
And should have stated from wikipedia en.wikipedia.org/wiki/Algorithms_for_calculating_variance At first I thought it would be impossible and there would be no such algorithm.
$endgroup$
– Yrogirg
Jan 24 '12 at 18:48
add a comment |
$begingroup$
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.
answered Jan 24 '12 at 16:53
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
oh, how could I forget this.
$endgroup$
– Yrogirg
Jan 24 '12 at 17:24
$begingroup$
And should have stated from wikipedia en.wikipedia.org/wiki/Algorithms_for_calculating_variance At first I thought it would be impossible and there would be no such algorithm.
$endgroup$
– Yrogirg
Jan 24 '12 at 18:48
add a comment |
$begingroup$
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.
answered Jan 24 '12 at 16:53
Ross MillikanRoss Millikan
297k23198371
$endgroup$
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.
answered Jan 24 '12 at 16:53
Ross MillikanRoss Millikan
297k23198371
answered Jan 24 '12 at 16:53
Ross MillikanRoss Millikan
297k23198371
answered Jan 24 '12 at 16:53
Ross MillikanRoss Millikan
297k23198371
answered Jan 24 '12 at 16:53
Ross MillikanRoss Millikan
297k23198371
297k23198371
$begingroup$
oh, how could I forget this.
$endgroup$
– Yrogirg
Jan 24 '12 at 17:24
$begingroup$
And should have stated from wikipedia en.wikipedia.org/wiki/Algorithms_for_calculating_variance At first I thought it would be impossible and there would be no such algorithm.
$endgroup$
– Yrogirg
Jan 24 '12 at 18:48
add a comment |
$begingroup$
oh, how could I forget this.
$endgroup$
– Yrogirg
Jan 24 '12 at 17:24
$begingroup$
And should have stated from wikipedia en.wikipedia.org/wiki/Algorithms_for_calculating_variance At first I thought it would be impossible and there would be no such algorithm.
$endgroup$
– Yrogirg
Jan 24 '12 at 18:48
$begingroup$
oh, how could I forget this.
$endgroup$
– Yrogirg
Jan 24 '12 at 17:24
$begingroup$
oh, how could I forget this.
$endgroup$
– Yrogirg
Jan 24 '12 at 17:24
$begingroup$
And should have stated from wikipedia en.wikipedia.org/wiki/Algorithms_for_calculating_variance At first I thought it would be impossible and there would be no such algorithm.
$endgroup$
– Yrogirg
Jan 24 '12 at 18:48
$begingroup$
And should have stated from wikipedia en.wikipedia.org/wiki/Algorithms_for_calculating_variance At first I thought it would be impossible and there would be no such algorithm.
$endgroup$
– Yrogirg
Jan 24 '12 at 18:48
add a comment |
$begingroup$
How about sum of squared pairwise differences ? Indeed, you can check by direct computation that
$$
2v_X = frac{1}{n(n-1)}sum_{1 le i < j le n}(x_i - x_j)^2.
$$
Disclaimer: This is not meant to be efficient, just another way to represent the variance, without first computing the mean.
answered Dec 14 '18 at 7:10
dohmatobdohmatob
3,672629
$endgroup$
add a comment |
$begingroup$
How about sum of squared pairwise differences ? Indeed, you can check by direct computation that
$$
2v_X = frac{1}{n(n-1)}sum_{1 le i < j le n}(x_i - x_j)^2.
$$
Disclaimer: This is not meant to be efficient, just another way to represent the variance, without first computing the mean.
answered Dec 14 '18 at 7:10
dohmatobdohmatob
3,672629
$endgroup$
add a comment |
$begingroup$
How about sum of squared pairwise differences ? Indeed, you can check by direct computation that
$$
2v_X = frac{1}{n(n-1)}sum_{1 le i < j le n}(x_i - x_j)^2.
$$
Disclaimer: This is not meant to be efficient, just another way to represent the variance, without first computing the mean.
answered Dec 14 '18 at 7:10
dohmatobdohmatob
3,672629
$endgroup$
How about sum of squared pairwise differences ? Indeed, you can check by direct computation that
$$
2v_X = frac{1}{n(n-1)}sum_{1 le i < j le n}(x_i - x_j)^2.
$$
Disclaimer: This is not meant to be efficient, just another way to represent the variance, without first computing the mean.
answered Dec 14 '18 at 7:10
dohmatobdohmatob
3,672629
edited Dec 14 '18 at 7:21
answered Dec 14 '18 at 7:10
dohmatobdohmatob
3,672629
answered Dec 14 '18 at 7:10
dohmatobdohmatob
3,672629
answered Dec 14 '18 at 7:10
dohmatobdohmatob
3,672629
3,672629
add a comment |
add a comment |
$begingroup$
The sample variance with mean is calculated as:
$$ v_{X}=frac{1}{n-1}sum_{i=1}^{n}(x_{i}-overline{x})^{2} $$
And the sample variance without mean as:
$$ v_{X}=frac{1}{n-1}left [ sum_{i=1}^{n}x_{i}^{2}-frac{1}{n}left ( sum_{i=1}^{n}x_{i} right ) ^{2}right ] $$
answered Sep 3 '17 at 16:34
FreemanFreeman
1184
$endgroup$
add a comment |
$begingroup$
The sample variance with mean is calculated as:
$$ v_{X}=frac{1}{n-1}sum_{i=1}^{n}(x_{i}-overline{x})^{2} $$
And the sample variance without mean as:
$$ v_{X}=frac{1}{n-1}left [ sum_{i=1}^{n}x_{i}^{2}-frac{1}{n}left ( sum_{i=1}^{n}x_{i} right ) ^{2}right ] $$
answered Sep 3 '17 at 16:34
FreemanFreeman
1184
$endgroup$
add a comment |
$begingroup$
The sample variance with mean is calculated as:
$$ v_{X}=frac{1}{n-1}sum_{i=1}^{n}(x_{i}-overline{x})^{2} $$
And the sample variance without mean as:
$$ v_{X}=frac{1}{n-1}left [ sum_{i=1}^{n}x_{i}^{2}-frac{1}{n}left ( sum_{i=1}^{n}x_{i} right ) ^{2}right ] $$
answered Sep 3 '17 at 16:34
FreemanFreeman
1184
$endgroup$
The sample variance with mean is calculated as:
$$ v_{X}=frac{1}{n-1}sum_{i=1}^{n}(x_{i}-overline{x})^{2} $$
And the sample variance without mean as:
$$ v_{X}=frac{1}{n-1}left [ sum_{i=1}^{n}x_{i}^{2}-frac{1}{n}left ( sum_{i=1}^{n}x_{i} right ) ^{2}right ] $$
answered Sep 3 '17 at 16:34
FreemanFreeman
1184
answered Sep 3 '17 at 16:34
FreemanFreeman
1184
answered Sep 3 '17 at 16:34
FreemanFreeman
1184
answered Sep 3 '17 at 16:34
FreemanFreeman
1184
1184
add a comment |
add a comment |
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