Is it mathematically possible to calculate the total standard deviation of a daily list of numbers from their...












0












$begingroup$


Say I have the following list of numbers, each corresponding to a single day:



list = [1,   6,    7,    12,   5]


I can calculate the daily, rolling standard deviations between each as:



stds = [NaN, 3.54, 0.71, 3.54, 4.95]


Is it now possible to calculate the standard deviation across all values of list, using only the values of stds?






standard-deviation







share|cite|improve this question











$endgroup$












  • $begingroup$
    Not sure I understand the values in stds. The standard deviation of the values $1,2$ is not $3.54$, whether you divide by $n$ or by $n-1$. What exactly is the $3.54$? Also...regardless of what your numbers mean, why isn't the last term in stds the desired value for the entire list?
    $endgroup$
    – lulu
    Jan 16 at 13:09












  • $begingroup$
    @lulu Sorry, the second number was supposed to be 6, not 2. stds contains the standard deviations between each successive pair of numbers. The first value is blank, since there is no number before 1. The second value is std(1, 6), the third value is std(6, 7), etc.
    $endgroup$
    – KOB
    Jan 16 at 13:17










  • $begingroup$
    Ok, but then the answer is "obviously not". The list $[1,6,7,12,19]$ has the same pairwise standard deviations but not the same total standard deviation.
    $endgroup$
    – lulu
    Jan 16 at 13:22










  • $begingroup$
    Note that you can calculate standard deviation incrementally using just three stored values: the number of data observed, the sum of the data, and the sum of their squares. This seems to me a much more useful trick, since storing the list of "daily rolling standard deviations" takes about the same resources as storing the numbers themselves.
    $endgroup$
    – David K
    Jan 17 at 6:11












  • $begingroup$
    @DavidK unfortunately the only data I receive is the pairwise standard deviations, and this cannot be changed as of now.
    $endgroup$
    – KOB
    Jan 17 at 7:26
















0












$begingroup$


Say I have the following list of numbers, each corresponding to a single day:



list = [1,   6,    7,    12,   5]


I can calculate the daily, rolling standard deviations between each as:



stds = [NaN, 3.54, 0.71, 3.54, 4.95]


Is it now possible to calculate the standard deviation across all values of list, using only the values of stds?






standard-deviation







share|cite|improve this question











$endgroup$












  • $begingroup$
    Not sure I understand the values in stds. The standard deviation of the values $1,2$ is not $3.54$, whether you divide by $n$ or by $n-1$. What exactly is the $3.54$? Also...regardless of what your numbers mean, why isn't the last term in stds the desired value for the entire list?
    $endgroup$
    – lulu
    Jan 16 at 13:09












  • $begingroup$
    @lulu Sorry, the second number was supposed to be 6, not 2. stds contains the standard deviations between each successive pair of numbers. The first value is blank, since there is no number before 1. The second value is std(1, 6), the third value is std(6, 7), etc.
    $endgroup$
    – KOB
    Jan 16 at 13:17










  • $begingroup$
    Ok, but then the answer is "obviously not". The list $[1,6,7,12,19]$ has the same pairwise standard deviations but not the same total standard deviation.
    $endgroup$
    – lulu
    Jan 16 at 13:22










  • $begingroup$
    Note that you can calculate standard deviation incrementally using just three stored values: the number of data observed, the sum of the data, and the sum of their squares. This seems to me a much more useful trick, since storing the list of "daily rolling standard deviations" takes about the same resources as storing the numbers themselves.
    $endgroup$
    – David K
    Jan 17 at 6:11












  • $begingroup$
    @DavidK unfortunately the only data I receive is the pairwise standard deviations, and this cannot be changed as of now.
    $endgroup$
    – KOB
    Jan 17 at 7:26














0












0








0





$begingroup$


Say I have the following list of numbers, each corresponding to a single day:



list = [1,   6,    7,    12,   5]


I can calculate the daily, rolling standard deviations between each as:



stds = [NaN, 3.54, 0.71, 3.54, 4.95]


Is it now possible to calculate the standard deviation across all values of list, using only the values of stds?






standard-deviation







share|cite|improve this question











$endgroup$




Say I have the following list of numbers, each corresponding to a single day:



list = [1,   6,    7,    12,   5]


I can calculate the daily, rolling standard deviations between each as:



stds = [NaN, 3.54, 0.71, 3.54, 4.95]


Is it now possible to calculate the standard deviation across all values of list, using only the values of stds?






standard-deviation




standard-deviation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 13:15







KOB

















asked Jan 16 at 12:56









KOBKOB

1919




1919












  • $begingroup$
    Not sure I understand the values in stds. The standard deviation of the values $1,2$ is not $3.54$, whether you divide by $n$ or by $n-1$. What exactly is the $3.54$? Also...regardless of what your numbers mean, why isn't the last term in stds the desired value for the entire list?
    $endgroup$
    – lulu
    Jan 16 at 13:09












  • $begingroup$
    @lulu Sorry, the second number was supposed to be 6, not 2. stds contains the standard deviations between each successive pair of numbers. The first value is blank, since there is no number before 1. The second value is std(1, 6), the third value is std(6, 7), etc.
    $endgroup$
    – KOB
    Jan 16 at 13:17










  • $begingroup$
    Ok, but then the answer is "obviously not". The list $[1,6,7,12,19]$ has the same pairwise standard deviations but not the same total standard deviation.
    $endgroup$
    – lulu
    Jan 16 at 13:22










  • $begingroup$
    Note that you can calculate standard deviation incrementally using just three stored values: the number of data observed, the sum of the data, and the sum of their squares. This seems to me a much more useful trick, since storing the list of "daily rolling standard deviations" takes about the same resources as storing the numbers themselves.
    $endgroup$
    – David K
    Jan 17 at 6:11












  • $begingroup$
    @DavidK unfortunately the only data I receive is the pairwise standard deviations, and this cannot be changed as of now.
    $endgroup$
    – KOB
    Jan 17 at 7:26


















  • $begingroup$
    Not sure I understand the values in stds. The standard deviation of the values $1,2$ is not $3.54$, whether you divide by $n$ or by $n-1$. What exactly is the $3.54$? Also...regardless of what your numbers mean, why isn't the last term in stds the desired value for the entire list?
    $endgroup$
    – lulu
    Jan 16 at 13:09












  • $begingroup$
    @lulu Sorry, the second number was supposed to be 6, not 2. stds contains the standard deviations between each successive pair of numbers. The first value is blank, since there is no number before 1. The second value is std(1, 6), the third value is std(6, 7), etc.
    $endgroup$
    – KOB
    Jan 16 at 13:17










  • $begingroup$
    Ok, but then the answer is "obviously not". The list $[1,6,7,12,19]$ has the same pairwise standard deviations but not the same total standard deviation.
    $endgroup$
    – lulu
    Jan 16 at 13:22










  • $begingroup$
    Note that you can calculate standard deviation incrementally using just three stored values: the number of data observed, the sum of the data, and the sum of their squares. This seems to me a much more useful trick, since storing the list of "daily rolling standard deviations" takes about the same resources as storing the numbers themselves.
    $endgroup$
    – David K
    Jan 17 at 6:11












  • $begingroup$
    @DavidK unfortunately the only data I receive is the pairwise standard deviations, and this cannot be changed as of now.
    $endgroup$
    – KOB
    Jan 17 at 7:26
















$begingroup$
Not sure I understand the values in stds. The standard deviation of the values $1,2$ is not $3.54$, whether you divide by $n$ or by $n-1$. What exactly is the $3.54$? Also...regardless of what your numbers mean, why isn't the last term in stds the desired value for the entire list?
$endgroup$
– lulu
Jan 16 at 13:09






$begingroup$
Not sure I understand the values in stds. The standard deviation of the values $1,2$ is not $3.54$, whether you divide by $n$ or by $n-1$. What exactly is the $3.54$? Also...regardless of what your numbers mean, why isn't the last term in stds the desired value for the entire list?
$endgroup$
– lulu
Jan 16 at 13:09














$begingroup$
@lulu Sorry, the second number was supposed to be 6, not 2. stds contains the standard deviations between each successive pair of numbers. The first value is blank, since there is no number before 1. The second value is std(1, 6), the third value is std(6, 7), etc.
$endgroup$
– KOB
Jan 16 at 13:17




$begingroup$
@lulu Sorry, the second number was supposed to be 6, not 2. stds contains the standard deviations between each successive pair of numbers. The first value is blank, since there is no number before 1. The second value is std(1, 6), the third value is std(6, 7), etc.
$endgroup$
– KOB
Jan 16 at 13:17












$begingroup$
Ok, but then the answer is "obviously not". The list $[1,6,7,12,19]$ has the same pairwise standard deviations but not the same total standard deviation.
$endgroup$
– lulu
Jan 16 at 13:22




$begingroup$
Ok, but then the answer is "obviously not". The list $[1,6,7,12,19]$ has the same pairwise standard deviations but not the same total standard deviation.
$endgroup$
– lulu
Jan 16 at 13:22












$begingroup$
Note that you can calculate standard deviation incrementally using just three stored values: the number of data observed, the sum of the data, and the sum of their squares. This seems to me a much more useful trick, since storing the list of "daily rolling standard deviations" takes about the same resources as storing the numbers themselves.
$endgroup$
– David K
Jan 17 at 6:11






$begingroup$
Note that you can calculate standard deviation incrementally using just three stored values: the number of data observed, the sum of the data, and the sum of their squares. This seems to me a much more useful trick, since storing the list of "daily rolling standard deviations" takes about the same resources as storing the numbers themselves.
$endgroup$
– David K
Jan 17 at 6:11














$begingroup$
@DavidK unfortunately the only data I receive is the pairwise standard deviations, and this cannot be changed as of now.
$endgroup$
– KOB
Jan 17 at 7:26




$begingroup$
@DavidK unfortunately the only data I receive is the pairwise standard deviations, and this cannot be changed as of now.
$endgroup$
– KOB
Jan 17 at 7:26










1 Answer
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$begingroup$

The rolling pairwise standard deviations don't give you enough information to get the total standard deviation. The two lists $$[0,1,0,1]quad &quad [0,1,2,3]$$ both have the pairwise standard deviations $$Big[frac 1{sqrt 2},,frac 1{sqrt 2},,frac 1{sqrt 2}Big]$$



But the total standard deviation of the first is $.577$ and the standard deviation of the second is $1.29$.



Note: in all cases we are using the sample standard deviation (i.e. we divide by $n-1$ not $n$).






share|cite|improve this answer









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    $begingroup$

    The rolling pairwise standard deviations don't give you enough information to get the total standard deviation. The two lists $$[0,1,0,1]quad &quad [0,1,2,3]$$ both have the pairwise standard deviations $$Big[frac 1{sqrt 2},,frac 1{sqrt 2},,frac 1{sqrt 2}Big]$$



    But the total standard deviation of the first is $.577$ and the standard deviation of the second is $1.29$.



    Note: in all cases we are using the sample standard deviation (i.e. we divide by $n-1$ not $n$).






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The rolling pairwise standard deviations don't give you enough information to get the total standard deviation. The two lists $$[0,1,0,1]quad &quad [0,1,2,3]$$ both have the pairwise standard deviations $$Big[frac 1{sqrt 2},,frac 1{sqrt 2},,frac 1{sqrt 2}Big]$$



      But the total standard deviation of the first is $.577$ and the standard deviation of the second is $1.29$.



      Note: in all cases we are using the sample standard deviation (i.e. we divide by $n-1$ not $n$).






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The rolling pairwise standard deviations don't give you enough information to get the total standard deviation. The two lists $$[0,1,0,1]quad &quad [0,1,2,3]$$ both have the pairwise standard deviations $$Big[frac 1{sqrt 2},,frac 1{sqrt 2},,frac 1{sqrt 2}Big]$$



        But the total standard deviation of the first is $.577$ and the standard deviation of the second is $1.29$.



        Note: in all cases we are using the sample standard deviation (i.e. we divide by $n-1$ not $n$).






        share|cite|improve this answer









        $endgroup$



        The rolling pairwise standard deviations don't give you enough information to get the total standard deviation. The two lists $$[0,1,0,1]quad &quad [0,1,2,3]$$ both have the pairwise standard deviations $$Big[frac 1{sqrt 2},,frac 1{sqrt 2},,frac 1{sqrt 2}Big]$$



        But the total standard deviation of the first is $.577$ and the standard deviation of the second is $1.29$.



        Note: in all cases we are using the sample standard deviation (i.e. we divide by $n-1$ not $n$).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 16 at 13:28









        lulululu

        43.3k25080




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