Why does the function VarianceMLE give a different result from Variance?












3












$begingroup$


Why does the function VarianceMLE give a different result from Variance?



And what is it in Mathematica 11.3?



enter image description here



Please see the picture above t665he MLE is 621 and the other is 665.






probability-or-statistics







share|improve this question











$endgroup$












  • $begingroup$
    In Mathematica 11.3, << Statistics` produces an error message and Variance@data // N gives 665.524
    $endgroup$
    – m_goldberg
    Jan 14 at 7:47








  • 2




    $begingroup$
    I'm going to guess that VarianceMLE is the maximum likelihood variance estimator rather than the unbiased one (assuming normally distributed data). The difference between the two is that Variance divides by N-1 (N == Length[data]) while the MLE estimator divides by N.
    $endgroup$
    – Sjoerd Smit
    Jan 14 at 9:22






  • 1




    $begingroup$
    And for future reference: please post copyable code in your question rather than a screenshot. This makes it much easier for someone else to copy your code and try things out.
    $endgroup$
    – Sjoerd Smit
    Jan 14 at 9:23












  • $begingroup$
    Which book did you see this in? It looks like a scan.
    $endgroup$
    – Szabolcs
    Jan 14 at 11:45
















3












$begingroup$


Why does the function VarianceMLE give a different result from Variance?



And what is it in Mathematica 11.3?



enter image description here



Please see the picture above t665he MLE is 621 and the other is 665.






probability-or-statistics







share|improve this question











$endgroup$












  • $begingroup$
    In Mathematica 11.3, << Statistics` produces an error message and Variance@data // N gives 665.524
    $endgroup$
    – m_goldberg
    Jan 14 at 7:47








  • 2




    $begingroup$
    I'm going to guess that VarianceMLE is the maximum likelihood variance estimator rather than the unbiased one (assuming normally distributed data). The difference between the two is that Variance divides by N-1 (N == Length[data]) while the MLE estimator divides by N.
    $endgroup$
    – Sjoerd Smit
    Jan 14 at 9:22






  • 1




    $begingroup$
    And for future reference: please post copyable code in your question rather than a screenshot. This makes it much easier for someone else to copy your code and try things out.
    $endgroup$
    – Sjoerd Smit
    Jan 14 at 9:23












  • $begingroup$
    Which book did you see this in? It looks like a scan.
    $endgroup$
    – Szabolcs
    Jan 14 at 11:45














3












3








3





$begingroup$


Why does the function VarianceMLE give a different result from Variance?



And what is it in Mathematica 11.3?



enter image description here



Please see the picture above t665he MLE is 621 and the other is 665.






probability-or-statistics







share|improve this question











$endgroup$




Why does the function VarianceMLE give a different result from Variance?



And what is it in Mathematica 11.3?



enter image description here



Please see the picture above t665he MLE is 621 and the other is 665.






probability-or-statistics




probability-or-statistics






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Jan 14 at 8:01









m_goldberg

85k872196




85k872196










asked Jan 14 at 7:24









FacetFacet

254




254












  • $begingroup$
    In Mathematica 11.3, << Statistics` produces an error message and Variance@data // N gives 665.524
    $endgroup$
    – m_goldberg
    Jan 14 at 7:47








  • 2




    $begingroup$
    I'm going to guess that VarianceMLE is the maximum likelihood variance estimator rather than the unbiased one (assuming normally distributed data). The difference between the two is that Variance divides by N-1 (N == Length[data]) while the MLE estimator divides by N.
    $endgroup$
    – Sjoerd Smit
    Jan 14 at 9:22






  • 1




    $begingroup$
    And for future reference: please post copyable code in your question rather than a screenshot. This makes it much easier for someone else to copy your code and try things out.
    $endgroup$
    – Sjoerd Smit
    Jan 14 at 9:23












  • $begingroup$
    Which book did you see this in? It looks like a scan.
    $endgroup$
    – Szabolcs
    Jan 14 at 11:45


















  • $begingroup$
    In Mathematica 11.3, << Statistics` produces an error message and Variance@data // N gives 665.524
    $endgroup$
    – m_goldberg
    Jan 14 at 7:47








  • 2




    $begingroup$
    I'm going to guess that VarianceMLE is the maximum likelihood variance estimator rather than the unbiased one (assuming normally distributed data). The difference between the two is that Variance divides by N-1 (N == Length[data]) while the MLE estimator divides by N.
    $endgroup$
    – Sjoerd Smit
    Jan 14 at 9:22






  • 1




    $begingroup$
    And for future reference: please post copyable code in your question rather than a screenshot. This makes it much easier for someone else to copy your code and try things out.
    $endgroup$
    – Sjoerd Smit
    Jan 14 at 9:23












  • $begingroup$
    Which book did you see this in? It looks like a scan.
    $endgroup$
    – Szabolcs
    Jan 14 at 11:45
















$begingroup$
In Mathematica 11.3, << Statistics` produces an error message and Variance@data // N gives 665.524
$endgroup$
– m_goldberg
Jan 14 at 7:47






$begingroup$
In Mathematica 11.3, << Statistics` produces an error message and Variance@data // N gives 665.524
$endgroup$
– m_goldberg
Jan 14 at 7:47






2




2




$begingroup$
I'm going to guess that VarianceMLE is the maximum likelihood variance estimator rather than the unbiased one (assuming normally distributed data). The difference between the two is that Variance divides by N-1 (N == Length[data]) while the MLE estimator divides by N.
$endgroup$
– Sjoerd Smit
Jan 14 at 9:22




$begingroup$
I'm going to guess that VarianceMLE is the maximum likelihood variance estimator rather than the unbiased one (assuming normally distributed data). The difference between the two is that Variance divides by N-1 (N == Length[data]) while the MLE estimator divides by N.
$endgroup$
– Sjoerd Smit
Jan 14 at 9:22




1




1




$begingroup$
And for future reference: please post copyable code in your question rather than a screenshot. This makes it much easier for someone else to copy your code and try things out.
$endgroup$
– Sjoerd Smit
Jan 14 at 9:23






$begingroup$
And for future reference: please post copyable code in your question rather than a screenshot. This makes it much easier for someone else to copy your code and try things out.
$endgroup$
– Sjoerd Smit
Jan 14 at 9:23














$begingroup$
Which book did you see this in? It looks like a scan.
$endgroup$
– Szabolcs
Jan 14 at 11:45




$begingroup$
Which book did you see this in? It looks like a scan.
$endgroup$
– Szabolcs
Jan 14 at 11:45










2 Answers
2






active

oldest

votes


















8












$begingroup$

I just checked my guess in my comment and I was right. VarianceMLE is the maximum likelihood variance estimator (see, e.g. here).



data = {34, 56, 28, 62, 32, 90, 20, 10, 12, 35, 63, 78, 12, 25, 68};

Variance[data]




13976/21





myVariance[lst_List] := Total[(lst - Mean[lst])^2]/(Length[lst] - 1);

myVarianceMLE[lst_List] := Total[(lst - Mean[lst])^2]/Length[lst];



myVariance[data]

myVarianceMLE[data]




13976/21



27952/45








share|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your kind help, I will remember to post code next time.
    $endgroup$
    – Facet
    Jan 14 at 14:47










  • $begingroup$
    Wolfram Research article also mentions a replacement for VarianceMLE: reference.wolfram.com/language/Compatibility/tutorial/… .
    $endgroup$
    – kirma
    Jan 18 at 17:31



















3












$begingroup$

VarianceMLE computes a biased, maximum likelihood estimate of the population variance. Variance computes an unbiased estimate of the population variance. It can be shown that VarianceMLE underestimates the variance of the population.



Let ${y_i : 1 leq i leq n}$ be a sample of $n$ values from a population. The variance (central second moment) of the sample is
$$ sigma_y^2 = frac{1}{n} sum_{i=1}^n (y_i - bar{y}) text{,} $$
where $bar{y} = frac{1}{n} sum_{i=1}^n y_i$ is the sample mean. This $sigma_y^2$ is computed by VarianceMLE.



If $sigma^2$ is the population variance, with some work, one can show that the expected value of $sigma_y^2$ is $frac{n-1}{n} sigma^2$, so the sample variance is a biased estimator of the population variance. We can make this an unbiased estimator via
$$ s^2 = frac{n}{n-1} sigma_y^2 = frac{1}{n-1}sum_{i=1}^n (y_i - bar{y}) text{.} $$
This $s^2$ is computed by Variance. From the documentation (in the Details):



"Variance[list] is equivalent to Total[(list-Mean[list])^2]/(Length[list]-1) for real-valued data."



The very sparse documentation for VarianceMLE indicates that it is implemented in terms of Variance:



"VarianceMLE[data_] := Variance[data] (Length[data] - 1)/Length[data]"






share|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    8












    $begingroup$

    I just checked my guess in my comment and I was right. VarianceMLE is the maximum likelihood variance estimator (see, e.g. here).



    data = {34, 56, 28, 62, 32, 90, 20, 10, 12, 35, 63, 78, 12, 25, 68};
    
Variance[data]




    13976/21





    myVariance[lst_List] := Total[(lst - Mean[lst])^2]/(Length[lst] - 1);
    
myVarianceMLE[lst_List] := Total[(lst - Mean[lst])^2]/Length[lst];
    

    
myVariance[data]
    
myVarianceMLE[data]




    13976/21



    27952/45








    share|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for your kind help, I will remember to post code next time.
      $endgroup$
      – Facet
      Jan 14 at 14:47










    • $begingroup$
      Wolfram Research article also mentions a replacement for VarianceMLE: reference.wolfram.com/language/Compatibility/tutorial/… .
      $endgroup$
      – kirma
      Jan 18 at 17:31
















    8












    $begingroup$

    I just checked my guess in my comment and I was right. VarianceMLE is the maximum likelihood variance estimator (see, e.g. here).



    data = {34, 56, 28, 62, 32, 90, 20, 10, 12, 35, 63, 78, 12, 25, 68};
    
Variance[data]




    13976/21





    myVariance[lst_List] := Total[(lst - Mean[lst])^2]/(Length[lst] - 1);
    
myVarianceMLE[lst_List] := Total[(lst - Mean[lst])^2]/Length[lst];
    

    
myVariance[data]
    
myVarianceMLE[data]




    13976/21



    27952/45








    share|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for your kind help, I will remember to post code next time.
      $endgroup$
      – Facet
      Jan 14 at 14:47










    • $begingroup$
      Wolfram Research article also mentions a replacement for VarianceMLE: reference.wolfram.com/language/Compatibility/tutorial/… .
      $endgroup$
      – kirma
      Jan 18 at 17:31














    8












    8








    8





    $begingroup$

    I just checked my guess in my comment and I was right. VarianceMLE is the maximum likelihood variance estimator (see, e.g. here).



    data = {34, 56, 28, 62, 32, 90, 20, 10, 12, 35, 63, 78, 12, 25, 68};
    
Variance[data]




    13976/21





    myVariance[lst_List] := Total[(lst - Mean[lst])^2]/(Length[lst] - 1);
    
myVarianceMLE[lst_List] := Total[(lst - Mean[lst])^2]/Length[lst];
    

    
myVariance[data]
    
myVarianceMLE[data]




    13976/21



    27952/45








    share|improve this answer









    $endgroup$



    I just checked my guess in my comment and I was right. VarianceMLE is the maximum likelihood variance estimator (see, e.g. here).



    data = {34, 56, 28, 62, 32, 90, 20, 10, 12, 35, 63, 78, 12, 25, 68};
    
Variance[data]




    13976/21





    myVariance[lst_List] := Total[(lst - Mean[lst])^2]/(Length[lst] - 1);
    
myVarianceMLE[lst_List] := Total[(lst - Mean[lst])^2]/Length[lst];
    

    
myVariance[data]
    
myVarianceMLE[data]




    13976/21



    27952/45









    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Jan 14 at 9:28









    Sjoerd SmitSjoerd Smit

    3,410715




    3,410715












    • $begingroup$
      Thank you for your kind help, I will remember to post code next time.
      $endgroup$
      – Facet
      Jan 14 at 14:47










    • $begingroup$
      Wolfram Research article also mentions a replacement for VarianceMLE: reference.wolfram.com/language/Compatibility/tutorial/… .
      $endgroup$
      – kirma
      Jan 18 at 17:31


















    • $begingroup$
      Thank you for your kind help, I will remember to post code next time.
      $endgroup$
      – Facet
      Jan 14 at 14:47










    • $begingroup$
      Wolfram Research article also mentions a replacement for VarianceMLE: reference.wolfram.com/language/Compatibility/tutorial/… .
      $endgroup$
      – kirma
      Jan 18 at 17:31
















    $begingroup$
    Thank you for your kind help, I will remember to post code next time.
    $endgroup$
    – Facet
    Jan 14 at 14:47




    $begingroup$
    Thank you for your kind help, I will remember to post code next time.
    $endgroup$
    – Facet
    Jan 14 at 14:47












    $begingroup$
    Wolfram Research article also mentions a replacement for VarianceMLE: reference.wolfram.com/language/Compatibility/tutorial/… .
    $endgroup$
    – kirma
    Jan 18 at 17:31




    $begingroup$
    Wolfram Research article also mentions a replacement for VarianceMLE: reference.wolfram.com/language/Compatibility/tutorial/… .
    $endgroup$
    – kirma
    Jan 18 at 17:31











    3












    $begingroup$

    VarianceMLE computes a biased, maximum likelihood estimate of the population variance. Variance computes an unbiased estimate of the population variance. It can be shown that VarianceMLE underestimates the variance of the population.



    Let ${y_i : 1 leq i leq n}$ be a sample of $n$ values from a population. The variance (central second moment) of the sample is
    $$ sigma_y^2 = frac{1}{n} sum_{i=1}^n (y_i - bar{y}) text{,} $$
    where $bar{y} = frac{1}{n} sum_{i=1}^n y_i$ is the sample mean. This $sigma_y^2$ is computed by VarianceMLE.



    If $sigma^2$ is the population variance, with some work, one can show that the expected value of $sigma_y^2$ is $frac{n-1}{n} sigma^2$, so the sample variance is a biased estimator of the population variance. We can make this an unbiased estimator via
    $$ s^2 = frac{n}{n-1} sigma_y^2 = frac{1}{n-1}sum_{i=1}^n (y_i - bar{y}) text{.} $$
    This $s^2$ is computed by Variance. From the documentation (in the Details):



    "Variance[list] is equivalent to Total[(list-Mean[list])^2]/(Length[list]-1) for real-valued data."



    The very sparse documentation for VarianceMLE indicates that it is implemented in terms of Variance:



    "VarianceMLE[data_] := Variance[data] (Length[data] - 1)/Length[data]"






    share|improve this answer









    $endgroup$


















      3












      $begingroup$

      VarianceMLE computes a biased, maximum likelihood estimate of the population variance. Variance computes an unbiased estimate of the population variance. It can be shown that VarianceMLE underestimates the variance of the population.



      Let ${y_i : 1 leq i leq n}$ be a sample of $n$ values from a population. The variance (central second moment) of the sample is
      $$ sigma_y^2 = frac{1}{n} sum_{i=1}^n (y_i - bar{y}) text{,} $$
      where $bar{y} = frac{1}{n} sum_{i=1}^n y_i$ is the sample mean. This $sigma_y^2$ is computed by VarianceMLE.



      If $sigma^2$ is the population variance, with some work, one can show that the expected value of $sigma_y^2$ is $frac{n-1}{n} sigma^2$, so the sample variance is a biased estimator of the population variance. We can make this an unbiased estimator via
      $$ s^2 = frac{n}{n-1} sigma_y^2 = frac{1}{n-1}sum_{i=1}^n (y_i - bar{y}) text{.} $$
      This $s^2$ is computed by Variance. From the documentation (in the Details):



      "Variance[list] is equivalent to Total[(list-Mean[list])^2]/(Length[list]-1) for real-valued data."



      The very sparse documentation for VarianceMLE indicates that it is implemented in terms of Variance:



      "VarianceMLE[data_] := Variance[data] (Length[data] - 1)/Length[data]"






      share|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        VarianceMLE computes a biased, maximum likelihood estimate of the population variance. Variance computes an unbiased estimate of the population variance. It can be shown that VarianceMLE underestimates the variance of the population.



        Let ${y_i : 1 leq i leq n}$ be a sample of $n$ values from a population. The variance (central second moment) of the sample is
        $$ sigma_y^2 = frac{1}{n} sum_{i=1}^n (y_i - bar{y}) text{,} $$
        where $bar{y} = frac{1}{n} sum_{i=1}^n y_i$ is the sample mean. This $sigma_y^2$ is computed by VarianceMLE.



        If $sigma^2$ is the population variance, with some work, one can show that the expected value of $sigma_y^2$ is $frac{n-1}{n} sigma^2$, so the sample variance is a biased estimator of the population variance. We can make this an unbiased estimator via
        $$ s^2 = frac{n}{n-1} sigma_y^2 = frac{1}{n-1}sum_{i=1}^n (y_i - bar{y}) text{.} $$
        This $s^2$ is computed by Variance. From the documentation (in the Details):



        "Variance[list] is equivalent to Total[(list-Mean[list])^2]/(Length[list]-1) for real-valued data."



        The very sparse documentation for VarianceMLE indicates that it is implemented in terms of Variance:



        "VarianceMLE[data_] := Variance[data] (Length[data] - 1)/Length[data]"






        share|improve this answer









        $endgroup$



        VarianceMLE computes a biased, maximum likelihood estimate of the population variance. Variance computes an unbiased estimate of the population variance. It can be shown that VarianceMLE underestimates the variance of the population.



        Let ${y_i : 1 leq i leq n}$ be a sample of $n$ values from a population. The variance (central second moment) of the sample is
        $$ sigma_y^2 = frac{1}{n} sum_{i=1}^n (y_i - bar{y}) text{,} $$
        where $bar{y} = frac{1}{n} sum_{i=1}^n y_i$ is the sample mean. This $sigma_y^2$ is computed by VarianceMLE.



        If $sigma^2$ is the population variance, with some work, one can show that the expected value of $sigma_y^2$ is $frac{n-1}{n} sigma^2$, so the sample variance is a biased estimator of the population variance. We can make this an unbiased estimator via
        $$ s^2 = frac{n}{n-1} sigma_y^2 = frac{1}{n-1}sum_{i=1}^n (y_i - bar{y}) text{.} $$
        This $s^2$ is computed by Variance. From the documentation (in the Details):



        "Variance[list] is equivalent to Total[(list-Mean[list])^2]/(Length[list]-1) for real-valued data."



        The very sparse documentation for VarianceMLE indicates that it is implemented in terms of Variance:



        "VarianceMLE[data_] := Variance[data] (Length[data] - 1)/Length[data]"







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Jan 14 at 15:15









        Eric TowersEric Towers

        2,306613




        2,306613






























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            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
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