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









New contributor




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$endgroup$












  • $begingroup$
    In Mathematica 11.3, << Statistics` produces an error message and Variance@data // N gives 665.524
    $endgroup$
    – m_goldberg
    21 hours ago








  • 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
    19 hours ago






  • 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
    19 hours ago












  • $begingroup$
    Which book did you see this in? It looks like a scan.
    $endgroup$
    – Szabolcs
    17 hours ago
















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









New contributor




Facet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    In Mathematica 11.3, << Statistics` produces an error message and Variance@data // N gives 665.524
    $endgroup$
    – m_goldberg
    21 hours ago








  • 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
    19 hours ago






  • 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
    19 hours ago












  • $begingroup$
    Which book did you see this in? It looks like a scan.
    $endgroup$
    – Szabolcs
    17 hours ago














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









New contributor




Facet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




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









New contributor




Facet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Facet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 20 hours ago









m_goldberg

84.7k872196




84.7k872196






New contributor




Facet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 21 hours ago









FacetFacet

254




254




New contributor




Facet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Facet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Facet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    In Mathematica 11.3, << Statistics` produces an error message and Variance@data // N gives 665.524
    $endgroup$
    – m_goldberg
    21 hours ago








  • 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
    19 hours ago






  • 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
    19 hours ago












  • $begingroup$
    Which book did you see this in? It looks like a scan.
    $endgroup$
    – Szabolcs
    17 hours ago


















  • $begingroup$
    In Mathematica 11.3, << Statistics` produces an error message and Variance@data // N gives 665.524
    $endgroup$
    – m_goldberg
    21 hours ago








  • 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
    19 hours ago






  • 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
    19 hours ago












  • $begingroup$
    Which book did you see this in? It looks like a scan.
    $endgroup$
    – Szabolcs
    17 hours ago
















$begingroup$
In Mathematica 11.3, << Statistics` produces an error message and Variance@data // N gives 665.524
$endgroup$
– m_goldberg
21 hours ago






$begingroup$
In Mathematica 11.3, << Statistics` produces an error message and Variance@data // N gives 665.524
$endgroup$
– m_goldberg
21 hours ago






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
19 hours ago




$begingroup$
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
19 hours ago




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
19 hours ago






$begingroup$
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
19 hours ago














$begingroup$
Which book did you see this in? It looks like a scan.
$endgroup$
– Szabolcs
17 hours ago




$begingroup$
Which book did you see this in? It looks like a scan.
$endgroup$
– Szabolcs
17 hours ago










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
    14 hours ago



















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
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    active

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    2 Answers
    2






    active

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    active

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    active

    oldest

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    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
      14 hours ago
















    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
      14 hours ago














    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 19 hours ago









    Sjoerd SmitSjoerd Smit

    3,410715




    3,410715












    • $begingroup$
      Thank you for your kind help, I will remember to post code next time.
      $endgroup$
      – Facet
      14 hours ago


















    • $begingroup$
      Thank you for your kind help, I will remember to post code next time.
      $endgroup$
      – Facet
      14 hours ago
















    $begingroup$
    Thank you for your kind help, I will remember to post code next time.
    $endgroup$
    – Facet
    14 hours ago




    $begingroup$
    Thank you for your kind help, I will remember to post code next time.
    $endgroup$
    – Facet
    14 hours ago











    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 13 hours ago









        Eric TowersEric Towers

        2,286613




        2,286613






















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            Муниципалитет Алькесар Alquézar Герб 42°10′16″ с. ш. 0°01′27″ в. д. H G Я O L Страна Испания   Испания Автономное сообщество Арагон Провинция Уэска Комарка Сомонтано-де-Барбастро Алькальд Хосе Марияно Альтемир Ласкорс (ИСРП) История и география Площадь 32,36 км² Высота НУМ 660 м Часовой пояс UTC+1, летом UTC+2 Население Население 313 человек ( 2010 ) Плотность 9,92 чел./км² Этнохороним alquezarano/-a Цифровые идентификаторы Телефонный код +34 974 Почтовые индексы 22112 Автомобильный код HU alquezar.org   (исп.) Показать/скрыть карты Алькесар Алькесар Алькесар  Аудио, фото и видео на Викискладе Алькесар (исп.  Alquézar ) — муниципалитет в Испании, входит в провинцию Уэска, в составе автономного сообщества Арагон. Муниципалитет находится в составе района (комарки) Сомонтано-де-Барбастро. Занимает площадь 32,36 км². Население — 321 челове
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            Josef Freinademetz

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            San Ujöp (Giuseppe/Josef) Freinademetz Josef Freinademetz in abiti cinesi   Religioso e missionario   Nascita 15 aprile 1852 Morte 28 gennaio 1908 Venerato da Chiesa cattolica Beatificazione 1975 Canonizzazione 5 ottobre 2003 Ricorrenza 28 gennaio Manuale Ujöp Freinademetz noto anche come Giuseppe o Josef Freinademetz (Oies, 15 aprile 1852 – Taikia, 28 gennaio 1908) è stato un presbitero austriaco, membro della Società del Verbo Divino, fu missionario in Cina; è venerato come santo dalla Chiesa cattolica. .mw-parser-output .citazione-table{margin-bottom:.5em;font-size:95%}.mw-parser-output .citazione-table td{padding:0 1.2em 0 2.4em}.mw-parser-output .citazione-lang{vertical-align:top}.mw-parser-output .citazione-lang td{width:50%}.mw-parser-output .citazione-lang td:first-child{padding:0 0 0 2.4em}.mw-parser-output .citazione-lang td:nth-child(2){padding:0 1.2em} «Io amo la Cina e i cinesi; voglio morire in mezzo a loro, e tra loro
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