Prove Standard deviation greater than or equal to Mean deviation












1












$begingroup$


Ho do we prove that the standard deviation is greater than or equal to the mean deviation about the arithmetic mean ?




$$
sqrtfrac{sum_{i=1}^{n}(x_i-bar{x})^2}{n}geqfrac{sum_{i=1}^{n}|x_i-bar{x}|}{n}
$$




and under what conditions we get the equality ?



I think i understand that it is because of the squaring in standard deviation which tends to give more weightage to the data far from the central tendency.






statistics standard-deviation







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    The Cauchy–Schwarz inequality will do.
    $endgroup$
    – Chappers
    Jul 20 '17 at 18:18










  • $begingroup$
    @Chappers could u help me how to proceed. i'm unable to find where to start inorder to prove it.
    $endgroup$
    – ss1729
    Jul 20 '17 at 18:41


















1












$begingroup$


Ho do we prove that the standard deviation is greater than or equal to the mean deviation about the arithmetic mean ?




$$
sqrtfrac{sum_{i=1}^{n}(x_i-bar{x})^2}{n}geqfrac{sum_{i=1}^{n}|x_i-bar{x}|}{n}
$$




and under what conditions we get the equality ?



I think i understand that it is because of the squaring in standard deviation which tends to give more weightage to the data far from the central tendency.






statistics standard-deviation







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    The Cauchy–Schwarz inequality will do.
    $endgroup$
    – Chappers
    Jul 20 '17 at 18:18










  • $begingroup$
    @Chappers could u help me how to proceed. i'm unable to find where to start inorder to prove it.
    $endgroup$
    – ss1729
    Jul 20 '17 at 18:41
















1












1








1





$begingroup$


Ho do we prove that the standard deviation is greater than or equal to the mean deviation about the arithmetic mean ?




$$
sqrtfrac{sum_{i=1}^{n}(x_i-bar{x})^2}{n}geqfrac{sum_{i=1}^{n}|x_i-bar{x}|}{n}
$$




and under what conditions we get the equality ?



I think i understand that it is because of the squaring in standard deviation which tends to give more weightage to the data far from the central tendency.






statistics standard-deviation







share|cite|improve this question











$endgroup$




Ho do we prove that the standard deviation is greater than or equal to the mean deviation about the arithmetic mean ?




$$
sqrtfrac{sum_{i=1}^{n}(x_i-bar{x})^2}{n}geqfrac{sum_{i=1}^{n}|x_i-bar{x}|}{n}
$$




and under what conditions we get the equality ?



I think i understand that it is because of the squaring in standard deviation which tends to give more weightage to the data far from the central tendency.






statistics standard-deviation




statistics standard-deviation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 20 '17 at 18:34







ss1729

















asked Jul 20 '17 at 18:12









ss1729ss1729

2,00311024




2,00311024








  • 3




    $begingroup$
    The Cauchy–Schwarz inequality will do.
    $endgroup$
    – Chappers
    Jul 20 '17 at 18:18










  • $begingroup$
    @Chappers could u help me how to proceed. i'm unable to find where to start inorder to prove it.
    $endgroup$
    – ss1729
    Jul 20 '17 at 18:41
















  • 3




    $begingroup$
    The Cauchy–Schwarz inequality will do.
    $endgroup$
    – Chappers
    Jul 20 '17 at 18:18










  • $begingroup$
    @Chappers could u help me how to proceed. i'm unable to find where to start inorder to prove it.
    $endgroup$
    – ss1729
    Jul 20 '17 at 18:41










3




3




$begingroup$
The Cauchy–Schwarz inequality will do.
$endgroup$
– Chappers
Jul 20 '17 at 18:18




$begingroup$
The Cauchy–Schwarz inequality will do.
$endgroup$
– Chappers
Jul 20 '17 at 18:18












$begingroup$
@Chappers could u help me how to proceed. i'm unable to find where to start inorder to prove it.
$endgroup$
– ss1729
Jul 20 '17 at 18:41






$begingroup$
@Chappers could u help me how to proceed. i'm unable to find where to start inorder to prove it.
$endgroup$
– ss1729
Jul 20 '17 at 18:41












2 Answers
2






active

oldest

votes


















1












$begingroup$

Let $v=left|vec{x} - bar{x}vec{1}right|$, where $|cdot|$ is component-wise.
Then:
begin{align}
frac{1}{n^2}left( sum_i |x_i - bar{x}| right)^2
&= frac{ left(vcdotvec{1}right)^2}{n^2}\
&leq frac{(vec{1}cdotvec{1})(vcdot v)}{n^2}\
&= frac{1}{n}sum_i|x_i - bar{x}|^2
end{align}
where we used the CS inequality for the second step.
Now take the root of the first and last terms:
$$
sqrt{frac{1}{n}sum_i|x_i - bar{x}|^2;} ,geq frac{1}{n}sum_i |x_i - bar{x}|
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How does $left|vec{x} - bar{x}vec{1}right|= sum_i left|x_i - bar{x} right|$ ?
    $endgroup$
    – ThePassenger
    Jul 21 '17 at 8:22



















0












$begingroup$

The Cauchy Schwarz inequality




$$bigg(sum_{i=1}^{n}a_{i}^2bigg).bigg(sum_{i=1}^{n}b_{i}^2bigg)ge bigg(sum_{i=1}^{n}a_ib_ibigg)^2
$$




taking $a_i=|x_i-bar{x}|$ and $b_i=1/n$,



$$
bigg(sum_{i=1}^{n}|x_i-bar{x}|^2bigg).bigg(sum_{i=1}^{n}tfrac{1}{n^2}bigg)ge bigg(sum_{i=1}^{n}|x_i-bar{x}|.tfrac{1}{n}bigg)^2\
bigg(sum_{i=1}^{n}(x_i-bar{x})^2bigg).bigg(n.tfrac{1}{n^2}bigg)ge bigg(frac{sum_{i=1}^{n}|x_i-bar{x}|}{{n}}bigg)^2\
frac{sum_{i=1}^n(x_i-bar{x})^2}{n}ge bigg(frac{sum_{i=1}^n|x_i-bar{x}|}{n}bigg)^2\
sqrtfrac{sum_{i=1}^n(x_i-bar{x})^2}{n}gefrac{sum_{i=1}^n|x_i-bar{x}|}{n}\
S.Dge M.D
$$






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Let $v=left|vec{x} - bar{x}vec{1}right|$, where $|cdot|$ is component-wise.
    Then:
    begin{align}
    frac{1}{n^2}left( sum_i |x_i - bar{x}| right)^2
    &= frac{ left(vcdotvec{1}right)^2}{n^2}\
    &leq frac{(vec{1}cdotvec{1})(vcdot v)}{n^2}\
    &= frac{1}{n}sum_i|x_i - bar{x}|^2
    end{align}
    where we used the CS inequality for the second step.
    Now take the root of the first and last terms:
    $$
    sqrt{frac{1}{n}sum_i|x_i - bar{x}|^2;} ,geq frac{1}{n}sum_i |x_i - bar{x}|
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How does $left|vec{x} - bar{x}vec{1}right|= sum_i left|x_i - bar{x} right|$ ?
      $endgroup$
      – ThePassenger
      Jul 21 '17 at 8:22
















    1












    $begingroup$

    Let $v=left|vec{x} - bar{x}vec{1}right|$, where $|cdot|$ is component-wise.
    Then:
    begin{align}
    frac{1}{n^2}left( sum_i |x_i - bar{x}| right)^2
    &= frac{ left(vcdotvec{1}right)^2}{n^2}\
    &leq frac{(vec{1}cdotvec{1})(vcdot v)}{n^2}\
    &= frac{1}{n}sum_i|x_i - bar{x}|^2
    end{align}
    where we used the CS inequality for the second step.
    Now take the root of the first and last terms:
    $$
    sqrt{frac{1}{n}sum_i|x_i - bar{x}|^2;} ,geq frac{1}{n}sum_i |x_i - bar{x}|
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How does $left|vec{x} - bar{x}vec{1}right|= sum_i left|x_i - bar{x} right|$ ?
      $endgroup$
      – ThePassenger
      Jul 21 '17 at 8:22














    1












    1








    1





    $begingroup$

    Let $v=left|vec{x} - bar{x}vec{1}right|$, where $|cdot|$ is component-wise.
    Then:
    begin{align}
    frac{1}{n^2}left( sum_i |x_i - bar{x}| right)^2
    &= frac{ left(vcdotvec{1}right)^2}{n^2}\
    &leq frac{(vec{1}cdotvec{1})(vcdot v)}{n^2}\
    &= frac{1}{n}sum_i|x_i - bar{x}|^2
    end{align}
    where we used the CS inequality for the second step.
    Now take the root of the first and last terms:
    $$
    sqrt{frac{1}{n}sum_i|x_i - bar{x}|^2;} ,geq frac{1}{n}sum_i |x_i - bar{x}|
    $$






    share|cite|improve this answer









    $endgroup$



    Let $v=left|vec{x} - bar{x}vec{1}right|$, where $|cdot|$ is component-wise.
    Then:
    begin{align}
    frac{1}{n^2}left( sum_i |x_i - bar{x}| right)^2
    &= frac{ left(vcdotvec{1}right)^2}{n^2}\
    &leq frac{(vec{1}cdotvec{1})(vcdot v)}{n^2}\
    &= frac{1}{n}sum_i|x_i - bar{x}|^2
    end{align}
    where we used the CS inequality for the second step.
    Now take the root of the first and last terms:
    $$
    sqrt{frac{1}{n}sum_i|x_i - bar{x}|^2;} ,geq frac{1}{n}sum_i |x_i - bar{x}|
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jul 20 '17 at 21:33









    user3658307user3658307

    4,8133946




    4,8133946












    • $begingroup$
      How does $left|vec{x} - bar{x}vec{1}right|= sum_i left|x_i - bar{x} right|$ ?
      $endgroup$
      – ThePassenger
      Jul 21 '17 at 8:22


















    • $begingroup$
      How does $left|vec{x} - bar{x}vec{1}right|= sum_i left|x_i - bar{x} right|$ ?
      $endgroup$
      – ThePassenger
      Jul 21 '17 at 8:22
















    $begingroup$
    How does $left|vec{x} - bar{x}vec{1}right|= sum_i left|x_i - bar{x} right|$ ?
    $endgroup$
    – ThePassenger
    Jul 21 '17 at 8:22




    $begingroup$
    How does $left|vec{x} - bar{x}vec{1}right|= sum_i left|x_i - bar{x} right|$ ?
    $endgroup$
    – ThePassenger
    Jul 21 '17 at 8:22











    0












    $begingroup$

    The Cauchy Schwarz inequality




    $$bigg(sum_{i=1}^{n}a_{i}^2bigg).bigg(sum_{i=1}^{n}b_{i}^2bigg)ge bigg(sum_{i=1}^{n}a_ib_ibigg)^2
    $$




    taking $a_i=|x_i-bar{x}|$ and $b_i=1/n$,



    $$
    bigg(sum_{i=1}^{n}|x_i-bar{x}|^2bigg).bigg(sum_{i=1}^{n}tfrac{1}{n^2}bigg)ge bigg(sum_{i=1}^{n}|x_i-bar{x}|.tfrac{1}{n}bigg)^2\
    bigg(sum_{i=1}^{n}(x_i-bar{x})^2bigg).bigg(n.tfrac{1}{n^2}bigg)ge bigg(frac{sum_{i=1}^{n}|x_i-bar{x}|}{{n}}bigg)^2\
    frac{sum_{i=1}^n(x_i-bar{x})^2}{n}ge bigg(frac{sum_{i=1}^n|x_i-bar{x}|}{n}bigg)^2\
    sqrtfrac{sum_{i=1}^n(x_i-bar{x})^2}{n}gefrac{sum_{i=1}^n|x_i-bar{x}|}{n}\
    S.Dge M.D
    $$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      The Cauchy Schwarz inequality




      $$bigg(sum_{i=1}^{n}a_{i}^2bigg).bigg(sum_{i=1}^{n}b_{i}^2bigg)ge bigg(sum_{i=1}^{n}a_ib_ibigg)^2
      $$




      taking $a_i=|x_i-bar{x}|$ and $b_i=1/n$,



      $$
      bigg(sum_{i=1}^{n}|x_i-bar{x}|^2bigg).bigg(sum_{i=1}^{n}tfrac{1}{n^2}bigg)ge bigg(sum_{i=1}^{n}|x_i-bar{x}|.tfrac{1}{n}bigg)^2\
      bigg(sum_{i=1}^{n}(x_i-bar{x})^2bigg).bigg(n.tfrac{1}{n^2}bigg)ge bigg(frac{sum_{i=1}^{n}|x_i-bar{x}|}{{n}}bigg)^2\
      frac{sum_{i=1}^n(x_i-bar{x})^2}{n}ge bigg(frac{sum_{i=1}^n|x_i-bar{x}|}{n}bigg)^2\
      sqrtfrac{sum_{i=1}^n(x_i-bar{x})^2}{n}gefrac{sum_{i=1}^n|x_i-bar{x}|}{n}\
      S.Dge M.D
      $$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        The Cauchy Schwarz inequality




        $$bigg(sum_{i=1}^{n}a_{i}^2bigg).bigg(sum_{i=1}^{n}b_{i}^2bigg)ge bigg(sum_{i=1}^{n}a_ib_ibigg)^2
        $$




        taking $a_i=|x_i-bar{x}|$ and $b_i=1/n$,



        $$
        bigg(sum_{i=1}^{n}|x_i-bar{x}|^2bigg).bigg(sum_{i=1}^{n}tfrac{1}{n^2}bigg)ge bigg(sum_{i=1}^{n}|x_i-bar{x}|.tfrac{1}{n}bigg)^2\
        bigg(sum_{i=1}^{n}(x_i-bar{x})^2bigg).bigg(n.tfrac{1}{n^2}bigg)ge bigg(frac{sum_{i=1}^{n}|x_i-bar{x}|}{{n}}bigg)^2\
        frac{sum_{i=1}^n(x_i-bar{x})^2}{n}ge bigg(frac{sum_{i=1}^n|x_i-bar{x}|}{n}bigg)^2\
        sqrtfrac{sum_{i=1}^n(x_i-bar{x})^2}{n}gefrac{sum_{i=1}^n|x_i-bar{x}|}{n}\
        S.Dge M.D
        $$






        share|cite|improve this answer











        $endgroup$



        The Cauchy Schwarz inequality




        $$bigg(sum_{i=1}^{n}a_{i}^2bigg).bigg(sum_{i=1}^{n}b_{i}^2bigg)ge bigg(sum_{i=1}^{n}a_ib_ibigg)^2
        $$




        taking $a_i=|x_i-bar{x}|$ and $b_i=1/n$,



        $$
        bigg(sum_{i=1}^{n}|x_i-bar{x}|^2bigg).bigg(sum_{i=1}^{n}tfrac{1}{n^2}bigg)ge bigg(sum_{i=1}^{n}|x_i-bar{x}|.tfrac{1}{n}bigg)^2\
        bigg(sum_{i=1}^{n}(x_i-bar{x})^2bigg).bigg(n.tfrac{1}{n^2}bigg)ge bigg(frac{sum_{i=1}^{n}|x_i-bar{x}|}{{n}}bigg)^2\
        frac{sum_{i=1}^n(x_i-bar{x})^2}{n}ge bigg(frac{sum_{i=1}^n|x_i-bar{x}|}{n}bigg)^2\
        sqrtfrac{sum_{i=1}^n(x_i-bar{x})^2}{n}gefrac{sum_{i=1}^n|x_i-bar{x}|}{n}\
        S.Dge M.D
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 25 '18 at 19:23

























        answered Jul 21 '17 at 10:27









        ss1729ss1729

        2,00311024




        2,00311024






























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