Explanation for Additive Property of Variance?












2












$begingroup$


I'm wondering why variance has additive property, as opposed to why this property doesn't extend to standard deviation? Additive property is defined as:



Var(A+B) = Var(A) + Var(B)



I imagine this as adding two distribution together which makes sense. But in that case SD should have similar property as well. Why does variance possess this magical property?






variance







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  • 2




    $begingroup$
    This is only the case if $A$ and $B$ are uncorrelated random variables. If this holds, then $text{Sd}(A+B)=sqrt{text{Var}(A)+text{Var}(B)}$, which doesn't equal $text{Sd}(A)+text{Sd}(B)$ simply because $sqrt{a+b}ne sqrt a+sqrt b$ in general.
    $endgroup$
    – StubbornAtom
    yesterday








  • 1




    $begingroup$
    @StubbornAtom you should make it an answer.
    $endgroup$
    – Tim♦
    yesterday
















2












$begingroup$


I'm wondering why variance has additive property, as opposed to why this property doesn't extend to standard deviation? Additive property is defined as:



Var(A+B) = Var(A) + Var(B)



I imagine this as adding two distribution together which makes sense. But in that case SD should have similar property as well. Why does variance possess this magical property?






variance







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    This is only the case if $A$ and $B$ are uncorrelated random variables. If this holds, then $text{Sd}(A+B)=sqrt{text{Var}(A)+text{Var}(B)}$, which doesn't equal $text{Sd}(A)+text{Sd}(B)$ simply because $sqrt{a+b}ne sqrt a+sqrt b$ in general.
    $endgroup$
    – StubbornAtom
    yesterday








  • 1




    $begingroup$
    @StubbornAtom you should make it an answer.
    $endgroup$
    – Tim♦
    yesterday














2












2








2





$begingroup$


I'm wondering why variance has additive property, as opposed to why this property doesn't extend to standard deviation? Additive property is defined as:



Var(A+B) = Var(A) + Var(B)



I imagine this as adding two distribution together which makes sense. But in that case SD should have similar property as well. Why does variance possess this magical property?






variance







share|cite|improve this question









$endgroup$




I'm wondering why variance has additive property, as opposed to why this property doesn't extend to standard deviation? Additive property is defined as:



Var(A+B) = Var(A) + Var(B)



I imagine this as adding two distribution together which makes sense. But in that case SD should have similar property as well. Why does variance possess this magical property?






variance




variance






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked yesterday









Fudge AruthFudge Aruth

161




161








  • 2




    $begingroup$
    This is only the case if $A$ and $B$ are uncorrelated random variables. If this holds, then $text{Sd}(A+B)=sqrt{text{Var}(A)+text{Var}(B)}$, which doesn't equal $text{Sd}(A)+text{Sd}(B)$ simply because $sqrt{a+b}ne sqrt a+sqrt b$ in general.
    $endgroup$
    – StubbornAtom
    yesterday








  • 1




    $begingroup$
    @StubbornAtom you should make it an answer.
    $endgroup$
    – Tim♦
    yesterday














  • 2




    $begingroup$
    This is only the case if $A$ and $B$ are uncorrelated random variables. If this holds, then $text{Sd}(A+B)=sqrt{text{Var}(A)+text{Var}(B)}$, which doesn't equal $text{Sd}(A)+text{Sd}(B)$ simply because $sqrt{a+b}ne sqrt a+sqrt b$ in general.
    $endgroup$
    – StubbornAtom
    yesterday








  • 1




    $begingroup$
    @StubbornAtom you should make it an answer.
    $endgroup$
    – Tim♦
    yesterday








2




2




$begingroup$
This is only the case if $A$ and $B$ are uncorrelated random variables. If this holds, then $text{Sd}(A+B)=sqrt{text{Var}(A)+text{Var}(B)}$, which doesn't equal $text{Sd}(A)+text{Sd}(B)$ simply because $sqrt{a+b}ne sqrt a+sqrt b$ in general.
$endgroup$
– StubbornAtom
yesterday






$begingroup$
This is only the case if $A$ and $B$ are uncorrelated random variables. If this holds, then $text{Sd}(A+B)=sqrt{text{Var}(A)+text{Var}(B)}$, which doesn't equal $text{Sd}(A)+text{Sd}(B)$ simply because $sqrt{a+b}ne sqrt a+sqrt b$ in general.
$endgroup$
– StubbornAtom
yesterday






1




1




$begingroup$
@StubbornAtom you should make it an answer.
$endgroup$
– Tim♦
yesterday




$begingroup$
@StubbornAtom you should make it an answer.
$endgroup$
– Tim♦
yesterday










2 Answers
2






active

oldest

votes


















3












$begingroup$

It doesn't!



In general:



Var(A+B) = Var(A) + Var(B) + Cov(A, B)


The additive property only holds if the two random variables have no covariation. This is almost a circular statement, since a legitimate definition of the covariation could be:



Cov(A, B) = Var(A) + Var(B) - Var(A + B)


This means that the covariance measures the failure of the additive property of variance.



This leads to the true heart of the matter, the covariance is bi-linear:



Cov(A_1 + A_2, B) = Cov(A_1, B) + Cov(A_2, B)

Cov(A, B_1 + B_2) = Cov(A, B_1) + Cov(A, B_2)


For an intuitive understanding of this, I'll link to the wonderful: How would you explain covariance to someone who understands only the mean?. In particular, see @whuber's answer.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    enter image description here



    The first thing to notice is that Var(A+B) equals VarA + Var B only when Cov(A,B)=0.



    To gain some intuition behind the relationship between sd(A+B) and sd(A)+sd(B), notice that in order to complete the square in this expression enter image description here



    Cov(A,B) would have to equal sd(A)*sd(B). The next question is whether that ever happens? Indeed it does. The Cauchy Schwartz inequality gives us the inequality below:



    enter image description here.



    Whenever the following equality holds
    enter image description here, we can complete the square and obtain that sd(A+B)=sd(A)+sd(B). However, in all other cases, sd(A+B) will not equal sd(A)+sd(B).






    share|cite|improve this answer











    $endgroup$









    • 3




      $begingroup$
      You know you can use MathJax here instead of copy-pasting images of math formulas, right?
      $endgroup$
      – Ilmari Karonen
      yesterday










    • $begingroup$
      I was actually looking for a solution on this front :) Thank you for the link.
      $endgroup$
      – ColorStatistics
      yesterday











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    It doesn't!



    In general:



    Var(A+B) = Var(A) + Var(B) + Cov(A, B)


    The additive property only holds if the two random variables have no covariation. This is almost a circular statement, since a legitimate definition of the covariation could be:



    Cov(A, B) = Var(A) + Var(B) - Var(A + B)


    This means that the covariance measures the failure of the additive property of variance.



    This leads to the true heart of the matter, the covariance is bi-linear:



    Cov(A_1 + A_2, B) = Cov(A_1, B) + Cov(A_2, B)
    
Cov(A, B_1 + B_2) = Cov(A, B_1) + Cov(A, B_2)


    For an intuitive understanding of this, I'll link to the wonderful: How would you explain covariance to someone who understands only the mean?. In particular, see @whuber's answer.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      It doesn't!



      In general:



      Var(A+B) = Var(A) + Var(B) + Cov(A, B)


      The additive property only holds if the two random variables have no covariation. This is almost a circular statement, since a legitimate definition of the covariation could be:



      Cov(A, B) = Var(A) + Var(B) - Var(A + B)


      This means that the covariance measures the failure of the additive property of variance.



      This leads to the true heart of the matter, the covariance is bi-linear:



      Cov(A_1 + A_2, B) = Cov(A_1, B) + Cov(A_2, B)
      
Cov(A, B_1 + B_2) = Cov(A, B_1) + Cov(A, B_2)


      For an intuitive understanding of this, I'll link to the wonderful: How would you explain covariance to someone who understands only the mean?. In particular, see @whuber's answer.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        It doesn't!



        In general:



        Var(A+B) = Var(A) + Var(B) + Cov(A, B)


        The additive property only holds if the two random variables have no covariation. This is almost a circular statement, since a legitimate definition of the covariation could be:



        Cov(A, B) = Var(A) + Var(B) - Var(A + B)


        This means that the covariance measures the failure of the additive property of variance.



        This leads to the true heart of the matter, the covariance is bi-linear:



        Cov(A_1 + A_2, B) = Cov(A_1, B) + Cov(A_2, B)
        
Cov(A, B_1 + B_2) = Cov(A, B_1) + Cov(A, B_2)


        For an intuitive understanding of this, I'll link to the wonderful: How would you explain covariance to someone who understands only the mean?. In particular, see @whuber's answer.






        share|cite|improve this answer









        $endgroup$



        It doesn't!



        In general:



        Var(A+B) = Var(A) + Var(B) + Cov(A, B)


        The additive property only holds if the two random variables have no covariation. This is almost a circular statement, since a legitimate definition of the covariation could be:



        Cov(A, B) = Var(A) + Var(B) - Var(A + B)


        This means that the covariance measures the failure of the additive property of variance.



        This leads to the true heart of the matter, the covariance is bi-linear:



        Cov(A_1 + A_2, B) = Cov(A_1, B) + Cov(A_2, B)
        
Cov(A, B_1 + B_2) = Cov(A, B_1) + Cov(A, B_2)


        For an intuitive understanding of this, I'll link to the wonderful: How would you explain covariance to someone who understands only the mean?. In particular, see @whuber's answer.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Matthew DruryMatthew Drury

        25.6k261102




        25.6k261102

























            0












            $begingroup$

            enter image description here



            The first thing to notice is that Var(A+B) equals VarA + Var B only when Cov(A,B)=0.



            To gain some intuition behind the relationship between sd(A+B) and sd(A)+sd(B), notice that in order to complete the square in this expression enter image description here



            Cov(A,B) would have to equal sd(A)*sd(B). The next question is whether that ever happens? Indeed it does. The Cauchy Schwartz inequality gives us the inequality below:



            enter image description here.



            Whenever the following equality holds
            enter image description here, we can complete the square and obtain that sd(A+B)=sd(A)+sd(B). However, in all other cases, sd(A+B) will not equal sd(A)+sd(B).






            share|cite|improve this answer











            $endgroup$









            • 3




              $begingroup$
              You know you can use MathJax here instead of copy-pasting images of math formulas, right?
              $endgroup$
              – Ilmari Karonen
              yesterday










            • $begingroup$
              I was actually looking for a solution on this front :) Thank you for the link.
              $endgroup$
              – ColorStatistics
              yesterday
















            0












            $begingroup$

            enter image description here



            The first thing to notice is that Var(A+B) equals VarA + Var B only when Cov(A,B)=0.



            To gain some intuition behind the relationship between sd(A+B) and sd(A)+sd(B), notice that in order to complete the square in this expression enter image description here



            Cov(A,B) would have to equal sd(A)*sd(B). The next question is whether that ever happens? Indeed it does. The Cauchy Schwartz inequality gives us the inequality below:



            enter image description here.



            Whenever the following equality holds
            enter image description here, we can complete the square and obtain that sd(A+B)=sd(A)+sd(B). However, in all other cases, sd(A+B) will not equal sd(A)+sd(B).






            share|cite|improve this answer











            $endgroup$









            • 3




              $begingroup$
              You know you can use MathJax here instead of copy-pasting images of math formulas, right?
              $endgroup$
              – Ilmari Karonen
              yesterday










            • $begingroup$
              I was actually looking for a solution on this front :) Thank you for the link.
              $endgroup$
              – ColorStatistics
              yesterday














            0












            0








            0





            $begingroup$

            enter image description here



            The first thing to notice is that Var(A+B) equals VarA + Var B only when Cov(A,B)=0.



            To gain some intuition behind the relationship between sd(A+B) and sd(A)+sd(B), notice that in order to complete the square in this expression enter image description here



            Cov(A,B) would have to equal sd(A)*sd(B). The next question is whether that ever happens? Indeed it does. The Cauchy Schwartz inequality gives us the inequality below:



            enter image description here.



            Whenever the following equality holds
            enter image description here, we can complete the square and obtain that sd(A+B)=sd(A)+sd(B). However, in all other cases, sd(A+B) will not equal sd(A)+sd(B).






            share|cite|improve this answer











            $endgroup$



            enter image description here



            The first thing to notice is that Var(A+B) equals VarA + Var B only when Cov(A,B)=0.



            To gain some intuition behind the relationship between sd(A+B) and sd(A)+sd(B), notice that in order to complete the square in this expression enter image description here



            Cov(A,B) would have to equal sd(A)*sd(B). The next question is whether that ever happens? Indeed it does. The Cauchy Schwartz inequality gives us the inequality below:



            enter image description here.



            Whenever the following equality holds
            enter image description here, we can complete the square and obtain that sd(A+B)=sd(A)+sd(B). However, in all other cases, sd(A+B) will not equal sd(A)+sd(B).







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited yesterday

























            answered yesterday









            ColorStatisticsColorStatistics

            595114




            595114








            • 3




              $begingroup$
              You know you can use MathJax here instead of copy-pasting images of math formulas, right?
              $endgroup$
              – Ilmari Karonen
              yesterday










            • $begingroup$
              I was actually looking for a solution on this front :) Thank you for the link.
              $endgroup$
              – ColorStatistics
              yesterday














            • 3




              $begingroup$
              You know you can use MathJax here instead of copy-pasting images of math formulas, right?
              $endgroup$
              – Ilmari Karonen
              yesterday










            • $begingroup$
              I was actually looking for a solution on this front :) Thank you for the link.
              $endgroup$
              – ColorStatistics
              yesterday








            3




            3




            $begingroup$
            You know you can use MathJax here instead of copy-pasting images of math formulas, right?
            $endgroup$
            – Ilmari Karonen
            yesterday




            $begingroup$
            You know you can use MathJax here instead of copy-pasting images of math formulas, right?
            $endgroup$
            – Ilmari Karonen
            yesterday












            $begingroup$
            I was actually looking for a solution on this front :) Thank you for the link.
            $endgroup$
            – ColorStatistics
            yesterday




            $begingroup$
            I was actually looking for a solution on this front :) Thank you for the link.
            $endgroup$
            – ColorStatistics
            yesterday


















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