Why does $ operatorname{Var}(X) = E[X^2] - (E[X])^2 $












8












$begingroup$


$ operatorname{Var}(X) = E[X^2] - (E[X])^2 $



I have seen and understand (mathematically) the proof for this. What I want to understand is: intuitively, why is this true? What does this formula tell us? From the formula, we see that if we subtract the square of expected value of x from the expected value of $ x^2 $, we get a measure of dispersion in the data (or in the case of standard deviation, the root of this value gets us a measure of dispersion in the data).



So it seems that there is some linkage between the expected value of $ x^2 $ and $ x $. How do I make sense of this formula? For example, the formula



$$ sigma^2 = frac 1n sum_{i = 1}^n (x_i - bar{x})^2 $$



makes perfect intuitive sense. It simply gives us the average of squares of deviations from the mean. What does the other formula tell us?










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$endgroup$








  • 1




    $begingroup$
    But... this is... a... definition, no?
    $endgroup$
    – Did
    Dec 4 '18 at 23:51
















8












$begingroup$


$ operatorname{Var}(X) = E[X^2] - (E[X])^2 $



I have seen and understand (mathematically) the proof for this. What I want to understand is: intuitively, why is this true? What does this formula tell us? From the formula, we see that if we subtract the square of expected value of x from the expected value of $ x^2 $, we get a measure of dispersion in the data (or in the case of standard deviation, the root of this value gets us a measure of dispersion in the data).



So it seems that there is some linkage between the expected value of $ x^2 $ and $ x $. How do I make sense of this formula? For example, the formula



$$ sigma^2 = frac 1n sum_{i = 1}^n (x_i - bar{x})^2 $$



makes perfect intuitive sense. It simply gives us the average of squares of deviations from the mean. What does the other formula tell us?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    But... this is... a... definition, no?
    $endgroup$
    – Did
    Dec 4 '18 at 23:51














8












8








8


1



$begingroup$


$ operatorname{Var}(X) = E[X^2] - (E[X])^2 $



I have seen and understand (mathematically) the proof for this. What I want to understand is: intuitively, why is this true? What does this formula tell us? From the formula, we see that if we subtract the square of expected value of x from the expected value of $ x^2 $, we get a measure of dispersion in the data (or in the case of standard deviation, the root of this value gets us a measure of dispersion in the data).



So it seems that there is some linkage between the expected value of $ x^2 $ and $ x $. How do I make sense of this formula? For example, the formula



$$ sigma^2 = frac 1n sum_{i = 1}^n (x_i - bar{x})^2 $$



makes perfect intuitive sense. It simply gives us the average of squares of deviations from the mean. What does the other formula tell us?










share|cite|improve this question











$endgroup$




$ operatorname{Var}(X) = E[X^2] - (E[X])^2 $



I have seen and understand (mathematically) the proof for this. What I want to understand is: intuitively, why is this true? What does this formula tell us? From the formula, we see that if we subtract the square of expected value of x from the expected value of $ x^2 $, we get a measure of dispersion in the data (or in the case of standard deviation, the root of this value gets us a measure of dispersion in the data).



So it seems that there is some linkage between the expected value of $ x^2 $ and $ x $. How do I make sense of this formula? For example, the formula



$$ sigma^2 = frac 1n sum_{i = 1}^n (x_i - bar{x})^2 $$



makes perfect intuitive sense. It simply gives us the average of squares of deviations from the mean. What does the other formula tell us?







probability statistics variance






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share|cite|improve this question













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share|cite|improve this question








edited Dec 4 '18 at 22:50









Foobaz John

22.3k41452




22.3k41452










asked Dec 4 '18 at 22:45









WorldGovWorldGov

324111




324111








  • 1




    $begingroup$
    But... this is... a... definition, no?
    $endgroup$
    – Did
    Dec 4 '18 at 23:51














  • 1




    $begingroup$
    But... this is... a... definition, no?
    $endgroup$
    – Did
    Dec 4 '18 at 23:51








1




1




$begingroup$
But... this is... a... definition, no?
$endgroup$
– Did
Dec 4 '18 at 23:51




$begingroup$
But... this is... a... definition, no?
$endgroup$
– Did
Dec 4 '18 at 23:51










4 Answers
4






active

oldest

votes


















1












$begingroup$

The other formula tells you exactly the same thing as the one that you have given with $x,x^2$ $&$ $n$. You say you understand this formula so I assume that you also get that variance is just the average of all the deviations squared.



Now,
$mathbb{E}(X)$ is just the average of of all $x’_is$, which is to say that it is the mean of all $x’_is$.



Let us now define a deviation using the expectation operator.
$$Deviation = D = (X-mathbb{E}(X))$$
And Deviation squared is,
$$D^2 = (X-mathbb{E}(X))^2$$



Now that we have deviation let’s find the variance.
Using the above mentioned definition of variance, you should be able to see that



$$Variance = mathbb{E}(D^2)$$
Since $mathbb{E}(X)$ is the average value of $X$,The above equation is just the average of deviations squared.



Putting the value of $D^2$, we get,
$$Var(X) = mathbb{E}(X-mathbb{E}(X))^2 = mathbb{E}(X^2+mathbb{E}(X)^2-2X*mathbb{E}(X)) = mathbb{E}(X^2)+mathbb{E}(X)^2-2mathbb{E}(X)^2 = mathbb{E}(X^2)-mathbb{E}(X)^2$$
Hope this helps.






share|cite|improve this answer









$endgroup$





















    5












    $begingroup$

    Easy! Expand by the definition. Variance is the mean squared deviation, i.e., $V(X) = E((X-mu)^2).$ Now:



    $$ (X-mu)^2 = X^2 - 2X mu + mu^2$$



    and use the fact that $E(cdot)$ is a linear function and that $mu$ (the mean) is a constant.



    The shortcut computes the same thing, but counts the difference in the mean of squares and the square of the mean.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How can one prove that the expected value is a linear function?
      $endgroup$
      – Zacky
      Dec 4 '18 at 22:57






    • 1




      $begingroup$
      It follows from writing it as a sum: $$E(kX + Y) = sum (kxP(X = x) + yP(Y = y)) = ksum xP(X = x) + sum yP(Y = y)$$
      $endgroup$
      – Sean Roberson
      Dec 4 '18 at 22:59








    • 1




      $begingroup$
      Just to add to this, and take this with a grain of salt since I don't know probability: That this is a good definition for variance follows from wanting to get a sense of the distance you expect values of your random variable to be from the mean, one might naively choose the absolute value, but squaring is better as a smooth operation.
      $endgroup$
      – qbert
      Dec 4 '18 at 23:50



















    4












    $begingroup$

    Some times ago, a professor showed me this right triangle:



    enter image description here



    The formula you reported can be seen as the application of the Phytagora's theorem:



    $$P = mathbb{E}[X^2] = text{Var}[X] + mathbb{E}^2[X].$$



    Here, $P = mathbb{E}^2[X]$ (which is the second uncentered moment of $X$) is read as "the power" of $X$. Indeed, there is a physical explanation.



    In physics, energy and power are related to the "square" of some quantity (i.e. $X$ can be velocity for kinetic energy, current for Joule law, etc.).



    Suppose that these quantities are random (indeed, $X$ is a random variable). Then, the power $P$ is the sum of two contribution:




    1. The square of the expected value of $X$;

    2. Its variance (i.e. how much it varies from the expected value).


    It is clear that, if $X$ is not random, then $text{Var}[X] = 0$ and $mathbb{E}^2[X] = X^2$, so that:



    $$P = X^2,$$



    which is a typical physical definition of energy/power. When randomness is present, the we must use the whole formula



    $$P = mathbb{E}[X^2] = text{Var}[X] + mathbb{E}^2[X]$$



    to evaluate the power of the signal.



    As a final remark, the power of $X$ can be seen as the length of the vector which components corresponds to the square of its expected value plus its variability.





    P.S.
    A further clarification... the values $P$, $text{Var}[X]$ and $mathbb{E}^2[X]$ represent the squares of the sides of the triangle, not their length...






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      +1, I love this interpretation! I never saw it before.
      $endgroup$
      – Sean Roberson
      Dec 5 '18 at 0:50



















    0












    $begingroup$

    One intuitive way of measuring the variation of $X$ would be to look at how far, on average, $X$ is from it’s mean, $E(X)=mu$. That is, we want to compute $E(X-mu)$. However, mathematically, it’s “inconvenient” to use $E(X-mu)$, so we use the more convenient $E((X-mu)^{2}))$.



    To add, the formula you gave above, $frac{1}{n}sum_{i=1}^{n}(x_{i}-bar{x})$ is what you would use when you have finite data points. There is nothing random once you have your data points. $Var(X)$ is for a random variable, that can take on finite values, infinite countable values, or values on an interval.






    share|cite|improve this answer











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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      The other formula tells you exactly the same thing as the one that you have given with $x,x^2$ $&$ $n$. You say you understand this formula so I assume that you also get that variance is just the average of all the deviations squared.



      Now,
      $mathbb{E}(X)$ is just the average of of all $x’_is$, which is to say that it is the mean of all $x’_is$.



      Let us now define a deviation using the expectation operator.
      $$Deviation = D = (X-mathbb{E}(X))$$
      And Deviation squared is,
      $$D^2 = (X-mathbb{E}(X))^2$$



      Now that we have deviation let’s find the variance.
      Using the above mentioned definition of variance, you should be able to see that



      $$Variance = mathbb{E}(D^2)$$
      Since $mathbb{E}(X)$ is the average value of $X$,The above equation is just the average of deviations squared.



      Putting the value of $D^2$, we get,
      $$Var(X) = mathbb{E}(X-mathbb{E}(X))^2 = mathbb{E}(X^2+mathbb{E}(X)^2-2X*mathbb{E}(X)) = mathbb{E}(X^2)+mathbb{E}(X)^2-2mathbb{E}(X)^2 = mathbb{E}(X^2)-mathbb{E}(X)^2$$
      Hope this helps.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        The other formula tells you exactly the same thing as the one that you have given with $x,x^2$ $&$ $n$. You say you understand this formula so I assume that you also get that variance is just the average of all the deviations squared.



        Now,
        $mathbb{E}(X)$ is just the average of of all $x’_is$, which is to say that it is the mean of all $x’_is$.



        Let us now define a deviation using the expectation operator.
        $$Deviation = D = (X-mathbb{E}(X))$$
        And Deviation squared is,
        $$D^2 = (X-mathbb{E}(X))^2$$



        Now that we have deviation let’s find the variance.
        Using the above mentioned definition of variance, you should be able to see that



        $$Variance = mathbb{E}(D^2)$$
        Since $mathbb{E}(X)$ is the average value of $X$,The above equation is just the average of deviations squared.



        Putting the value of $D^2$, we get,
        $$Var(X) = mathbb{E}(X-mathbb{E}(X))^2 = mathbb{E}(X^2+mathbb{E}(X)^2-2X*mathbb{E}(X)) = mathbb{E}(X^2)+mathbb{E}(X)^2-2mathbb{E}(X)^2 = mathbb{E}(X^2)-mathbb{E}(X)^2$$
        Hope this helps.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          The other formula tells you exactly the same thing as the one that you have given with $x,x^2$ $&$ $n$. You say you understand this formula so I assume that you also get that variance is just the average of all the deviations squared.



          Now,
          $mathbb{E}(X)$ is just the average of of all $x’_is$, which is to say that it is the mean of all $x’_is$.



          Let us now define a deviation using the expectation operator.
          $$Deviation = D = (X-mathbb{E}(X))$$
          And Deviation squared is,
          $$D^2 = (X-mathbb{E}(X))^2$$



          Now that we have deviation let’s find the variance.
          Using the above mentioned definition of variance, you should be able to see that



          $$Variance = mathbb{E}(D^2)$$
          Since $mathbb{E}(X)$ is the average value of $X$,The above equation is just the average of deviations squared.



          Putting the value of $D^2$, we get,
          $$Var(X) = mathbb{E}(X-mathbb{E}(X))^2 = mathbb{E}(X^2+mathbb{E}(X)^2-2X*mathbb{E}(X)) = mathbb{E}(X^2)+mathbb{E}(X)^2-2mathbb{E}(X)^2 = mathbb{E}(X^2)-mathbb{E}(X)^2$$
          Hope this helps.






          share|cite|improve this answer









          $endgroup$



          The other formula tells you exactly the same thing as the one that you have given with $x,x^2$ $&$ $n$. You say you understand this formula so I assume that you also get that variance is just the average of all the deviations squared.



          Now,
          $mathbb{E}(X)$ is just the average of of all $x’_is$, which is to say that it is the mean of all $x’_is$.



          Let us now define a deviation using the expectation operator.
          $$Deviation = D = (X-mathbb{E}(X))$$
          And Deviation squared is,
          $$D^2 = (X-mathbb{E}(X))^2$$



          Now that we have deviation let’s find the variance.
          Using the above mentioned definition of variance, you should be able to see that



          $$Variance = mathbb{E}(D^2)$$
          Since $mathbb{E}(X)$ is the average value of $X$,The above equation is just the average of deviations squared.



          Putting the value of $D^2$, we get,
          $$Var(X) = mathbb{E}(X-mathbb{E}(X))^2 = mathbb{E}(X^2+mathbb{E}(X)^2-2X*mathbb{E}(X)) = mathbb{E}(X^2)+mathbb{E}(X)^2-2mathbb{E}(X)^2 = mathbb{E}(X^2)-mathbb{E}(X)^2$$
          Hope this helps.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 0:02









          user601297user601297

          37119




          37119























              5












              $begingroup$

              Easy! Expand by the definition. Variance is the mean squared deviation, i.e., $V(X) = E((X-mu)^2).$ Now:



              $$ (X-mu)^2 = X^2 - 2X mu + mu^2$$



              and use the fact that $E(cdot)$ is a linear function and that $mu$ (the mean) is a constant.



              The shortcut computes the same thing, but counts the difference in the mean of squares and the square of the mean.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                How can one prove that the expected value is a linear function?
                $endgroup$
                – Zacky
                Dec 4 '18 at 22:57






              • 1




                $begingroup$
                It follows from writing it as a sum: $$E(kX + Y) = sum (kxP(X = x) + yP(Y = y)) = ksum xP(X = x) + sum yP(Y = y)$$
                $endgroup$
                – Sean Roberson
                Dec 4 '18 at 22:59








              • 1




                $begingroup$
                Just to add to this, and take this with a grain of salt since I don't know probability: That this is a good definition for variance follows from wanting to get a sense of the distance you expect values of your random variable to be from the mean, one might naively choose the absolute value, but squaring is better as a smooth operation.
                $endgroup$
                – qbert
                Dec 4 '18 at 23:50
















              5












              $begingroup$

              Easy! Expand by the definition. Variance is the mean squared deviation, i.e., $V(X) = E((X-mu)^2).$ Now:



              $$ (X-mu)^2 = X^2 - 2X mu + mu^2$$



              and use the fact that $E(cdot)$ is a linear function and that $mu$ (the mean) is a constant.



              The shortcut computes the same thing, but counts the difference in the mean of squares and the square of the mean.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                How can one prove that the expected value is a linear function?
                $endgroup$
                – Zacky
                Dec 4 '18 at 22:57






              • 1




                $begingroup$
                It follows from writing it as a sum: $$E(kX + Y) = sum (kxP(X = x) + yP(Y = y)) = ksum xP(X = x) + sum yP(Y = y)$$
                $endgroup$
                – Sean Roberson
                Dec 4 '18 at 22:59








              • 1




                $begingroup$
                Just to add to this, and take this with a grain of salt since I don't know probability: That this is a good definition for variance follows from wanting to get a sense of the distance you expect values of your random variable to be from the mean, one might naively choose the absolute value, but squaring is better as a smooth operation.
                $endgroup$
                – qbert
                Dec 4 '18 at 23:50














              5












              5








              5





              $begingroup$

              Easy! Expand by the definition. Variance is the mean squared deviation, i.e., $V(X) = E((X-mu)^2).$ Now:



              $$ (X-mu)^2 = X^2 - 2X mu + mu^2$$



              and use the fact that $E(cdot)$ is a linear function and that $mu$ (the mean) is a constant.



              The shortcut computes the same thing, but counts the difference in the mean of squares and the square of the mean.






              share|cite|improve this answer









              $endgroup$



              Easy! Expand by the definition. Variance is the mean squared deviation, i.e., $V(X) = E((X-mu)^2).$ Now:



              $$ (X-mu)^2 = X^2 - 2X mu + mu^2$$



              and use the fact that $E(cdot)$ is a linear function and that $mu$ (the mean) is a constant.



              The shortcut computes the same thing, but counts the difference in the mean of squares and the square of the mean.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 4 '18 at 22:50









              Sean RobersonSean Roberson

              6,39031327




              6,39031327












              • $begingroup$
                How can one prove that the expected value is a linear function?
                $endgroup$
                – Zacky
                Dec 4 '18 at 22:57






              • 1




                $begingroup$
                It follows from writing it as a sum: $$E(kX + Y) = sum (kxP(X = x) + yP(Y = y)) = ksum xP(X = x) + sum yP(Y = y)$$
                $endgroup$
                – Sean Roberson
                Dec 4 '18 at 22:59








              • 1




                $begingroup$
                Just to add to this, and take this with a grain of salt since I don't know probability: That this is a good definition for variance follows from wanting to get a sense of the distance you expect values of your random variable to be from the mean, one might naively choose the absolute value, but squaring is better as a smooth operation.
                $endgroup$
                – qbert
                Dec 4 '18 at 23:50


















              • $begingroup$
                How can one prove that the expected value is a linear function?
                $endgroup$
                – Zacky
                Dec 4 '18 at 22:57






              • 1




                $begingroup$
                It follows from writing it as a sum: $$E(kX + Y) = sum (kxP(X = x) + yP(Y = y)) = ksum xP(X = x) + sum yP(Y = y)$$
                $endgroup$
                – Sean Roberson
                Dec 4 '18 at 22:59








              • 1




                $begingroup$
                Just to add to this, and take this with a grain of salt since I don't know probability: That this is a good definition for variance follows from wanting to get a sense of the distance you expect values of your random variable to be from the mean, one might naively choose the absolute value, but squaring is better as a smooth operation.
                $endgroup$
                – qbert
                Dec 4 '18 at 23:50
















              $begingroup$
              How can one prove that the expected value is a linear function?
              $endgroup$
              – Zacky
              Dec 4 '18 at 22:57




              $begingroup$
              How can one prove that the expected value is a linear function?
              $endgroup$
              – Zacky
              Dec 4 '18 at 22:57




              1




              1




              $begingroup$
              It follows from writing it as a sum: $$E(kX + Y) = sum (kxP(X = x) + yP(Y = y)) = ksum xP(X = x) + sum yP(Y = y)$$
              $endgroup$
              – Sean Roberson
              Dec 4 '18 at 22:59






              $begingroup$
              It follows from writing it as a sum: $$E(kX + Y) = sum (kxP(X = x) + yP(Y = y)) = ksum xP(X = x) + sum yP(Y = y)$$
              $endgroup$
              – Sean Roberson
              Dec 4 '18 at 22:59






              1




              1




              $begingroup$
              Just to add to this, and take this with a grain of salt since I don't know probability: That this is a good definition for variance follows from wanting to get a sense of the distance you expect values of your random variable to be from the mean, one might naively choose the absolute value, but squaring is better as a smooth operation.
              $endgroup$
              – qbert
              Dec 4 '18 at 23:50




              $begingroup$
              Just to add to this, and take this with a grain of salt since I don't know probability: That this is a good definition for variance follows from wanting to get a sense of the distance you expect values of your random variable to be from the mean, one might naively choose the absolute value, but squaring is better as a smooth operation.
              $endgroup$
              – qbert
              Dec 4 '18 at 23:50











              4












              $begingroup$

              Some times ago, a professor showed me this right triangle:



              enter image description here



              The formula you reported can be seen as the application of the Phytagora's theorem:



              $$P = mathbb{E}[X^2] = text{Var}[X] + mathbb{E}^2[X].$$



              Here, $P = mathbb{E}^2[X]$ (which is the second uncentered moment of $X$) is read as "the power" of $X$. Indeed, there is a physical explanation.



              In physics, energy and power are related to the "square" of some quantity (i.e. $X$ can be velocity for kinetic energy, current for Joule law, etc.).



              Suppose that these quantities are random (indeed, $X$ is a random variable). Then, the power $P$ is the sum of two contribution:




              1. The square of the expected value of $X$;

              2. Its variance (i.e. how much it varies from the expected value).


              It is clear that, if $X$ is not random, then $text{Var}[X] = 0$ and $mathbb{E}^2[X] = X^2$, so that:



              $$P = X^2,$$



              which is a typical physical definition of energy/power. When randomness is present, the we must use the whole formula



              $$P = mathbb{E}[X^2] = text{Var}[X] + mathbb{E}^2[X]$$



              to evaluate the power of the signal.



              As a final remark, the power of $X$ can be seen as the length of the vector which components corresponds to the square of its expected value plus its variability.





              P.S.
              A further clarification... the values $P$, $text{Var}[X]$ and $mathbb{E}^2[X]$ represent the squares of the sides of the triangle, not their length...






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                +1, I love this interpretation! I never saw it before.
                $endgroup$
                – Sean Roberson
                Dec 5 '18 at 0:50
















              4












              $begingroup$

              Some times ago, a professor showed me this right triangle:



              enter image description here



              The formula you reported can be seen as the application of the Phytagora's theorem:



              $$P = mathbb{E}[X^2] = text{Var}[X] + mathbb{E}^2[X].$$



              Here, $P = mathbb{E}^2[X]$ (which is the second uncentered moment of $X$) is read as "the power" of $X$. Indeed, there is a physical explanation.



              In physics, energy and power are related to the "square" of some quantity (i.e. $X$ can be velocity for kinetic energy, current for Joule law, etc.).



              Suppose that these quantities are random (indeed, $X$ is a random variable). Then, the power $P$ is the sum of two contribution:




              1. The square of the expected value of $X$;

              2. Its variance (i.e. how much it varies from the expected value).


              It is clear that, if $X$ is not random, then $text{Var}[X] = 0$ and $mathbb{E}^2[X] = X^2$, so that:



              $$P = X^2,$$



              which is a typical physical definition of energy/power. When randomness is present, the we must use the whole formula



              $$P = mathbb{E}[X^2] = text{Var}[X] + mathbb{E}^2[X]$$



              to evaluate the power of the signal.



              As a final remark, the power of $X$ can be seen as the length of the vector which components corresponds to the square of its expected value plus its variability.





              P.S.
              A further clarification... the values $P$, $text{Var}[X]$ and $mathbb{E}^2[X]$ represent the squares of the sides of the triangle, not their length...






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                +1, I love this interpretation! I never saw it before.
                $endgroup$
                – Sean Roberson
                Dec 5 '18 at 0:50














              4












              4








              4





              $begingroup$

              Some times ago, a professor showed me this right triangle:



              enter image description here



              The formula you reported can be seen as the application of the Phytagora's theorem:



              $$P = mathbb{E}[X^2] = text{Var}[X] + mathbb{E}^2[X].$$



              Here, $P = mathbb{E}^2[X]$ (which is the second uncentered moment of $X$) is read as "the power" of $X$. Indeed, there is a physical explanation.



              In physics, energy and power are related to the "square" of some quantity (i.e. $X$ can be velocity for kinetic energy, current for Joule law, etc.).



              Suppose that these quantities are random (indeed, $X$ is a random variable). Then, the power $P$ is the sum of two contribution:




              1. The square of the expected value of $X$;

              2. Its variance (i.e. how much it varies from the expected value).


              It is clear that, if $X$ is not random, then $text{Var}[X] = 0$ and $mathbb{E}^2[X] = X^2$, so that:



              $$P = X^2,$$



              which is a typical physical definition of energy/power. When randomness is present, the we must use the whole formula



              $$P = mathbb{E}[X^2] = text{Var}[X] + mathbb{E}^2[X]$$



              to evaluate the power of the signal.



              As a final remark, the power of $X$ can be seen as the length of the vector which components corresponds to the square of its expected value plus its variability.





              P.S.
              A further clarification... the values $P$, $text{Var}[X]$ and $mathbb{E}^2[X]$ represent the squares of the sides of the triangle, not their length...






              share|cite|improve this answer











              $endgroup$



              Some times ago, a professor showed me this right triangle:



              enter image description here



              The formula you reported can be seen as the application of the Phytagora's theorem:



              $$P = mathbb{E}[X^2] = text{Var}[X] + mathbb{E}^2[X].$$



              Here, $P = mathbb{E}^2[X]$ (which is the second uncentered moment of $X$) is read as "the power" of $X$. Indeed, there is a physical explanation.



              In physics, energy and power are related to the "square" of some quantity (i.e. $X$ can be velocity for kinetic energy, current for Joule law, etc.).



              Suppose that these quantities are random (indeed, $X$ is a random variable). Then, the power $P$ is the sum of two contribution:




              1. The square of the expected value of $X$;

              2. Its variance (i.e. how much it varies from the expected value).


              It is clear that, if $X$ is not random, then $text{Var}[X] = 0$ and $mathbb{E}^2[X] = X^2$, so that:



              $$P = X^2,$$



              which is a typical physical definition of energy/power. When randomness is present, the we must use the whole formula



              $$P = mathbb{E}[X^2] = text{Var}[X] + mathbb{E}^2[X]$$



              to evaluate the power of the signal.



              As a final remark, the power of $X$ can be seen as the length of the vector which components corresponds to the square of its expected value plus its variability.





              P.S.
              A further clarification... the values $P$, $text{Var}[X]$ and $mathbb{E}^2[X]$ represent the squares of the sides of the triangle, not their length...







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 4 '18 at 23:52

























              answered Dec 4 '18 at 23:47









              the_candymanthe_candyman

              8,97832145




              8,97832145












              • $begingroup$
                +1, I love this interpretation! I never saw it before.
                $endgroup$
                – Sean Roberson
                Dec 5 '18 at 0:50


















              • $begingroup$
                +1, I love this interpretation! I never saw it before.
                $endgroup$
                – Sean Roberson
                Dec 5 '18 at 0:50
















              $begingroup$
              +1, I love this interpretation! I never saw it before.
              $endgroup$
              – Sean Roberson
              Dec 5 '18 at 0:50




              $begingroup$
              +1, I love this interpretation! I never saw it before.
              $endgroup$
              – Sean Roberson
              Dec 5 '18 at 0:50











              0












              $begingroup$

              One intuitive way of measuring the variation of $X$ would be to look at how far, on average, $X$ is from it’s mean, $E(X)=mu$. That is, we want to compute $E(X-mu)$. However, mathematically, it’s “inconvenient” to use $E(X-mu)$, so we use the more convenient $E((X-mu)^{2}))$.



              To add, the formula you gave above, $frac{1}{n}sum_{i=1}^{n}(x_{i}-bar{x})$ is what you would use when you have finite data points. There is nothing random once you have your data points. $Var(X)$ is for a random variable, that can take on finite values, infinite countable values, or values on an interval.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                One intuitive way of measuring the variation of $X$ would be to look at how far, on average, $X$ is from it’s mean, $E(X)=mu$. That is, we want to compute $E(X-mu)$. However, mathematically, it’s “inconvenient” to use $E(X-mu)$, so we use the more convenient $E((X-mu)^{2}))$.



                To add, the formula you gave above, $frac{1}{n}sum_{i=1}^{n}(x_{i}-bar{x})$ is what you would use when you have finite data points. There is nothing random once you have your data points. $Var(X)$ is for a random variable, that can take on finite values, infinite countable values, or values on an interval.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  One intuitive way of measuring the variation of $X$ would be to look at how far, on average, $X$ is from it’s mean, $E(X)=mu$. That is, we want to compute $E(X-mu)$. However, mathematically, it’s “inconvenient” to use $E(X-mu)$, so we use the more convenient $E((X-mu)^{2}))$.



                  To add, the formula you gave above, $frac{1}{n}sum_{i=1}^{n}(x_{i}-bar{x})$ is what you would use when you have finite data points. There is nothing random once you have your data points. $Var(X)$ is for a random variable, that can take on finite values, infinite countable values, or values on an interval.






                  share|cite|improve this answer











                  $endgroup$



                  One intuitive way of measuring the variation of $X$ would be to look at how far, on average, $X$ is from it’s mean, $E(X)=mu$. That is, we want to compute $E(X-mu)$. However, mathematically, it’s “inconvenient” to use $E(X-mu)$, so we use the more convenient $E((X-mu)^{2}))$.



                  To add, the formula you gave above, $frac{1}{n}sum_{i=1}^{n}(x_{i}-bar{x})$ is what you would use when you have finite data points. There is nothing random once you have your data points. $Var(X)$ is for a random variable, that can take on finite values, infinite countable values, or values on an interval.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 4 '18 at 23:18

























                  answered Dec 4 '18 at 23:10









                  Live Free or π HardLive Free or π Hard

                  479213




                  479213






























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