Expected value and variance of a piecewise function with the integral












0












$begingroup$


I have the following stepwise function:
f(x) = 1/3 for -1 <= x < 0, 2/3 for 0 <=x<=1, 0 otherwise



I wonder how I can derive E(X) and Var(X) of a stepwise uniform function using the integral. If I draw it I would say the mean is at x = 1/4 (as then the area below the function equals 0.5). But when I am integrating it I get 1/6.



Can someony help me and show how I properly solve this question?



Thanks a lot!






integration statistics means variance







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Perhaps we could be more helpful if you explained how your got $1/6,$ and why you think it is wrong.
    $endgroup$
    – BruceET
    Sep 24 '17 at 6:07












  • $begingroup$
    What about applying the formula for of E(X) in terms of the PDF?
    $endgroup$
    – Did
    Sep 24 '17 at 18:22
















0












$begingroup$


I have the following stepwise function:
f(x) = 1/3 for -1 <= x < 0, 2/3 for 0 <=x<=1, 0 otherwise



I wonder how I can derive E(X) and Var(X) of a stepwise uniform function using the integral. If I draw it I would say the mean is at x = 1/4 (as then the area below the function equals 0.5). But when I am integrating it I get 1/6.



Can someony help me and show how I properly solve this question?



Thanks a lot!






integration statistics means variance







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Perhaps we could be more helpful if you explained how your got $1/6,$ and why you think it is wrong.
    $endgroup$
    – BruceET
    Sep 24 '17 at 6:07












  • $begingroup$
    What about applying the formula for of E(X) in terms of the PDF?
    $endgroup$
    – Did
    Sep 24 '17 at 18:22














0












0








0





$begingroup$


I have the following stepwise function:
f(x) = 1/3 for -1 <= x < 0, 2/3 for 0 <=x<=1, 0 otherwise



I wonder how I can derive E(X) and Var(X) of a stepwise uniform function using the integral. If I draw it I would say the mean is at x = 1/4 (as then the area below the function equals 0.5). But when I am integrating it I get 1/6.



Can someony help me and show how I properly solve this question?



Thanks a lot!






integration statistics means variance







share|cite|improve this question









$endgroup$




I have the following stepwise function:
f(x) = 1/3 for -1 <= x < 0, 2/3 for 0 <=x<=1, 0 otherwise



I wonder how I can derive E(X) and Var(X) of a stepwise uniform function using the integral. If I draw it I would say the mean is at x = 1/4 (as then the area below the function equals 0.5). But when I am integrating it I get 1/6.



Can someony help me and show how I properly solve this question?



Thanks a lot!






integration statistics means variance




integration statistics means variance






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 24 '17 at 4:51









Severin LinderSeverin Linder

6




6








  • 1




    $begingroup$
    Perhaps we could be more helpful if you explained how your got $1/6,$ and why you think it is wrong.
    $endgroup$
    – BruceET
    Sep 24 '17 at 6:07












  • $begingroup$
    What about applying the formula for of E(X) in terms of the PDF?
    $endgroup$
    – Did
    Sep 24 '17 at 18:22














  • 1




    $begingroup$
    Perhaps we could be more helpful if you explained how your got $1/6,$ and why you think it is wrong.
    $endgroup$
    – BruceET
    Sep 24 '17 at 6:07












  • $begingroup$
    What about applying the formula for of E(X) in terms of the PDF?
    $endgroup$
    – Did
    Sep 24 '17 at 18:22








1




1




$begingroup$
Perhaps we could be more helpful if you explained how your got $1/6,$ and why you think it is wrong.
$endgroup$
– BruceET
Sep 24 '17 at 6:07






$begingroup$
Perhaps we could be more helpful if you explained how your got $1/6,$ and why you think it is wrong.
$endgroup$
– BruceET
Sep 24 '17 at 6:07














$begingroup$
What about applying the formula for of E(X) in terms of the PDF?
$endgroup$
– Did
Sep 24 '17 at 18:22




$begingroup$
What about applying the formula for of E(X) in terms of the PDF?
$endgroup$
– Did
Sep 24 '17 at 18:22










2 Answers
2






active

oldest

votes


















0












$begingroup$

Here is a start:



$$mu = E(X) = int_{-1}^1 xf(x),dx = int_{-1}^0 x(1/3),dx
+ int_0^1 x(2/3),dx = dots,.$$
Find $E(X^2)$ similarly, and use $Var(X) = E(X^2) - mu^2.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you already! So, then $$mu=E[X]=int_{-1}^0x(1/3)dx + int_0^1x(2/3)dx = (0-{1over6})+({1over3}-0)={1over6}$$ and $$Var[X]=int_{-1}^0x^2(1/3)dx + int_0^1x^2(2/3)dx - mu^2=(0+{1over9})+({2over9}-0)-({1over6})^2={11over36}$$ Is this correct?
    $endgroup$
    – Severin Linder
    Sep 24 '17 at 17:30



















0












$begingroup$

Thank you already!



So, then



$$mu=E[X]=int_{-1}^0x(1/3)dx + int_0^1x(2/3)dx = (0-{1over6})+({1over3}-0)={1over6}$$



and



$$Var[X]=int_{-1}^0x^2(1/3)dx + int_0^1x^2(2/3)dx - mu^2=(0+{1over9})+({2over9}-0)-({1over6})^2={11over36}$$



Is this correct?






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Here is a start:



    $$mu = E(X) = int_{-1}^1 xf(x),dx = int_{-1}^0 x(1/3),dx
    + int_0^1 x(2/3),dx = dots,.$$
    Find $E(X^2)$ similarly, and use $Var(X) = E(X^2) - mu^2.$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you already! So, then $$mu=E[X]=int_{-1}^0x(1/3)dx + int_0^1x(2/3)dx = (0-{1over6})+({1over3}-0)={1over6}$$ and $$Var[X]=int_{-1}^0x^2(1/3)dx + int_0^1x^2(2/3)dx - mu^2=(0+{1over9})+({2over9}-0)-({1over6})^2={11over36}$$ Is this correct?
      $endgroup$
      – Severin Linder
      Sep 24 '17 at 17:30
















    0












    $begingroup$

    Here is a start:



    $$mu = E(X) = int_{-1}^1 xf(x),dx = int_{-1}^0 x(1/3),dx
    + int_0^1 x(2/3),dx = dots,.$$
    Find $E(X^2)$ similarly, and use $Var(X) = E(X^2) - mu^2.$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you already! So, then $$mu=E[X]=int_{-1}^0x(1/3)dx + int_0^1x(2/3)dx = (0-{1over6})+({1over3}-0)={1over6}$$ and $$Var[X]=int_{-1}^0x^2(1/3)dx + int_0^1x^2(2/3)dx - mu^2=(0+{1over9})+({2over9}-0)-({1over6})^2={11over36}$$ Is this correct?
      $endgroup$
      – Severin Linder
      Sep 24 '17 at 17:30














    0












    0








    0





    $begingroup$

    Here is a start:



    $$mu = E(X) = int_{-1}^1 xf(x),dx = int_{-1}^0 x(1/3),dx
    + int_0^1 x(2/3),dx = dots,.$$
    Find $E(X^2)$ similarly, and use $Var(X) = E(X^2) - mu^2.$






    share|cite|improve this answer









    $endgroup$



    Here is a start:



    $$mu = E(X) = int_{-1}^1 xf(x),dx = int_{-1}^0 x(1/3),dx
    + int_0^1 x(2/3),dx = dots,.$$
    Find $E(X^2)$ similarly, and use $Var(X) = E(X^2) - mu^2.$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Sep 24 '17 at 6:12









    BruceETBruceET

    35.7k71440




    35.7k71440












    • $begingroup$
      Thank you already! So, then $$mu=E[X]=int_{-1}^0x(1/3)dx + int_0^1x(2/3)dx = (0-{1over6})+({1over3}-0)={1over6}$$ and $$Var[X]=int_{-1}^0x^2(1/3)dx + int_0^1x^2(2/3)dx - mu^2=(0+{1over9})+({2over9}-0)-({1over6})^2={11over36}$$ Is this correct?
      $endgroup$
      – Severin Linder
      Sep 24 '17 at 17:30


















    • $begingroup$
      Thank you already! So, then $$mu=E[X]=int_{-1}^0x(1/3)dx + int_0^1x(2/3)dx = (0-{1over6})+({1over3}-0)={1over6}$$ and $$Var[X]=int_{-1}^0x^2(1/3)dx + int_0^1x^2(2/3)dx - mu^2=(0+{1over9})+({2over9}-0)-({1over6})^2={11over36}$$ Is this correct?
      $endgroup$
      – Severin Linder
      Sep 24 '17 at 17:30
















    $begingroup$
    Thank you already! So, then $$mu=E[X]=int_{-1}^0x(1/3)dx + int_0^1x(2/3)dx = (0-{1over6})+({1over3}-0)={1over6}$$ and $$Var[X]=int_{-1}^0x^2(1/3)dx + int_0^1x^2(2/3)dx - mu^2=(0+{1over9})+({2over9}-0)-({1over6})^2={11over36}$$ Is this correct?
    $endgroup$
    – Severin Linder
    Sep 24 '17 at 17:30




    $begingroup$
    Thank you already! So, then $$mu=E[X]=int_{-1}^0x(1/3)dx + int_0^1x(2/3)dx = (0-{1over6})+({1over3}-0)={1over6}$$ and $$Var[X]=int_{-1}^0x^2(1/3)dx + int_0^1x^2(2/3)dx - mu^2=(0+{1over9})+({2over9}-0)-({1over6})^2={11over36}$$ Is this correct?
    $endgroup$
    – Severin Linder
    Sep 24 '17 at 17:30











    0












    $begingroup$

    Thank you already!



    So, then



    $$mu=E[X]=int_{-1}^0x(1/3)dx + int_0^1x(2/3)dx = (0-{1over6})+({1over3}-0)={1over6}$$



    and



    $$Var[X]=int_{-1}^0x^2(1/3)dx + int_0^1x^2(2/3)dx - mu^2=(0+{1over9})+({2over9}-0)-({1over6})^2={11over36}$$



    Is this correct?






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Thank you already!



      So, then



      $$mu=E[X]=int_{-1}^0x(1/3)dx + int_0^1x(2/3)dx = (0-{1over6})+({1over3}-0)={1over6}$$



      and



      $$Var[X]=int_{-1}^0x^2(1/3)dx + int_0^1x^2(2/3)dx - mu^2=(0+{1over9})+({2over9}-0)-({1over6})^2={11over36}$$



      Is this correct?






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Thank you already!



        So, then



        $$mu=E[X]=int_{-1}^0x(1/3)dx + int_0^1x(2/3)dx = (0-{1over6})+({1over3}-0)={1over6}$$



        and



        $$Var[X]=int_{-1}^0x^2(1/3)dx + int_0^1x^2(2/3)dx - mu^2=(0+{1over9})+({2over9}-0)-({1over6})^2={11over36}$$



        Is this correct?






        share|cite|improve this answer









        $endgroup$



        Thank you already!



        So, then



        $$mu=E[X]=int_{-1}^0x(1/3)dx + int_0^1x(2/3)dx = (0-{1over6})+({1over3}-0)={1over6}$$



        and



        $$Var[X]=int_{-1}^0x^2(1/3)dx + int_0^1x^2(2/3)dx - mu^2=(0+{1over9})+({2over9}-0)-({1over6})^2={11over36}$$



        Is this correct?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 24 '17 at 17:29









        Severin LinderSeverin Linder

        6




        6






























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