Expected value of half normal using gamma function












0














Let $X$ be $N(0,1)$. I want to compute $E(|X|)$. This is what I tried:



$$E(|X|)=frac{1}{sqrt{2pi}}int_{-infty}^infty|x|e^{frac{-x^2}{2}}dx=frac{2}{sqrt{2pi}}int_{0}^infty xe^{frac{-x^2}{2}}dx$$



In order to aim for the gamma function, I do the change of variables $x^2/2=y.$ This results in



$$frac{2}{sqrt{2pi}}int_{0}^infty xe^{frac{-x^2}{2}}dx=frac{2}{sqrt{pi}}int_{0}^infty y^{1/2}e^{-y}dy=frac{2}{sqrt{pi}}Gamma(1/2+1)=frac{2}{sqrt{pi}}frac{sqrt pi}{2}=1$$



This is obviously wrong, but I cannot find my mistake.






probability-distributions moment-problem







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    0














    Let $X$ be $N(0,1)$. I want to compute $E(|X|)$. This is what I tried:



    $$E(|X|)=frac{1}{sqrt{2pi}}int_{-infty}^infty|x|e^{frac{-x^2}{2}}dx=frac{2}{sqrt{2pi}}int_{0}^infty xe^{frac{-x^2}{2}}dx$$



    In order to aim for the gamma function, I do the change of variables $x^2/2=y.$ This results in



    $$frac{2}{sqrt{2pi}}int_{0}^infty xe^{frac{-x^2}{2}}dx=frac{2}{sqrt{pi}}int_{0}^infty y^{1/2}e^{-y}dy=frac{2}{sqrt{pi}}Gamma(1/2+1)=frac{2}{sqrt{pi}}frac{sqrt pi}{2}=1$$



    This is obviously wrong, but I cannot find my mistake.






    probability-distributions moment-problem







    share|cite|improve this question



























      0












      0








      0







      Let $X$ be $N(0,1)$. I want to compute $E(|X|)$. This is what I tried:



      $$E(|X|)=frac{1}{sqrt{2pi}}int_{-infty}^infty|x|e^{frac{-x^2}{2}}dx=frac{2}{sqrt{2pi}}int_{0}^infty xe^{frac{-x^2}{2}}dx$$



      In order to aim for the gamma function, I do the change of variables $x^2/2=y.$ This results in



      $$frac{2}{sqrt{2pi}}int_{0}^infty xe^{frac{-x^2}{2}}dx=frac{2}{sqrt{pi}}int_{0}^infty y^{1/2}e^{-y}dy=frac{2}{sqrt{pi}}Gamma(1/2+1)=frac{2}{sqrt{pi}}frac{sqrt pi}{2}=1$$



      This is obviously wrong, but I cannot find my mistake.






      probability-distributions moment-problem







      share|cite|improve this question















      Let $X$ be $N(0,1)$. I want to compute $E(|X|)$. This is what I tried:



      $$E(|X|)=frac{1}{sqrt{2pi}}int_{-infty}^infty|x|e^{frac{-x^2}{2}}dx=frac{2}{sqrt{2pi}}int_{0}^infty xe^{frac{-x^2}{2}}dx$$



      In order to aim for the gamma function, I do the change of variables $x^2/2=y.$ This results in



      $$frac{2}{sqrt{2pi}}int_{0}^infty xe^{frac{-x^2}{2}}dx=frac{2}{sqrt{pi}}int_{0}^infty y^{1/2}e^{-y}dy=frac{2}{sqrt{pi}}Gamma(1/2+1)=frac{2}{sqrt{pi}}frac{sqrt pi}{2}=1$$



      This is obviously wrong, but I cannot find my mistake.






      probability-distributions moment-problem




      probability-distributions moment-problem






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 26 at 17:28









      mrtaurho

      3,3982932




      3,3982932










      asked Nov 26 at 17:25









      julian.marr

      349112




      349112






















          2 Answers
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          Its just a slight mistake within your substitution. In fact by replacing $x^2/2$ with $y$ you have to consider that your differential changes to $xdx=dy$ and not only to $dy$. Therefore the single $x$ is absorbed within the $dy$ instead of remaining as $y^{1/2}$. This finally leads to



          $$int_0^{infty}e^{-frac{x^2}2}(xdx)=int_0^{infty}e^{-y}dy$$



          where the last integral turns out to equal $1$ instead of $Gamma(1/2+1)$. Therefore your evaluation has to be finished as



          $$E(|X|)=frac{2}{sqrt{2pi}}int_{0}^infty xe^{frac{-x^2}{2}}dx=frac{2}{sqrt{2pi}}int_{0}^infty e^{-y}dy=sqrt{frac{2}{pi}}$$






          share|cite|improve this answer























          • I see. Now it all makes sense.
            – julian.marr
            Nov 26 at 17:34






          • 1




            @julian.marr In this case you can mark my answer as accepted to show that you are satisfied with the given :)
            – mrtaurho
            Nov 26 at 17:35



















          0














          Change of variable is wrong, you should have $int (ldots) frac{dx}{dy} , dy$ and you have omitted the (jacobian) $frac{dx}{dy}$.






          share|cite|improve this answer





















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            2 Answers
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            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            Its just a slight mistake within your substitution. In fact by replacing $x^2/2$ with $y$ you have to consider that your differential changes to $xdx=dy$ and not only to $dy$. Therefore the single $x$ is absorbed within the $dy$ instead of remaining as $y^{1/2}$. This finally leads to



            $$int_0^{infty}e^{-frac{x^2}2}(xdx)=int_0^{infty}e^{-y}dy$$



            where the last integral turns out to equal $1$ instead of $Gamma(1/2+1)$. Therefore your evaluation has to be finished as



            $$E(|X|)=frac{2}{sqrt{2pi}}int_{0}^infty xe^{frac{-x^2}{2}}dx=frac{2}{sqrt{2pi}}int_{0}^infty e^{-y}dy=sqrt{frac{2}{pi}}$$






            share|cite|improve this answer























            • I see. Now it all makes sense.
              – julian.marr
              Nov 26 at 17:34






            • 1




              @julian.marr In this case you can mark my answer as accepted to show that you are satisfied with the given :)
              – mrtaurho
              Nov 26 at 17:35
















            1














            Its just a slight mistake within your substitution. In fact by replacing $x^2/2$ with $y$ you have to consider that your differential changes to $xdx=dy$ and not only to $dy$. Therefore the single $x$ is absorbed within the $dy$ instead of remaining as $y^{1/2}$. This finally leads to



            $$int_0^{infty}e^{-frac{x^2}2}(xdx)=int_0^{infty}e^{-y}dy$$



            where the last integral turns out to equal $1$ instead of $Gamma(1/2+1)$. Therefore your evaluation has to be finished as



            $$E(|X|)=frac{2}{sqrt{2pi}}int_{0}^infty xe^{frac{-x^2}{2}}dx=frac{2}{sqrt{2pi}}int_{0}^infty e^{-y}dy=sqrt{frac{2}{pi}}$$






            share|cite|improve this answer























            • I see. Now it all makes sense.
              – julian.marr
              Nov 26 at 17:34






            • 1




              @julian.marr In this case you can mark my answer as accepted to show that you are satisfied with the given :)
              – mrtaurho
              Nov 26 at 17:35














            1












            1








            1






            Its just a slight mistake within your substitution. In fact by replacing $x^2/2$ with $y$ you have to consider that your differential changes to $xdx=dy$ and not only to $dy$. Therefore the single $x$ is absorbed within the $dy$ instead of remaining as $y^{1/2}$. This finally leads to



            $$int_0^{infty}e^{-frac{x^2}2}(xdx)=int_0^{infty}e^{-y}dy$$



            where the last integral turns out to equal $1$ instead of $Gamma(1/2+1)$. Therefore your evaluation has to be finished as



            $$E(|X|)=frac{2}{sqrt{2pi}}int_{0}^infty xe^{frac{-x^2}{2}}dx=frac{2}{sqrt{2pi}}int_{0}^infty e^{-y}dy=sqrt{frac{2}{pi}}$$






            share|cite|improve this answer














            Its just a slight mistake within your substitution. In fact by replacing $x^2/2$ with $y$ you have to consider that your differential changes to $xdx=dy$ and not only to $dy$. Therefore the single $x$ is absorbed within the $dy$ instead of remaining as $y^{1/2}$. This finally leads to



            $$int_0^{infty}e^{-frac{x^2}2}(xdx)=int_0^{infty}e^{-y}dy$$



            where the last integral turns out to equal $1$ instead of $Gamma(1/2+1)$. Therefore your evaluation has to be finished as



            $$E(|X|)=frac{2}{sqrt{2pi}}int_{0}^infty xe^{frac{-x^2}{2}}dx=frac{2}{sqrt{2pi}}int_{0}^infty e^{-y}dy=sqrt{frac{2}{pi}}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 26 at 17:34

























            answered Nov 26 at 17:32









            mrtaurho

            3,3982932




            3,3982932












            • I see. Now it all makes sense.
              – julian.marr
              Nov 26 at 17:34






            • 1




              @julian.marr In this case you can mark my answer as accepted to show that you are satisfied with the given :)
              – mrtaurho
              Nov 26 at 17:35


















            • I see. Now it all makes sense.
              – julian.marr
              Nov 26 at 17:34






            • 1




              @julian.marr In this case you can mark my answer as accepted to show that you are satisfied with the given :)
              – mrtaurho
              Nov 26 at 17:35
















            I see. Now it all makes sense.
            – julian.marr
            Nov 26 at 17:34




            I see. Now it all makes sense.
            – julian.marr
            Nov 26 at 17:34




            1




            1




            @julian.marr In this case you can mark my answer as accepted to show that you are satisfied with the given :)
            – mrtaurho
            Nov 26 at 17:35




            @julian.marr In this case you can mark my answer as accepted to show that you are satisfied with the given :)
            – mrtaurho
            Nov 26 at 17:35











            0














            Change of variable is wrong, you should have $int (ldots) frac{dx}{dy} , dy$ and you have omitted the (jacobian) $frac{dx}{dy}$.






            share|cite|improve this answer


























              0














              Change of variable is wrong, you should have $int (ldots) frac{dx}{dy} , dy$ and you have omitted the (jacobian) $frac{dx}{dy}$.






              share|cite|improve this answer
























                0












                0








                0






                Change of variable is wrong, you should have $int (ldots) frac{dx}{dy} , dy$ and you have omitted the (jacobian) $frac{dx}{dy}$.






                share|cite|improve this answer












                Change of variable is wrong, you should have $int (ldots) frac{dx}{dy} , dy$ and you have omitted the (jacobian) $frac{dx}{dy}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 26 at 17:32









                Richard Martin

                1,63618




                1,63618






























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