Question arising from the proof of the existence of $mathbb E[X]$ when $X$ ~ $mathcal{N}(mu, sigma ^2)$












0












$begingroup$


$mathbb E[X]$ when $X$ ~ $mathcal{N}(mu, sigma ^2)$



I know that $mathbb E[X]$ exists iff $mathbb E[|X|]<infty$



$mathbb E[|X|]=int_{mathbb R}|x|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx$



My Professor then goes into the next step saying:
$int_{mathbb R}|x|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dxleqint_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx+mu$



and this is where I get confused.



I recognize that he is trying to get to a substitution $y=x-mu$, and $mathbb E[|X|]=mathbb E[|X-mu+mu|]leqmathbb E[|X-mu|]+mu$



But why is $mathbb E[|X-mu|]=int_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx$ and not equal $int_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-2mu)}{2sigma^2}}dx$. I mean surely the density function $frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-2mu)}{2sigma^2}}$ shifts to the left if $mu$ is subtracted.



Additional Question out of interest:
If I can prove $mathbb E[X] in mathbb R$, can I then automatically assume $mathbb E[X]$ exists? In other words, why is $mathbb E[|X|]<infty$ neccessary rather than $mathbb E[X]$? I realize that it has something to do with the Lebesgue Integral that only looks at positive functions, but I am not sure.






probability normal-distribution expected-value







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    $mathbb E[X]$ when $X$ ~ $mathcal{N}(mu, sigma ^2)$



    I know that $mathbb E[X]$ exists iff $mathbb E[|X|]<infty$



    $mathbb E[|X|]=int_{mathbb R}|x|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx$



    My Professor then goes into the next step saying:
    $int_{mathbb R}|x|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dxleqint_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx+mu$



    and this is where I get confused.



    I recognize that he is trying to get to a substitution $y=x-mu$, and $mathbb E[|X|]=mathbb E[|X-mu+mu|]leqmathbb E[|X-mu|]+mu$



    But why is $mathbb E[|X-mu|]=int_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx$ and not equal $int_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-2mu)}{2sigma^2}}dx$. I mean surely the density function $frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-2mu)}{2sigma^2}}$ shifts to the left if $mu$ is subtracted.



    Additional Question out of interest:
    If I can prove $mathbb E[X] in mathbb R$, can I then automatically assume $mathbb E[X]$ exists? In other words, why is $mathbb E[|X|]<infty$ neccessary rather than $mathbb E[X]$? I realize that it has something to do with the Lebesgue Integral that only looks at positive functions, but I am not sure.






    probability normal-distribution expected-value







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      $mathbb E[X]$ when $X$ ~ $mathcal{N}(mu, sigma ^2)$



      I know that $mathbb E[X]$ exists iff $mathbb E[|X|]<infty$



      $mathbb E[|X|]=int_{mathbb R}|x|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx$



      My Professor then goes into the next step saying:
      $int_{mathbb R}|x|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dxleqint_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx+mu$



      and this is where I get confused.



      I recognize that he is trying to get to a substitution $y=x-mu$, and $mathbb E[|X|]=mathbb E[|X-mu+mu|]leqmathbb E[|X-mu|]+mu$



      But why is $mathbb E[|X-mu|]=int_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx$ and not equal $int_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-2mu)}{2sigma^2}}dx$. I mean surely the density function $frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-2mu)}{2sigma^2}}$ shifts to the left if $mu$ is subtracted.



      Additional Question out of interest:
      If I can prove $mathbb E[X] in mathbb R$, can I then automatically assume $mathbb E[X]$ exists? In other words, why is $mathbb E[|X|]<infty$ neccessary rather than $mathbb E[X]$? I realize that it has something to do with the Lebesgue Integral that only looks at positive functions, but I am not sure.






      probability normal-distribution expected-value







      share|cite|improve this question









      $endgroup$




      $mathbb E[X]$ when $X$ ~ $mathcal{N}(mu, sigma ^2)$



      I know that $mathbb E[X]$ exists iff $mathbb E[|X|]<infty$



      $mathbb E[|X|]=int_{mathbb R}|x|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx$



      My Professor then goes into the next step saying:
      $int_{mathbb R}|x|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dxleqint_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx+mu$



      and this is where I get confused.



      I recognize that he is trying to get to a substitution $y=x-mu$, and $mathbb E[|X|]=mathbb E[|X-mu+mu|]leqmathbb E[|X-mu|]+mu$



      But why is $mathbb E[|X-mu|]=int_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx$ and not equal $int_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-2mu)}{2sigma^2}}dx$. I mean surely the density function $frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-2mu)}{2sigma^2}}$ shifts to the left if $mu$ is subtracted.



      Additional Question out of interest:
      If I can prove $mathbb E[X] in mathbb R$, can I then automatically assume $mathbb E[X]$ exists? In other words, why is $mathbb E[|X|]<infty$ neccessary rather than $mathbb E[X]$? I realize that it has something to do with the Lebesgue Integral that only looks at positive functions, but I am not sure.






      probability normal-distribution expected-value




      probability normal-distribution expected-value






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 14 '18 at 16:46









      SABOYSABOY

      699311




      699311






















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          $begingroup$

          What your professor did is simply applying the triangle inequality (assuming $mugeq 0$)
          $$
          |x|leq|x-mu|+mu,
          $$
          which implies that
          begin{align}
          int_{mathbb R}|x|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx
          &leq
          int_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx
          +int_{mathbb R}mufrac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx\
          &=
          int_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx
          +mu.
          end{align}



          For the additional question: lots of authors define $EX$ only for absolutely integrable random variables (i.e., $E|X|<infty$) and leave $EX$ undefined when $E|X|=infty$. Whenever $EX$ is defined, it must be a real number (in the context of real random variables).






          share|cite|improve this answer











          $endgroup$













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            2












            $begingroup$

            What your professor did is simply applying the triangle inequality (assuming $mugeq 0$)
            $$
            |x|leq|x-mu|+mu,
            $$
            which implies that
            begin{align}
            int_{mathbb R}|x|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx
            &leq
            int_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx
            +int_{mathbb R}mufrac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx\
            &=
            int_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx
            +mu.
            end{align}



            For the additional question: lots of authors define $EX$ only for absolutely integrable random variables (i.e., $E|X|<infty$) and leave $EX$ undefined when $E|X|=infty$. Whenever $EX$ is defined, it must be a real number (in the context of real random variables).






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              What your professor did is simply applying the triangle inequality (assuming $mugeq 0$)
              $$
              |x|leq|x-mu|+mu,
              $$
              which implies that
              begin{align}
              int_{mathbb R}|x|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx
              &leq
              int_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx
              +int_{mathbb R}mufrac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx\
              &=
              int_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx
              +mu.
              end{align}



              For the additional question: lots of authors define $EX$ only for absolutely integrable random variables (i.e., $E|X|<infty$) and leave $EX$ undefined when $E|X|=infty$. Whenever $EX$ is defined, it must be a real number (in the context of real random variables).






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                What your professor did is simply applying the triangle inequality (assuming $mugeq 0$)
                $$
                |x|leq|x-mu|+mu,
                $$
                which implies that
                begin{align}
                int_{mathbb R}|x|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx
                &leq
                int_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx
                +int_{mathbb R}mufrac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx\
                &=
                int_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx
                +mu.
                end{align}



                For the additional question: lots of authors define $EX$ only for absolutely integrable random variables (i.e., $E|X|<infty$) and leave $EX$ undefined when $E|X|=infty$. Whenever $EX$ is defined, it must be a real number (in the context of real random variables).






                share|cite|improve this answer











                $endgroup$



                What your professor did is simply applying the triangle inequality (assuming $mugeq 0$)
                $$
                |x|leq|x-mu|+mu,
                $$
                which implies that
                begin{align}
                int_{mathbb R}|x|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx
                &leq
                int_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx
                +int_{mathbb R}mufrac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx\
                &=
                int_{mathbb R}|x-mu|frac{1}{sqrt{2pi sigma^2}}e^{-frac{(x-mu)}{2sigma^2}}dx
                +mu.
                end{align}



                For the additional question: lots of authors define $EX$ only for absolutely integrable random variables (i.e., $E|X|<infty$) and leave $EX$ undefined when $E|X|=infty$. Whenever $EX$ is defined, it must be a real number (in the context of real random variables).







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 14 '18 at 17:14

























                answered Dec 14 '18 at 17:05









                user587192user587192

                2,064415




                2,064415






























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