Why can we treat MGF in this way












1












$begingroup$


For the standard proof that if $Z sim N(0,1)$ than $Z^{2} sim chi^{2}_{1}$



We write



$$M_{Z^{2}}(t)=int_{mathbb{R}}exp(tz^{2})frac{1}{sqrt{2pi}}exp(frac{-z^{2}}{2}) dz$$



That is , we use the standard normal density to compute the moment generating function.



Why do we not need to use the density of $Z^{2}$ and instead are able to use the density of just $Z$, I assume because it is a continuous one to one function?






mgf







share|cite|edit









$endgroup$

















    1












    $begingroup$


    For the standard proof that if $Z sim N(0,1)$ than $Z^{2} sim chi^{2}_{1}$



    We write



    $$M_{Z^{2}}(t)=int_{mathbb{R}}exp(tz^{2})frac{1}{sqrt{2pi}}exp(frac{-z^{2}}{2}) dz$$



    That is , we use the standard normal density to compute the moment generating function.



    Why do we not need to use the density of $Z^{2}$ and instead are able to use the density of just $Z$, I assume because it is a continuous one to one function?






    mgf







    share|cite|edit









    $endgroup$















      1












      1








      1





      $begingroup$


      For the standard proof that if $Z sim N(0,1)$ than $Z^{2} sim chi^{2}_{1}$



      We write



      $$M_{Z^{2}}(t)=int_{mathbb{R}}exp(tz^{2})frac{1}{sqrt{2pi}}exp(frac{-z^{2}}{2}) dz$$



      That is , we use the standard normal density to compute the moment generating function.



      Why do we not need to use the density of $Z^{2}$ and instead are able to use the density of just $Z$, I assume because it is a continuous one to one function?






      mgf







      share|cite|edit









      $endgroup$




      For the standard proof that if $Z sim N(0,1)$ than $Z^{2} sim chi^{2}_{1}$



      We write



      $$M_{Z^{2}}(t)=int_{mathbb{R}}exp(tz^{2})frac{1}{sqrt{2pi}}exp(frac{-z^{2}}{2}) dz$$



      That is , we use the standard normal density to compute the moment generating function.



      Why do we not need to use the density of $Z^{2}$ and instead are able to use the density of just $Z$, I assume because it is a continuous one to one function?






      mgf




      mgf






      share|cite|edit













      share|cite|edit











      share|cite|edit




      share|cite|edit










      asked 5 hours ago









      QualityQuality

      24919




      24919






















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          $begingroup$

          In principle, you don't know the density of $Z$, so you can't use it.



          What's being done when you use the density of $X$ is an application of the law of the unconscious statistician.






          share|cite|improve this answer









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            1 Answer
            1






            active

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            active

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            active

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            2












            $begingroup$

            In principle, you don't know the density of $Z$, so you can't use it.



            What's being done when you use the density of $X$ is an application of the law of the unconscious statistician.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              In principle, you don't know the density of $Z$, so you can't use it.



              What's being done when you use the density of $X$ is an application of the law of the unconscious statistician.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                In principle, you don't know the density of $Z$, so you can't use it.



                What's being done when you use the density of $X$ is an application of the law of the unconscious statistician.






                share|cite|improve this answer









                $endgroup$



                In principle, you don't know the density of $Z$, so you can't use it.



                What's being done when you use the density of $X$ is an application of the law of the unconscious statistician.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 4 hours ago









                Lucas FariasLucas Farias

                469417




                469417






























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