Maximum of standard normals












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$begingroup$


Let $X_1,...,X_n$ be independent standard normals. Let $Z=max_iX_i$. Then



$F_Z(z)=F_X(z)^n$ and $f_Z(z)=nF_X(z)^{n-1}f_X(z)$



Now let's take the expectation of $Z$.



$E(Z)=int_{-infty}^{infty}znF_X(z)^{n-1}f_X(z)dz$



Since $F_X(z)^{n-1}leq1$ for every $z$, we can bound this as



$E(Z)=int_{-infty}^{infty}znF_X(z)^{n-1}f_X(z)dzleq int_{-infty}^{infty}znf_X(z)dz=0.$



So I get that $E(Z)leq 0$ which makes no sense, but I cannot figure out where is my mistake.






probability-distributions normal-distribution







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    1












    $begingroup$


    Let $X_1,...,X_n$ be independent standard normals. Let $Z=max_iX_i$. Then



    $F_Z(z)=F_X(z)^n$ and $f_Z(z)=nF_X(z)^{n-1}f_X(z)$



    Now let's take the expectation of $Z$.



    $E(Z)=int_{-infty}^{infty}znF_X(z)^{n-1}f_X(z)dz$



    Since $F_X(z)^{n-1}leq1$ for every $z$, we can bound this as



    $E(Z)=int_{-infty}^{infty}znF_X(z)^{n-1}f_X(z)dzleq int_{-infty}^{infty}znf_X(z)dz=0.$



    So I get that $E(Z)leq 0$ which makes no sense, but I cannot figure out where is my mistake.






    probability-distributions normal-distribution







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $X_1,...,X_n$ be independent standard normals. Let $Z=max_iX_i$. Then



      $F_Z(z)=F_X(z)^n$ and $f_Z(z)=nF_X(z)^{n-1}f_X(z)$



      Now let's take the expectation of $Z$.



      $E(Z)=int_{-infty}^{infty}znF_X(z)^{n-1}f_X(z)dz$



      Since $F_X(z)^{n-1}leq1$ for every $z$, we can bound this as



      $E(Z)=int_{-infty}^{infty}znF_X(z)^{n-1}f_X(z)dzleq int_{-infty}^{infty}znf_X(z)dz=0.$



      So I get that $E(Z)leq 0$ which makes no sense, but I cannot figure out where is my mistake.






      probability-distributions normal-distribution







      share|cite|improve this question









      $endgroup$




      Let $X_1,...,X_n$ be independent standard normals. Let $Z=max_iX_i$. Then



      $F_Z(z)=F_X(z)^n$ and $f_Z(z)=nF_X(z)^{n-1}f_X(z)$



      Now let's take the expectation of $Z$.



      $E(Z)=int_{-infty}^{infty}znF_X(z)^{n-1}f_X(z)dz$



      Since $F_X(z)^{n-1}leq1$ for every $z$, we can bound this as



      $E(Z)=int_{-infty}^{infty}znF_X(z)^{n-1}f_X(z)dzleq int_{-infty}^{infty}znf_X(z)dz=0.$



      So I get that $E(Z)leq 0$ which makes no sense, but I cannot figure out where is my mistake.






      probability-distributions normal-distribution




      probability-distributions normal-distribution






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      asked Dec 6 '18 at 20:18









      julian.marrjulian.marr

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          You cannot bound the expectation by substituting 1 for $F_X(z)^{n-1}$, because not all the values in the integral are positive hence, it isn't guaranteed to be bigger than expectation.



          Intuitively, by substituting 1, you are effectively giving more weight to negative values than they have in expectation, which should result in a lower than expected value.






          share|cite|improve this answer









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            $begingroup$

            You cannot bound the expectation by substituting 1 for $F_X(z)^{n-1}$, because not all the values in the integral are positive hence, it isn't guaranteed to be bigger than expectation.



            Intuitively, by substituting 1, you are effectively giving more weight to negative values than they have in expectation, which should result in a lower than expected value.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              You cannot bound the expectation by substituting 1 for $F_X(z)^{n-1}$, because not all the values in the integral are positive hence, it isn't guaranteed to be bigger than expectation.



              Intuitively, by substituting 1, you are effectively giving more weight to negative values than they have in expectation, which should result in a lower than expected value.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                You cannot bound the expectation by substituting 1 for $F_X(z)^{n-1}$, because not all the values in the integral are positive hence, it isn't guaranteed to be bigger than expectation.



                Intuitively, by substituting 1, you are effectively giving more weight to negative values than they have in expectation, which should result in a lower than expected value.






                share|cite|improve this answer









                $endgroup$



                You cannot bound the expectation by substituting 1 for $F_X(z)^{n-1}$, because not all the values in the integral are positive hence, it isn't guaranteed to be bigger than expectation.



                Intuitively, by substituting 1, you are effectively giving more weight to negative values than they have in expectation, which should result in a lower than expected value.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 6 '18 at 21:01









                OfyaOfya

                5048




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