How to prove this limit is $1/4$












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$$underset{nto infty }{mathop{lim }},int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{left( frac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n} right)}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}$$



I start thinking first in the Lebesgue monotone convergent theorem
but this leads to closed road
is there any shortcut to solve this problem ??






real-analysis integration lebesgue-integral







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    0












    $begingroup$


    $$underset{nto infty }{mathop{lim }},int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{left( frac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n} right)}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}$$



    I start thinking first in the Lebesgue monotone convergent theorem
    but this leads to closed road
    is there any shortcut to solve this problem ??






    real-analysis integration lebesgue-integral







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$


      $$underset{nto infty }{mathop{lim }},int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{left( frac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n} right)}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}$$



      I start thinking first in the Lebesgue monotone convergent theorem
      but this leads to closed road
      is there any shortcut to solve this problem ??






      real-analysis integration lebesgue-integral







      share|cite|improve this question









      $endgroup$




      $$underset{nto infty }{mathop{lim }},int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{left( frac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n} right)}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}$$



      I start thinking first in the Lebesgue monotone convergent theorem
      but this leads to closed road
      is there any shortcut to solve this problem ??






      real-analysis integration lebesgue-integral




      real-analysis integration lebesgue-integral






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 13 at 19:51









      Ramez HindiRamez Hindi

      1437




      1437






















          3 Answers
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          $begingroup$

          The integral has an equivalent expression
          $$
          Eleft[left(frac{1}{n}sum_{i=1}^n U_iright)^2right]
          $$           where $U_i$, $ile n$ are independently and uniformly distributed random variables on $[0,1]$. This gives
          $$begin{eqnarray}
          Eleft[left(frac{1}{n}sum_{i=1}^n U_iright)^2right]&=&frac{1}{n^2}sum_{i,j=1}^nEleft[U_iU_jright]\
          &=&frac{1}{n^2}sum_{i=1}^nEleft[U_i^2right] +frac{1}{n^2}sum_{ine j}Eleft[U_iU_jright]\
          &=&frac{n}{n^2}frac{1}{3}+frac{n(n-1)}{n^2}frac{1}{4}to frac{1}{4}
          end{eqnarray}$$
          since $E[U_i]=int_0^1 xdx=frac{1}{2}$, $E[U_i^2]=int_0^1 x^2dx =frac{1}{3}$ and $E[U_iU_j]=E[U_i]E[U_j]=frac{1}{2}cdotfrac{1}{2}=frac{1}{4}$ for all $ine j$.






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            3












            $begingroup$

            We have $$(x_1+x_2+cdots +x_n)^2=sum_i x_i^2+sum_{i,jne i} x_ix_j$$first of all note that $$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{x_i}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}=int_0^1 x_i^2dx_i={1over 3}$$and $$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{x_ix_j}{}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}={1over 4}$$therefore$$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{left( {{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{} right)}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}=ncdot {1over 3}+(n^2-n){1over 4}$$and by substitution we obtain $$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{left( frac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n} right)}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}={1over 3n}+{n-1over 4n}$$which obviously shows that the limit is $1over 4$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Sorry first line is not clear to if we say $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$ by induction you will get the following result , ${{left( sumnolimits_{i=1}^{n}{{{x}_{i}}} right)}^{2}}=sumnolimits_{i=1}^{n}{x_{i}^{2}+2sumnolimits_{ine j}^{n}{{{x}_{i}}{{x}_{j}}}}$
              $endgroup$
              – Ramez Hindi
              Jan 13 at 21:34












            • $begingroup$
              In fact you get $(sum_i x_i)^2=x_1^2+cdots + x_n^2+2sum_{i<j}x_ix_j$. The multiplicity has been considered in my equation since $2x_ix_j=x_ix_j+x_jx_i$ whenever $ine j$
              $endgroup$
              – Mostafa Ayaz
              Jan 13 at 21:38



















            2












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            Suppose $X_1,X_2,ldots$ are independent random variables having the uniform distribution on $[0,1]$. Then the common expectation of these variables exist and equals $mu=1/2$.



            Define $$overline X_n=frac{1}{n}sum_{k=1}^n X_k$$



            By Khintchine's weak law of large numbers, $$overline X_nstackrel{P}{longrightarrow}muquadtext{ as }quad ntoinfty$$



            And by the continuous mapping theorem, $$overline X_n^2stackrel{P}{longrightarrow}mu^2quadtext{ as }quad ntoinftytag{1}$$



            Moreover, $$0le X_1,ldots,X_nle 1implies 0le overline X_nle 1implies 0le overline X_n^2le 1tag{2}$$



            $(1)$ and $(2)$ together imply $$int_{[0,1]^n}left(frac{x_1+cdots+x_n}{n}right)^2mathrm{d}x_1ldotsmathrm{d}x_n = Eleft(overline X_n^2right)stackrel{ntoinfty}{longrightarrow}frac{1}{4}$$






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            • $begingroup$
              It’s always nice to see an easygoing probability solution to what appears to be a monstrous analytic problem. +1
              $endgroup$
              – LoveTooNap29
              Jan 13 at 20:08










            • $begingroup$
              there must be some confusion. I upvoted both yours and Song’s answer for the use of probability.
              $endgroup$
              – LoveTooNap29
              Jan 13 at 20:33










            • $begingroup$
              @LoveTooNap29 I wasn't replying to you.
              $endgroup$
              – StubbornAtom
              Jan 13 at 20:34










            • $begingroup$
              $X_nto X$ in probability isn't enough to conclude that $mathbb{E}[X_n]tomathbb{E}[X]$. It would be better to use the strong law of large numbers, plus dominated convergence.
              $endgroup$
              – carmichael561
              Jan 15 at 0:27










            • $begingroup$
              @carmichael561 Thank you. What if I add the fact that $|X_n|<1$ to my answer? I think that salvages the argument.
              $endgroup$
              – StubbornAtom
              Jan 15 at 4:59











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            3 Answers
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            3 Answers
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            4












            $begingroup$

            The integral has an equivalent expression
            $$
            Eleft[left(frac{1}{n}sum_{i=1}^n U_iright)^2right]
            $$             where $U_i$, $ile n$ are independently and uniformly distributed random variables on $[0,1]$. This gives
            $$begin{eqnarray}
            Eleft[left(frac{1}{n}sum_{i=1}^n U_iright)^2right]&=&frac{1}{n^2}sum_{i,j=1}^nEleft[U_iU_jright]\
            &=&frac{1}{n^2}sum_{i=1}^nEleft[U_i^2right] +frac{1}{n^2}sum_{ine j}Eleft[U_iU_jright]\
            &=&frac{n}{n^2}frac{1}{3}+frac{n(n-1)}{n^2}frac{1}{4}to frac{1}{4}
            end{eqnarray}$$
            since $E[U_i]=int_0^1 xdx=frac{1}{2}$, $E[U_i^2]=int_0^1 x^2dx =frac{1}{3}$ and $E[U_iU_j]=E[U_i]E[U_j]=frac{1}{2}cdotfrac{1}{2}=frac{1}{4}$ for all $ine j$.






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              The integral has an equivalent expression
              $$
              Eleft[left(frac{1}{n}sum_{i=1}^n U_iright)^2right]
              $$               where $U_i$, $ile n$ are independently and uniformly distributed random variables on $[0,1]$. This gives
              $$begin{eqnarray}
              Eleft[left(frac{1}{n}sum_{i=1}^n U_iright)^2right]&=&frac{1}{n^2}sum_{i,j=1}^nEleft[U_iU_jright]\
              &=&frac{1}{n^2}sum_{i=1}^nEleft[U_i^2right] +frac{1}{n^2}sum_{ine j}Eleft[U_iU_jright]\
              &=&frac{n}{n^2}frac{1}{3}+frac{n(n-1)}{n^2}frac{1}{4}to frac{1}{4}
              end{eqnarray}$$
              since $E[U_i]=int_0^1 xdx=frac{1}{2}$, $E[U_i^2]=int_0^1 x^2dx =frac{1}{3}$ and $E[U_iU_j]=E[U_i]E[U_j]=frac{1}{2}cdotfrac{1}{2}=frac{1}{4}$ for all $ine j$.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                The integral has an equivalent expression
                $$
                Eleft[left(frac{1}{n}sum_{i=1}^n U_iright)^2right]
                $$                 where $U_i$, $ile n$ are independently and uniformly distributed random variables on $[0,1]$. This gives
                $$begin{eqnarray}
                Eleft[left(frac{1}{n}sum_{i=1}^n U_iright)^2right]&=&frac{1}{n^2}sum_{i,j=1}^nEleft[U_iU_jright]\
                &=&frac{1}{n^2}sum_{i=1}^nEleft[U_i^2right] +frac{1}{n^2}sum_{ine j}Eleft[U_iU_jright]\
                &=&frac{n}{n^2}frac{1}{3}+frac{n(n-1)}{n^2}frac{1}{4}to frac{1}{4}
                end{eqnarray}$$
                since $E[U_i]=int_0^1 xdx=frac{1}{2}$, $E[U_i^2]=int_0^1 x^2dx =frac{1}{3}$ and $E[U_iU_j]=E[U_i]E[U_j]=frac{1}{2}cdotfrac{1}{2}=frac{1}{4}$ for all $ine j$.






                share|cite|improve this answer









                $endgroup$



                The integral has an equivalent expression
                $$
                Eleft[left(frac{1}{n}sum_{i=1}^n U_iright)^2right]
                $$                 where $U_i$, $ile n$ are independently and uniformly distributed random variables on $[0,1]$. This gives
                $$begin{eqnarray}
                Eleft[left(frac{1}{n}sum_{i=1}^n U_iright)^2right]&=&frac{1}{n^2}sum_{i,j=1}^nEleft[U_iU_jright]\
                &=&frac{1}{n^2}sum_{i=1}^nEleft[U_i^2right] +frac{1}{n^2}sum_{ine j}Eleft[U_iU_jright]\
                &=&frac{n}{n^2}frac{1}{3}+frac{n(n-1)}{n^2}frac{1}{4}to frac{1}{4}
                end{eqnarray}$$
                since $E[U_i]=int_0^1 xdx=frac{1}{2}$, $E[U_i^2]=int_0^1 x^2dx =frac{1}{3}$ and $E[U_iU_j]=E[U_i]E[U_j]=frac{1}{2}cdotfrac{1}{2}=frac{1}{4}$ for all $ine j$.







                share|cite|improve this answer












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                share|cite|improve this answer










                answered Jan 13 at 19:59









                SongSong

                11.2k628




                11.2k628























                    3












                    $begingroup$

                    We have $$(x_1+x_2+cdots +x_n)^2=sum_i x_i^2+sum_{i,jne i} x_ix_j$$first of all note that $$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{x_i}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}=int_0^1 x_i^2dx_i={1over 3}$$and $$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{x_ix_j}{}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}={1over 4}$$therefore$$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{left( {{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{} right)}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}=ncdot {1over 3}+(n^2-n){1over 4}$$and by substitution we obtain $$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{left( frac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n} right)}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}={1over 3n}+{n-1over 4n}$$which obviously shows that the limit is $1over 4$.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Sorry first line is not clear to if we say $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$ by induction you will get the following result , ${{left( sumnolimits_{i=1}^{n}{{{x}_{i}}} right)}^{2}}=sumnolimits_{i=1}^{n}{x_{i}^{2}+2sumnolimits_{ine j}^{n}{{{x}_{i}}{{x}_{j}}}}$
                      $endgroup$
                      – Ramez Hindi
                      Jan 13 at 21:34












                    • $begingroup$
                      In fact you get $(sum_i x_i)^2=x_1^2+cdots + x_n^2+2sum_{i<j}x_ix_j$. The multiplicity has been considered in my equation since $2x_ix_j=x_ix_j+x_jx_i$ whenever $ine j$
                      $endgroup$
                      – Mostafa Ayaz
                      Jan 13 at 21:38
















                    3












                    $begingroup$

                    We have $$(x_1+x_2+cdots +x_n)^2=sum_i x_i^2+sum_{i,jne i} x_ix_j$$first of all note that $$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{x_i}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}=int_0^1 x_i^2dx_i={1over 3}$$and $$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{x_ix_j}{}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}={1over 4}$$therefore$$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{left( {{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{} right)}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}=ncdot {1over 3}+(n^2-n){1over 4}$$and by substitution we obtain $$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{left( frac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n} right)}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}={1over 3n}+{n-1over 4n}$$which obviously shows that the limit is $1over 4$.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Sorry first line is not clear to if we say $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$ by induction you will get the following result , ${{left( sumnolimits_{i=1}^{n}{{{x}_{i}}} right)}^{2}}=sumnolimits_{i=1}^{n}{x_{i}^{2}+2sumnolimits_{ine j}^{n}{{{x}_{i}}{{x}_{j}}}}$
                      $endgroup$
                      – Ramez Hindi
                      Jan 13 at 21:34












                    • $begingroup$
                      In fact you get $(sum_i x_i)^2=x_1^2+cdots + x_n^2+2sum_{i<j}x_ix_j$. The multiplicity has been considered in my equation since $2x_ix_j=x_ix_j+x_jx_i$ whenever $ine j$
                      $endgroup$
                      – Mostafa Ayaz
                      Jan 13 at 21:38














                    3












                    3








                    3





                    $begingroup$

                    We have $$(x_1+x_2+cdots +x_n)^2=sum_i x_i^2+sum_{i,jne i} x_ix_j$$first of all note that $$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{x_i}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}=int_0^1 x_i^2dx_i={1over 3}$$and $$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{x_ix_j}{}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}={1over 4}$$therefore$$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{left( {{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{} right)}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}=ncdot {1over 3}+(n^2-n){1over 4}$$and by substitution we obtain $$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{left( frac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n} right)}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}={1over 3n}+{n-1over 4n}$$which obviously shows that the limit is $1over 4$.






                    share|cite|improve this answer









                    $endgroup$



                    We have $$(x_1+x_2+cdots +x_n)^2=sum_i x_i^2+sum_{i,jne i} x_ix_j$$first of all note that $$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{x_i}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}=int_0^1 x_i^2dx_i={1over 3}$$and $$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{x_ix_j}{}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}={1over 4}$$therefore$$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{left( {{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{} right)}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}=ncdot {1over 3}+(n^2-n){1over 4}$$and by substitution we obtain $$int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{left( frac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n} right)}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}={1over 3n}+{n-1over 4n}$$which obviously shows that the limit is $1over 4$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 13 at 20:19









                    Mostafa AyazMostafa Ayaz

                    15.3k3939




                    15.3k3939












                    • $begingroup$
                      Sorry first line is not clear to if we say $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$ by induction you will get the following result , ${{left( sumnolimits_{i=1}^{n}{{{x}_{i}}} right)}^{2}}=sumnolimits_{i=1}^{n}{x_{i}^{2}+2sumnolimits_{ine j}^{n}{{{x}_{i}}{{x}_{j}}}}$
                      $endgroup$
                      – Ramez Hindi
                      Jan 13 at 21:34












                    • $begingroup$
                      In fact you get $(sum_i x_i)^2=x_1^2+cdots + x_n^2+2sum_{i<j}x_ix_j$. The multiplicity has been considered in my equation since $2x_ix_j=x_ix_j+x_jx_i$ whenever $ine j$
                      $endgroup$
                      – Mostafa Ayaz
                      Jan 13 at 21:38


















                    • $begingroup$
                      Sorry first line is not clear to if we say $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$ by induction you will get the following result , ${{left( sumnolimits_{i=1}^{n}{{{x}_{i}}} right)}^{2}}=sumnolimits_{i=1}^{n}{x_{i}^{2}+2sumnolimits_{ine j}^{n}{{{x}_{i}}{{x}_{j}}}}$
                      $endgroup$
                      – Ramez Hindi
                      Jan 13 at 21:34












                    • $begingroup$
                      In fact you get $(sum_i x_i)^2=x_1^2+cdots + x_n^2+2sum_{i<j}x_ix_j$. The multiplicity has been considered in my equation since $2x_ix_j=x_ix_j+x_jx_i$ whenever $ine j$
                      $endgroup$
                      – Mostafa Ayaz
                      Jan 13 at 21:38
















                    $begingroup$
                    Sorry first line is not clear to if we say $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$ by induction you will get the following result , ${{left( sumnolimits_{i=1}^{n}{{{x}_{i}}} right)}^{2}}=sumnolimits_{i=1}^{n}{x_{i}^{2}+2sumnolimits_{ine j}^{n}{{{x}_{i}}{{x}_{j}}}}$
                    $endgroup$
                    – Ramez Hindi
                    Jan 13 at 21:34






                    $begingroup$
                    Sorry first line is not clear to if we say $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$ by induction you will get the following result , ${{left( sumnolimits_{i=1}^{n}{{{x}_{i}}} right)}^{2}}=sumnolimits_{i=1}^{n}{x_{i}^{2}+2sumnolimits_{ine j}^{n}{{{x}_{i}}{{x}_{j}}}}$
                    $endgroup$
                    – Ramez Hindi
                    Jan 13 at 21:34














                    $begingroup$
                    In fact you get $(sum_i x_i)^2=x_1^2+cdots + x_n^2+2sum_{i<j}x_ix_j$. The multiplicity has been considered in my equation since $2x_ix_j=x_ix_j+x_jx_i$ whenever $ine j$
                    $endgroup$
                    – Mostafa Ayaz
                    Jan 13 at 21:38




                    $begingroup$
                    In fact you get $(sum_i x_i)^2=x_1^2+cdots + x_n^2+2sum_{i<j}x_ix_j$. The multiplicity has been considered in my equation since $2x_ix_j=x_ix_j+x_jx_i$ whenever $ine j$
                    $endgroup$
                    – Mostafa Ayaz
                    Jan 13 at 21:38











                    2












                    $begingroup$

                    Suppose $X_1,X_2,ldots$ are independent random variables having the uniform distribution on $[0,1]$. Then the common expectation of these variables exist and equals $mu=1/2$.



                    Define $$overline X_n=frac{1}{n}sum_{k=1}^n X_k$$



                    By Khintchine's weak law of large numbers, $$overline X_nstackrel{P}{longrightarrow}muquadtext{ as }quad ntoinfty$$



                    And by the continuous mapping theorem, $$overline X_n^2stackrel{P}{longrightarrow}mu^2quadtext{ as }quad ntoinftytag{1}$$



                    Moreover, $$0le X_1,ldots,X_nle 1implies 0le overline X_nle 1implies 0le overline X_n^2le 1tag{2}$$



                    $(1)$ and $(2)$ together imply $$int_{[0,1]^n}left(frac{x_1+cdots+x_n}{n}right)^2mathrm{d}x_1ldotsmathrm{d}x_n = Eleft(overline X_n^2right)stackrel{ntoinfty}{longrightarrow}frac{1}{4}$$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      It’s always nice to see an easygoing probability solution to what appears to be a monstrous analytic problem. +1
                      $endgroup$
                      – LoveTooNap29
                      Jan 13 at 20:08










                    • $begingroup$
                      there must be some confusion. I upvoted both yours and Song’s answer for the use of probability.
                      $endgroup$
                      – LoveTooNap29
                      Jan 13 at 20:33










                    • $begingroup$
                      @LoveTooNap29 I wasn't replying to you.
                      $endgroup$
                      – StubbornAtom
                      Jan 13 at 20:34










                    • $begingroup$
                      $X_nto X$ in probability isn't enough to conclude that $mathbb{E}[X_n]tomathbb{E}[X]$. It would be better to use the strong law of large numbers, plus dominated convergence.
                      $endgroup$
                      – carmichael561
                      Jan 15 at 0:27










                    • $begingroup$
                      @carmichael561 Thank you. What if I add the fact that $|X_n|<1$ to my answer? I think that salvages the argument.
                      $endgroup$
                      – StubbornAtom
                      Jan 15 at 4:59
















                    2












                    $begingroup$

                    Suppose $X_1,X_2,ldots$ are independent random variables having the uniform distribution on $[0,1]$. Then the common expectation of these variables exist and equals $mu=1/2$.



                    Define $$overline X_n=frac{1}{n}sum_{k=1}^n X_k$$



                    By Khintchine's weak law of large numbers, $$overline X_nstackrel{P}{longrightarrow}muquadtext{ as }quad ntoinfty$$



                    And by the continuous mapping theorem, $$overline X_n^2stackrel{P}{longrightarrow}mu^2quadtext{ as }quad ntoinftytag{1}$$



                    Moreover, $$0le X_1,ldots,X_nle 1implies 0le overline X_nle 1implies 0le overline X_n^2le 1tag{2}$$



                    $(1)$ and $(2)$ together imply $$int_{[0,1]^n}left(frac{x_1+cdots+x_n}{n}right)^2mathrm{d}x_1ldotsmathrm{d}x_n = Eleft(overline X_n^2right)stackrel{ntoinfty}{longrightarrow}frac{1}{4}$$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      It’s always nice to see an easygoing probability solution to what appears to be a monstrous analytic problem. +1
                      $endgroup$
                      – LoveTooNap29
                      Jan 13 at 20:08










                    • $begingroup$
                      there must be some confusion. I upvoted both yours and Song’s answer for the use of probability.
                      $endgroup$
                      – LoveTooNap29
                      Jan 13 at 20:33










                    • $begingroup$
                      @LoveTooNap29 I wasn't replying to you.
                      $endgroup$
                      – StubbornAtom
                      Jan 13 at 20:34










                    • $begingroup$
                      $X_nto X$ in probability isn't enough to conclude that $mathbb{E}[X_n]tomathbb{E}[X]$. It would be better to use the strong law of large numbers, plus dominated convergence.
                      $endgroup$
                      – carmichael561
                      Jan 15 at 0:27










                    • $begingroup$
                      @carmichael561 Thank you. What if I add the fact that $|X_n|<1$ to my answer? I think that salvages the argument.
                      $endgroup$
                      – StubbornAtom
                      Jan 15 at 4:59














                    2












                    2








                    2





                    $begingroup$

                    Suppose $X_1,X_2,ldots$ are independent random variables having the uniform distribution on $[0,1]$. Then the common expectation of these variables exist and equals $mu=1/2$.



                    Define $$overline X_n=frac{1}{n}sum_{k=1}^n X_k$$



                    By Khintchine's weak law of large numbers, $$overline X_nstackrel{P}{longrightarrow}muquadtext{ as }quad ntoinfty$$



                    And by the continuous mapping theorem, $$overline X_n^2stackrel{P}{longrightarrow}mu^2quadtext{ as }quad ntoinftytag{1}$$



                    Moreover, $$0le X_1,ldots,X_nle 1implies 0le overline X_nle 1implies 0le overline X_n^2le 1tag{2}$$



                    $(1)$ and $(2)$ together imply $$int_{[0,1]^n}left(frac{x_1+cdots+x_n}{n}right)^2mathrm{d}x_1ldotsmathrm{d}x_n = Eleft(overline X_n^2right)stackrel{ntoinfty}{longrightarrow}frac{1}{4}$$






                    share|cite|improve this answer











                    $endgroup$



                    Suppose $X_1,X_2,ldots$ are independent random variables having the uniform distribution on $[0,1]$. Then the common expectation of these variables exist and equals $mu=1/2$.



                    Define $$overline X_n=frac{1}{n}sum_{k=1}^n X_k$$



                    By Khintchine's weak law of large numbers, $$overline X_nstackrel{P}{longrightarrow}muquadtext{ as }quad ntoinfty$$



                    And by the continuous mapping theorem, $$overline X_n^2stackrel{P}{longrightarrow}mu^2quadtext{ as }quad ntoinftytag{1}$$



                    Moreover, $$0le X_1,ldots,X_nle 1implies 0le overline X_nle 1implies 0le overline X_n^2le 1tag{2}$$



                    $(1)$ and $(2)$ together imply $$int_{[0,1]^n}left(frac{x_1+cdots+x_n}{n}right)^2mathrm{d}x_1ldotsmathrm{d}x_n = Eleft(overline X_n^2right)stackrel{ntoinfty}{longrightarrow}frac{1}{4}$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 15 at 7:08

























                    answered Jan 13 at 20:00









                    StubbornAtomStubbornAtom

                    5,75611138




                    5,75611138












                    • $begingroup$
                      It’s always nice to see an easygoing probability solution to what appears to be a monstrous analytic problem. +1
                      $endgroup$
                      – LoveTooNap29
                      Jan 13 at 20:08










                    • $begingroup$
                      there must be some confusion. I upvoted both yours and Song’s answer for the use of probability.
                      $endgroup$
                      – LoveTooNap29
                      Jan 13 at 20:33










                    • $begingroup$
                      @LoveTooNap29 I wasn't replying to you.
                      $endgroup$
                      – StubbornAtom
                      Jan 13 at 20:34










                    • $begingroup$
                      $X_nto X$ in probability isn't enough to conclude that $mathbb{E}[X_n]tomathbb{E}[X]$. It would be better to use the strong law of large numbers, plus dominated convergence.
                      $endgroup$
                      – carmichael561
                      Jan 15 at 0:27










                    • $begingroup$
                      @carmichael561 Thank you. What if I add the fact that $|X_n|<1$ to my answer? I think that salvages the argument.
                      $endgroup$
                      – StubbornAtom
                      Jan 15 at 4:59


















                    • $begingroup$
                      It’s always nice to see an easygoing probability solution to what appears to be a monstrous analytic problem. +1
                      $endgroup$
                      – LoveTooNap29
                      Jan 13 at 20:08










                    • $begingroup$
                      there must be some confusion. I upvoted both yours and Song’s answer for the use of probability.
                      $endgroup$
                      – LoveTooNap29
                      Jan 13 at 20:33










                    • $begingroup$
                      @LoveTooNap29 I wasn't replying to you.
                      $endgroup$
                      – StubbornAtom
                      Jan 13 at 20:34










                    • $begingroup$
                      $X_nto X$ in probability isn't enough to conclude that $mathbb{E}[X_n]tomathbb{E}[X]$. It would be better to use the strong law of large numbers, plus dominated convergence.
                      $endgroup$
                      – carmichael561
                      Jan 15 at 0:27










                    • $begingroup$
                      @carmichael561 Thank you. What if I add the fact that $|X_n|<1$ to my answer? I think that salvages the argument.
                      $endgroup$
                      – StubbornAtom
                      Jan 15 at 4:59
















                    $begingroup$
                    It’s always nice to see an easygoing probability solution to what appears to be a monstrous analytic problem. +1
                    $endgroup$
                    – LoveTooNap29
                    Jan 13 at 20:08




                    $begingroup$
                    It’s always nice to see an easygoing probability solution to what appears to be a monstrous analytic problem. +1
                    $endgroup$
                    – LoveTooNap29
                    Jan 13 at 20:08












                    $begingroup$
                    there must be some confusion. I upvoted both yours and Song’s answer for the use of probability.
                    $endgroup$
                    – LoveTooNap29
                    Jan 13 at 20:33




                    $begingroup$
                    there must be some confusion. I upvoted both yours and Song’s answer for the use of probability.
                    $endgroup$
                    – LoveTooNap29
                    Jan 13 at 20:33












                    $begingroup$
                    @LoveTooNap29 I wasn't replying to you.
                    $endgroup$
                    – StubbornAtom
                    Jan 13 at 20:34




                    $begingroup$
                    @LoveTooNap29 I wasn't replying to you.
                    $endgroup$
                    – StubbornAtom
                    Jan 13 at 20:34












                    $begingroup$
                    $X_nto X$ in probability isn't enough to conclude that $mathbb{E}[X_n]tomathbb{E}[X]$. It would be better to use the strong law of large numbers, plus dominated convergence.
                    $endgroup$
                    – carmichael561
                    Jan 15 at 0:27




                    $begingroup$
                    $X_nto X$ in probability isn't enough to conclude that $mathbb{E}[X_n]tomathbb{E}[X]$. It would be better to use the strong law of large numbers, plus dominated convergence.
                    $endgroup$
                    – carmichael561
                    Jan 15 at 0:27












                    $begingroup$
                    @carmichael561 Thank you. What if I add the fact that $|X_n|<1$ to my answer? I think that salvages the argument.
                    $endgroup$
                    – StubbornAtom
                    Jan 15 at 4:59




                    $begingroup$
                    @carmichael561 Thank you. What if I add the fact that $|X_n|<1$ to my answer? I think that salvages the argument.
                    $endgroup$
                    – StubbornAtom
                    Jan 15 at 4:59


















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