Mixing
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
asked Jan 13 at 19:51
Ramez HindiRamez Hindi
1437
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add a comment |
$begingroup$
$$underset{nto infty }{mathop{lim }},int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{left( frac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n} right)}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}$$
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
asked Jan 13 at 19:51
Ramez HindiRamez Hindi
1437
$endgroup$
add a comment |
$begingroup$
$$underset{nto infty }{mathop{lim }},int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{left( frac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n} right)}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}$$
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
asked Jan 13 at 19:51
Ramez HindiRamez Hindi
1437
$endgroup$
$$underset{nto infty }{mathop{lim }},int_{0}^{1}{int_{0}^{1}{cdots int_{0}^{1}{{{left( frac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n} right)}^{2}}d{{x}_{1}}d{{x}_{2}}cdots d{{x}_{n}}}}}$$
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
asked Jan 13 at 19:51
Ramez HindiRamez Hindi
1437
asked Jan 13 at 19:51
Ramez HindiRamez Hindi
1437
asked Jan 13 at 19:51
Ramez HindiRamez Hindi
1437
asked Jan 13 at 19:51
Ramez HindiRamez Hindi
1437
asked Jan 13 at 19:51
Ramez HindiRamez Hindi
1437
1437
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 $ine j$.
answered Jan 13 at 19:59
SongSong
13.1k632
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add a comment |
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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 $1over 4$.
answered Jan 13 at 20:19
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $$int_{[0,1]^n}left(frac{x_1+cdots+x_n}{n}right)^2mathrm{d}x_1ldotsmathrm{d}x_n = Eleft(overline X_n^2right)stackrel{ntoinfty}{longrightarrow}frac{1}{4}$$
answered Jan 13 at 20:00
StubbornAtomStubbornAtom
5,98811238
$endgroup$
$begingroup$
It’s always nice to see an easygoing probability solution to what appears to be a monstrous analytic problem. +1
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– LoveTooNap29
Jan 13 at 20:08
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there must be some confusion. I upvoted both yours and Song’s answer for the use of probability.
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– LoveTooNap29
Jan 13 at 20:33
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@LoveTooNap29 I wasn't replying to you.
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– StubbornAtom
Jan 13 at 20:34
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$X_nto X$ 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
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@carmichael561 Thank you. What if I add the fact that $|X_n|<1$ to my answer? I think that salvages the argument.
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– StubbornAtom
Jan 15 at 4:59
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3 Answers
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active
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $ine j$.
answered Jan 13 at 19:59
SongSong
13.1k632
$endgroup$
add a comment |
$begingroup$
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 $ine j$.
answered Jan 13 at 19:59
SongSong
13.1k632
$endgroup$
add a comment |
$begingroup$
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 $ine j$.
answered Jan 13 at 19:59
SongSong
13.1k632
$endgroup$
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 $ine j$.
answered Jan 13 at 19:59
SongSong
13.1k632
answered Jan 13 at 19:59
SongSong
13.1k632
answered Jan 13 at 19:59
SongSong
13.1k632
answered Jan 13 at 19:59
SongSong
13.1k632
13.1k632
add a comment |
add a comment |
$begingroup$
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 $1over 4$.
answered Jan 13 at 20:19
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $1over 4$.
answered Jan 13 at 20:19
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $1over 4$.
answered Jan 13 at 20:19
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
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 $1over 4$.
answered Jan 13 at 20:19
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 13 at 20:19
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 13 at 20:19
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 13 at 20:19
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 $$int_{[0,1]^n}left(frac{x_1+cdots+x_n}{n}right)^2mathrm{d}x_1ldotsmathrm{d}x_n = Eleft(overline X_n^2right)stackrel{ntoinfty}{longrightarrow}frac{1}{4}$$
answered Jan 13 at 20:00
StubbornAtomStubbornAtom
5,98811238
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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 $$int_{[0,1]^n}left(frac{x_1+cdots+x_n}{n}right)^2mathrm{d}x_1ldotsmathrm{d}x_n = Eleft(overline X_n^2right)stackrel{ntoinfty}{longrightarrow}frac{1}{4}$$
answered Jan 13 at 20:00
StubbornAtomStubbornAtom
5,98811238
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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 $$int_{[0,1]^n}left(frac{x_1+cdots+x_n}{n}right)^2mathrm{d}x_1ldotsmathrm{d}x_n = Eleft(overline X_n^2right)stackrel{ntoinfty}{longrightarrow}frac{1}{4}$$
answered Jan 13 at 20:00
StubbornAtomStubbornAtom
5,98811238
$endgroup$
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 $$int_{[0,1]^n}left(frac{x_1+cdots+x_n}{n}right)^2mathrm{d}x_1ldotsmathrm{d}x_n = Eleft(overline X_n^2right)stackrel{ntoinfty}{longrightarrow}frac{1}{4}$$
answered Jan 13 at 20:00
StubbornAtomStubbornAtom
5,98811238
edited Jan 15 at 7:08
answered Jan 13 at 20:00
StubbornAtomStubbornAtom
5,98811238
answered Jan 13 at 20:00
StubbornAtomStubbornAtom
5,98811238
answered Jan 13 at 20:00
StubbornAtomStubbornAtom
5,98811238
5,98811238
$begingroup$
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
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