Sequence of integrals
Letting $finC^0[0,1]$, compute the limit of the sequence:
$a_n=int_0^1int_0^1...int_0^1 prod_(i=1)^n (2x_i)*f((sum_(i=1)^n x_i)/n) dx_1dx_2...dx_n $
$a_n=int_0^1int_0^1...int_0^1 prod_(i=1)^n (2x_i)*f((sum_(i=1)^n x_i)/n) dx_1dx_2...dx_n $
Risposte
well... after a long time today I was thinking about this problem once again, in order to try to solve the problem before the end of the year 
... so I write the solution...
this problem was solved a long time ago applying techniqus coming from the probability theory...
$int_([0,1]^n)f(sum_ig(x_i)/n)->f(int_0^1g(x)dx)$
now there are only analitic passages (I don't know why I wasn't able to conclude last time, bah)... let's put $g(x)=sqrt(x)$ a real funtion... with this change of variables
$y_i=sqrt(x_i),i=1...n$
now the vector $(y_1,y_2,...,y_n)$ varies again in $[0,1]^n$, but the jacobian of the transformation is
$2^nprod_i y_i$, so the integral becomes
$int_([0,1]^n)f(sum_i (y_i)/n)prod_i(2y_i)$
and this is the function of the original problem...
So the solution is
$f(int_0^1g(x)dx)=f(2/3)$
right?
... so I write the solution...this problem was solved a long time ago applying techniqus coming from the probability theory...
$int_([0,1]^n)f(sum_ig(x_i)/n)->f(int_0^1g(x)dx)$
now there are only analitic passages (I don't know why I wasn't able to conclude last time, bah)... let's put $g(x)=sqrt(x)$ a real funtion... with this change of variables
$y_i=sqrt(x_i),i=1...n$
now the vector $(y_1,y_2,...,y_n)$ varies again in $[0,1]^n$, but the jacobian of the transformation is
$2^nprod_i y_i$, so the integral becomes
$int_([0,1]^n)f(sum_i (y_i)/n)prod_i(2y_i)$
and this is the function of the original problem...
So the solution is
$f(int_0^1g(x)dx)=f(2/3)$
right?
... give me a useful piece of advice, luca... I had already tried doing what u said...
bye
bye
My problem is right there 
... and here I need help... I've already tried to do what you say... I don't know how to get rid of the variables outside f...
... and here I need help... I've already tried to do what you say... I don't know how to get rid of the variables outside f...
"Thomas":
your last integral tends to $f(E[g(x)])=f(int_0^1g(x)dx)$
Well done.
Now you are very close to the solution... try to write the original problem in this form:
$int_([0,1]^n) f(sum_ig(x_i)/n) $
your last integral tends to $f(E[g(x)])=f(int_0^1g(x)dx)$, but I can't see how this can be useful to solve the problem....
I have already tried to "change the problem" in a nicer way using logarithms and change of variables but didn't succeed... I'm not able to pu the variables into the f
...
I have already tried to "change the problem" in a nicer way using logarithms and change of variables but didn't succeed... I'm not able to pu the variables into the f
...
"Thomas":
I'm stuck an other time... I'd like to estimate $V[X,Y]$ where $X=prod(2x_i)$ and $Y=sumf((x_j))/n$... the result in fact I think it should be $f(1/2)+V[X,Y]$...
I know the relation $V[X,Y]<=\sigma^2(X)\sigma^2(Y)$, but a calculation led me to the conclusion that this product diverges with n... and so it's quite useless... anyway, I'd need also an estimate from below of $V[X,Y]$....
boh.........help... do I miss some knowledge an other time or what?
Here I am.
This last procedure is too complicated. Till now you have observed that, for the big numbers law, it holds:
$int_([0,1]^n) f((x_1+x_2+...+x_n)/n) dx_1dx_2...dx_n$ converges to $f(1/2)$.
The next step could be to find:
$int_([0,1]^n) f((g(x_1)+g(x_2)+...+g(x_n))/n) dx_1dx_2...dx_n$
where $g(x)$ is a continuous function.
nobody helps, so I'm obliged leave this problem unsolved an other time... I've got other things to do... pherpaps when I'll study somehing that could be useful to solve this problem, I'll come back 
...
but can u at least tell me if the formula I wrote is correct?
...but can u at least tell me if the formula I wrote is correct?
so??????????????????????????????????????
I'm stuck an other time... I'd like to estimate $V[X,Y]$ where $X=prod(2x_i)$ and $Y=sumf((x_j))/n$... the result in fact I think it should be $f(1/2)+V[X,Y]$...
I know the relation $V[X,Y]<=\sigma^2(X)\sigma^2(Y)$, but a calculation led me to the conclusion that this product diverges with n... and so it's quite useless... anyway, I'd need also an estimate from below of $V[X,Y]$....
boh.........help... do I miss some knowledge an other time or what?
I know the relation $V[X,Y]<=\sigma^2(X)\sigma^2(Y)$, but a calculation led me to the conclusion that this product diverges with n... and so it's quite useless... anyway, I'd need also an estimate from below of $V[X,Y]$....
boh.........help... do I miss some knowledge an other time or what?
sorry... i forgot to write in english!!!!!!!
"Thomas":
studiando un pò di probabilità elementare mi sono reso conto che il problema chiede di trovare a cosa tende:
$E[prod(2x_i)sum(f(x_j))]$
se non ci fosse la produttoria con la legge dei grandi numeri si concluderebbe che tende a $f(1/2)$... forse...
la produttoria dà fastidio in quanto le variabili $(x_1,x_2,...,x_1+x_2+...+x_n)$ non sono indipendenti.... se qualcuno conosce un qualche modo per stimare la loro covarianza magari...
very close to the solution
studiando un pò di probabilità elementare mi sono reso conto che il problema chiede di trovare a cosa tende:
$E[prod(2x_i)sum(f(x_j))]$
se non ci fosse la produttoria con la legge dei grandi numeri si concluderebbe che tende a $f(1/2)$... forse...
la produttoria dà fastidio in quanto le variabili $(x_1,x_2,...,x_1+x_2+...+x_n)$ non sono indipendenti.... se qualcuno conosce un qualche modo per stimare la loro covarianza magari...
chissà...magari con la probabilità asi arriva a qualcosa... che dici?!
$E[prod(2x_i)sum(f(x_j))]$
se non ci fosse la produttoria con la legge dei grandi numeri si concluderebbe che tende a $f(1/2)$... forse...
la produttoria dà fastidio in quanto le variabili $(x_1,x_2,...,x_1+x_2+...+x_n)$ non sono indipendenti.... se qualcuno conosce un qualche modo per stimare la loro covarianza magari...
chissà...magari con la probabilità asi arriva a qualcosa... che dici?!
I didn't think something else... I'm studing a lot and I have no time to deep the problem.
I just thought that a density approach could solve the problem, but maybe I'm wrong.
I just thought that a density approach could solve the problem, but maybe I'm wrong.
why do you extimate $|a_n-a|$ and not $|a^k-a|$? I thought u had the intention to calculate the limit of the sequence for $f$ as the limit of the values calculated for a sequence of polyinomials converging to the function... 
... are u trying to do something else?
... are u trying to do something else?
I don't sure that it works..
Let $a$ be the limit of your conjecture of $a_n$, $f_k$ a Stone-Weierstrass approximation of $f$ and $a_n^k$ the relative succession and $a^k$ the limit.
Then
$||a_n-a||\le||a_n-a_n^k||+||a_n^k-a^k||+||a^k-a||
The last term is less then $\epsi$, by S.W., and I think that also the others are less then $epsi$, but I don't calculate them.
Let $a$ be the limit of your conjecture of $a_n$, $f_k$ a Stone-Weierstrass approximation of $f$ and $a_n^k$ the relative succession and $a^k$ the limit.
Then
$||a_n-a||\le||a_n-a_n^k||+||a_n^k-a^k||+||a^k-a||
The last term is less then $\epsi$, by S.W., and I think that also the others are less then $epsi$, but I don't calculate them.
- u mean using the density of polynomials in the sup norm? but how?
can u explain your idea better?
can u explain your idea better?
To pass from polynomials to generic continuos function isn't it sufficient Stone Weierstrass theorem?
as u wish 
, can I only ask why? ........
ps1: in any case, I guess that there will be a time when I'll try to complete the calculation for every polinomial (summer????
)(even if I don't know if it is easy to pass from polynomial to a generic f... I'd like to know why u suggested me to proceed with polynomial f?)... just for information the result is a closed simple formula?...
ps2: a long time ago, u didn't tell me if the calculation I posted for a second degree polynomial is correct (just check it with the generic formula, plz)
, can I only ask why? ........ps1: in any case, I guess that there will be a time when I'll try to complete the calculation for every polinomial (summer????
)(even if I don't know if it is easy to pass from polynomial to a generic f... I'd like to know why u suggested me to proceed with polynomial f?)... just for information the result is a closed simple formula?...ps2: a long time ago, u didn't tell me if the calculation I posted for a second degree polynomial is correct (just check it with the generic formula, plz)
mmmh, I won't post the solution
actually I didn't try again.... but luca, I guess it's time to post the solution... nobody has tried for a long time....
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