Convergence in distribution and asymptotic distribution
Clash Royale CLAN TAG#URR8PPP
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I am having a problem with proving convergence in distribution (or by law).
Consider that the sequence $X_n$ of random variables are IID and that $E[X_n]=0$ and $V[X_n]=1$.
Now define the variable $U_N$ as:
$$U_N= frac1sqrtNsum_n=1^N X_ncdot sinleft(fracnpiNright).$$
For $Nrightarrow infty$, I want to show that $U_N$ converges by distribution, furthermore, I also want to determine the asymptotic distribution of $U_N$.
For the first part, I have tried to show convergence in probability, because this implies conv. in distribution (by law), but this was not possible since I dont have the asymptotic distribution of the $X_n$.
For the second part, I tried with the delta-method, but this did not work because of the summation of in the expression for $U_N$.
Does someone have an idea to this?
probability convergence probability-limit-theorems
edited Aug 14 at 11:40
Davide Giraudo
121k15147250
asked Aug 13 at 9:32
Jonathan Kiersch
659
 |Â
show 4 more comments
up vote
1
down vote
favorite
I am having a problem with proving convergence in distribution (or by law).
Consider that the sequence $X_n$ of random variables are IID and that $E[X_n]=0$ and $V[X_n]=1$.
Now define the variable $U_N$ as:
$$U_N= frac1sqrtNsum_n=1^N X_ncdot sinleft(fracnpiNright).$$
For $Nrightarrow infty$, I want to show that $U_N$ converges by distribution, furthermore, I also want to determine the asymptotic distribution of $U_N$.
For the first part, I have tried to show convergence in probability, because this implies conv. in distribution (by law), but this was not possible since I dont have the asymptotic distribution of the $X_n$.
For the second part, I tried with the delta-method, but this did not work because of the summation of in the expression for $U_N$.
Does someone have an idea to this?
probability convergence probability-limit-theorems
edited Aug 14 at 11:40
Davide Giraudo
121k15147250
asked Aug 13 at 9:32
Jonathan Kiersch
659
$E[U_N] = 0$. Do you know what $Var(U_N)$ is or at least what it converges to?
– Henry
Aug 13 at 9:37
Var(U_N) should be 1/Nsum(sin(npi)/N), which also converges to zero for N->infty
– Jonathan Kiersch
Aug 13 at 9:45
I suspect $Var(U_N) to intlimits_0^1 sin^2(pi x), dx = frac12$
– Henry
Aug 13 at 11:10
How did you find that?
– Jonathan Kiersch
Aug 13 at 11:57
A combination of thinking that without the $sin(fracnpiN)$ term the variance would be $1$ as in the central limit theorem, plus simulation and basic calculus.
– Henry
Aug 13 at 12:01
 |Â
show 4 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am having a problem with proving convergence in distribution (or by law).
Consider that the sequence $X_n$ of random variables are IID and that $E[X_n]=0$ and $V[X_n]=1$.
Now define the variable $U_N$ as:
$$U_N= frac1sqrtNsum_n=1^N X_ncdot sinleft(fracnpiNright).$$
For $Nrightarrow infty$, I want to show that $U_N$ converges by distribution, furthermore, I also want to determine the asymptotic distribution of $U_N$.
For the first part, I have tried to show convergence in probability, because this implies conv. in distribution (by law), but this was not possible since I dont have the asymptotic distribution of the $X_n$.
For the second part, I tried with the delta-method, but this did not work because of the summation of in the expression for $U_N$.
Does someone have an idea to this?
probability convergence probability-limit-theorems
edited Aug 14 at 11:40
Davide Giraudo
121k15147250
asked Aug 13 at 9:32
Jonathan Kiersch
659
I am having a problem with proving convergence in distribution (or by law).
Consider that the sequence $X_n$ of random variables are IID and that $E[X_n]=0$ and $V[X_n]=1$.
Now define the variable $U_N$ as:
$$U_N= frac1sqrtNsum_n=1^N X_ncdot sinleft(fracnpiNright).$$
For $Nrightarrow infty$, I want to show that $U_N$ converges by distribution, furthermore, I also want to determine the asymptotic distribution of $U_N$.
For the first part, I have tried to show convergence in probability, because this implies conv. in distribution (by law), but this was not possible since I dont have the asymptotic distribution of the $X_n$.
For the second part, I tried with the delta-method, but this did not work because of the summation of in the expression for $U_N$.
Does someone have an idea to this?
probability convergence probability-limit-theorems
edited Aug 14 at 11:40
Davide Giraudo
121k15147250
asked Aug 13 at 9:32
Jonathan Kiersch
659
edited Aug 14 at 11:40
Davide Giraudo
121k15147250
edited Aug 14 at 11:40
Davide Giraudo
121k15147250
edited Aug 14 at 11:40
Davide Giraudo
121k15147250
121k15147250
asked Aug 13 at 9:32
Jonathan Kiersch
659
asked Aug 13 at 9:32
Jonathan Kiersch
659
asked Aug 13 at 9:32
Jonathan Kiersch
659
659
$E[U_N] = 0$. Do you know what $Var(U_N)$ is or at least what it converges to?
– Henry
Aug 13 at 9:37
Var(U_N) should be 1/Nsum(sin(npi)/N), which also converges to zero for N->infty
– Jonathan Kiersch
Aug 13 at 9:45
I suspect $Var(U_N) to intlimits_0^1 sin^2(pi x), dx = frac12$
– Henry
Aug 13 at 11:10
How did you find that?
– Jonathan Kiersch
Aug 13 at 11:57
A combination of thinking that without the $sin(fracnpiN)$ term the variance would be $1$ as in the central limit theorem, plus simulation and basic calculus.
– Henry
Aug 13 at 12:01
 |Â
show 4 more comments
$E[U_N] = 0$. Do you know what $Var(U_N)$ is or at least what it converges to?
– Henry
Aug 13 at 9:37
Var(U_N) should be 1/Nsum(sin(npi)/N), which also converges to zero for N->infty
– Jonathan Kiersch
Aug 13 at 9:45
I suspect $Var(U_N) to intlimits_0^1 sin^2(pi x), dx = frac12$
– Henry
Aug 13 at 11:10
How did you find that?
– Jonathan Kiersch
Aug 13 at 11:57
A combination of thinking that without the $sin(fracnpiN)$ term the variance would be $1$ as in the central limit theorem, plus simulation and basic calculus.
– Henry
Aug 13 at 12:01
$E[U_N] = 0$. Do you know what $Var(U_N)$ is or at least what it converges to?
– Henry
Aug 13 at 9:37
$E[U_N] = 0$. Do you know what $Var(U_N)$ is or at least what it converges to?
– Henry
Aug 13 at 9:37
Var(U_N) should be 1/Nsum(sin(npi)/N), which also converges to zero for N->infty
– Jonathan Kiersch
Aug 13 at 9:45
Var(U_N) should be 1/Nsum(sin(npi)/N), which also converges to zero for N->infty
– Jonathan Kiersch
Aug 13 at 9:45
I suspect $Var(U_N) to intlimits_0^1 sin^2(pi x), dx = frac12$
– Henry
Aug 13 at 11:10
I suspect $Var(U_N) to intlimits_0^1 sin^2(pi x), dx = frac12$
– Henry
Aug 13 at 11:10
How did you find that?
– Jonathan Kiersch
Aug 13 at 11:57
How did you find that?
– Jonathan Kiersch
Aug 13 at 11:57
A combination of thinking that without the $sin(fracnpiN)$ term the variance would be $1$ as in the central limit theorem, plus simulation and basic calculus.
– Henry
Aug 13 at 12:01
A combination of thinking that without the $sin(fracnpiN)$ term the variance would be $1$ as in the central limit theorem, plus simulation and basic calculus.
– Henry
Aug 13 at 12:01
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
One can try to check Lindeberg's condition with $X_N,i:= X_isinleft(ipi/Nright)$, using the following three facts:
- the limits $lim_Nto +inftyN^-1sum_n=1^Nsin^2left(npi/Nright)$ exists and is computable, as a limit of Riemann sums.
- $s_N=sum_i=1^NoperatornameVarleft(X_N,iright)sim csqrt N$.
- $$mathbb Eleft[X_N,i^2mathbf 1leftleftlvert X_N,irightvertgtvarepsilon s_N rightright]=mathbb Eleft[X_1^2sin^2left(ipi/Nright)mathbf 1leftleftlvert sinleft(ipi/Nright) X_1rightvertgtvarepsilon s_Nrightright]leqslantmathbb Eleft[X_1^2sin^2left(ipi/Nright)mathbf 1leftleftlvert X_1rightvertgtvarepsilon s_N rightright].$$
answered Aug 13 at 12:25
Davide Giraudo
121k15147250
I see, I was not familiar with the Lindeberg condition for CLT. So if I have understood you correct, I use the Lindeberg condition on X_n,N to find that U_N will converge to a standard normal distribution for N->infinity
– Jonathan Kiersch
Aug 13 at 12:44
Yes, the limit will be a normal distribution but I have not looked whether the variance is one or not.
– Davide Giraudo
Aug 13 at 13:20
@DavideGiraudo - the limiting variance is $frac12$ (it would have been $1$ without the sine term)
– Henry
Aug 13 at 14:04
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
One can try to check Lindeberg's condition with $X_N,i:= X_isinleft(ipi/Nright)$, using the following three facts:
- the limits $lim_Nto +inftyN^-1sum_n=1^Nsin^2left(npi/Nright)$ exists and is computable, as a limit of Riemann sums.
- $s_N=sum_i=1^NoperatornameVarleft(X_N,iright)sim csqrt N$.
- $$mathbb Eleft[X_N,i^2mathbf 1leftleftlvert X_N,irightvertgtvarepsilon s_N rightright]=mathbb Eleft[X_1^2sin^2left(ipi/Nright)mathbf 1leftleftlvert sinleft(ipi/Nright) X_1rightvertgtvarepsilon s_Nrightright]leqslantmathbb Eleft[X_1^2sin^2left(ipi/Nright)mathbf 1leftleftlvert X_1rightvertgtvarepsilon s_N rightright].$$
answered Aug 13 at 12:25
Davide Giraudo
121k15147250
I see, I was not familiar with the Lindeberg condition for CLT. So if I have understood you correct, I use the Lindeberg condition on X_n,N to find that U_N will converge to a standard normal distribution for N->infinity
– Jonathan Kiersch
Aug 13 at 12:44
Yes, the limit will be a normal distribution but I have not looked whether the variance is one or not.
– Davide Giraudo
Aug 13 at 13:20
@DavideGiraudo - the limiting variance is $frac12$ (it would have been $1$ without the sine term)
– Henry
Aug 13 at 14:04
add a comment |Â
up vote
0
down vote
One can try to check Lindeberg's condition with $X_N,i:= X_isinleft(ipi/Nright)$, using the following three facts:
- the limits $lim_Nto +inftyN^-1sum_n=1^Nsin^2left(npi/Nright)$ exists and is computable, as a limit of Riemann sums.
- $s_N=sum_i=1^NoperatornameVarleft(X_N,iright)sim csqrt N$.
- $$mathbb Eleft[X_N,i^2mathbf 1leftleftlvert X_N,irightvertgtvarepsilon s_N rightright]=mathbb Eleft[X_1^2sin^2left(ipi/Nright)mathbf 1leftleftlvert sinleft(ipi/Nright) X_1rightvertgtvarepsilon s_Nrightright]leqslantmathbb Eleft[X_1^2sin^2left(ipi/Nright)mathbf 1leftleftlvert X_1rightvertgtvarepsilon s_N rightright].$$
answered Aug 13 at 12:25
Davide Giraudo
121k15147250
I see, I was not familiar with the Lindeberg condition for CLT. So if I have understood you correct, I use the Lindeberg condition on X_n,N to find that U_N will converge to a standard normal distribution for N->infinity
– Jonathan Kiersch
Aug 13 at 12:44
Yes, the limit will be a normal distribution but I have not looked whether the variance is one or not.
– Davide Giraudo
Aug 13 at 13:20
@DavideGiraudo - the limiting variance is $frac12$ (it would have been $1$ without the sine term)
– Henry
Aug 13 at 14:04
add a comment |Â
up vote
0
down vote
up vote
0
down vote
One can try to check Lindeberg's condition with $X_N,i:= X_isinleft(ipi/Nright)$, using the following three facts:
- the limits $lim_Nto +inftyN^-1sum_n=1^Nsin^2left(npi/Nright)$ exists and is computable, as a limit of Riemann sums.
- $s_N=sum_i=1^NoperatornameVarleft(X_N,iright)sim csqrt N$.
- $$mathbb Eleft[X_N,i^2mathbf 1leftleftlvert X_N,irightvertgtvarepsilon s_N rightright]=mathbb Eleft[X_1^2sin^2left(ipi/Nright)mathbf 1leftleftlvert sinleft(ipi/Nright) X_1rightvertgtvarepsilon s_Nrightright]leqslantmathbb Eleft[X_1^2sin^2left(ipi/Nright)mathbf 1leftleftlvert X_1rightvertgtvarepsilon s_N rightright].$$
answered Aug 13 at 12:25
Davide Giraudo
121k15147250
One can try to check Lindeberg's condition with $X_N,i:= X_isinleft(ipi/Nright)$, using the following three facts:
- the limits $lim_Nto +inftyN^-1sum_n=1^Nsin^2left(npi/Nright)$ exists and is computable, as a limit of Riemann sums.
- $s_N=sum_i=1^NoperatornameVarleft(X_N,iright)sim csqrt N$.
- $$mathbb Eleft[X_N,i^2mathbf 1leftleftlvert X_N,irightvertgtvarepsilon s_N rightright]=mathbb Eleft[X_1^2sin^2left(ipi/Nright)mathbf 1leftleftlvert sinleft(ipi/Nright) X_1rightvertgtvarepsilon s_Nrightright]leqslantmathbb Eleft[X_1^2sin^2left(ipi/Nright)mathbf 1leftleftlvert X_1rightvertgtvarepsilon s_N rightright].$$
answered Aug 13 at 12:25
Davide Giraudo
121k15147250
answered Aug 13 at 12:25
Davide Giraudo
121k15147250
answered Aug 13 at 12:25
Davide Giraudo
121k15147250
answered Aug 13 at 12:25
Davide Giraudo
121k15147250
121k15147250
I see, I was not familiar with the Lindeberg condition for CLT. So if I have understood you correct, I use the Lindeberg condition on X_n,N to find that U_N will converge to a standard normal distribution for N->infinity
– Jonathan Kiersch
Aug 13 at 12:44
Yes, the limit will be a normal distribution but I have not looked whether the variance is one or not.
– Davide Giraudo
Aug 13 at 13:20
@DavideGiraudo - the limiting variance is $frac12$ (it would have been $1$ without the sine term)
– Henry
Aug 13 at 14:04
add a comment |Â
I see, I was not familiar with the Lindeberg condition for CLT. So if I have understood you correct, I use the Lindeberg condition on X_n,N to find that U_N will converge to a standard normal distribution for N->infinity
– Jonathan Kiersch
Aug 13 at 12:44
Yes, the limit will be a normal distribution but I have not looked whether the variance is one or not.
– Davide Giraudo
Aug 13 at 13:20
@DavideGiraudo - the limiting variance is $frac12$ (it would have been $1$ without the sine term)
– Henry
Aug 13 at 14:04
I see, I was not familiar with the Lindeberg condition for CLT. So if I have understood you correct, I use the Lindeberg condition on X_n,N to find that U_N will converge to a standard normal distribution for N->infinity
– Jonathan Kiersch
Aug 13 at 12:44
I see, I was not familiar with the Lindeberg condition for CLT. So if I have understood you correct, I use the Lindeberg condition on X_n,N to find that U_N will converge to a standard normal distribution for N->infinity
– Jonathan Kiersch
Aug 13 at 12:44
Yes, the limit will be a normal distribution but I have not looked whether the variance is one or not.
– Davide Giraudo
Aug 13 at 13:20
Yes, the limit will be a normal distribution but I have not looked whether the variance is one or not.
– Davide Giraudo
Aug 13 at 13:20
@DavideGiraudo - the limiting variance is $frac12$ (it would have been $1$ without the sine term)
– Henry
Aug 13 at 14:04
@DavideGiraudo - the limiting variance is $frac12$ (it would have been $1$ without the sine term)
– Henry
Aug 13 at 14:04
add a comment |Â
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$E[U_N] = 0$. Do you know what $Var(U_N)$ is or at least what it converges to?
– Henry
Aug 13 at 9:37
Var(U_N) should be 1/Nsum(sin(npi)/N), which also converges to zero for N->infty
– Jonathan Kiersch
Aug 13 at 9:45
I suspect $Var(U_N) to intlimits_0^1 sin^2(pi x), dx = frac12$
– Henry
Aug 13 at 11:10
How did you find that?
– Jonathan Kiersch
Aug 13 at 11:57
A combination of thinking that without the $sin(fracnpiN)$ term the variance would be $1$ as in the central limit theorem, plus simulation and basic calculus.
– Henry
Aug 13 at 12:01