A central limit theorem for dependent random variable.
$begingroup$
Suppose that $u_{j}$ is a sequence of iid standard Gaussian random variable, i.e.
$$
u_jstackrel{d}{=}text{N}(0,1).
$$
Call $mu_r=mathbb{E}[|u_j|^r]$. I need to find the asymptotic distribution of the random variable
$$
sqrt{n},left(frac{1}{n}sum_{j=1}^{n}u_j^2-frac{mu_1^{-2}}{n}sum_{j=1}^{n}|u_j|,|u_{j+1}|right)=n^{-1/2},sum_{j=1}^n|u_j|left(|u_j|-mu_1^{-2},|u_{j+1}|right)=n^{-1/2},sum_{j=1}^nxi_j,
$$
where $xi_j=|u_j|left(|u_j|-mu_1^{-2},|u_{j+1}|right)$. However, since the $xi_j$ are dependent, I do not know how to proceed.
probability-distributions central-limit-theorem
asked Jan 2 at 15:05
AlmostSureUserAlmostSureUser
326418
$endgroup$
add a comment |
$begingroup$
Suppose that $u_{j}$ is a sequence of iid standard Gaussian random variable, i.e.
$$
u_jstackrel{d}{=}text{N}(0,1).
$$
Call $mu_r=mathbb{E}[|u_j|^r]$. I need to find the asymptotic distribution of the random variable
$$
sqrt{n},left(frac{1}{n}sum_{j=1}^{n}u_j^2-frac{mu_1^{-2}}{n}sum_{j=1}^{n}|u_j|,|u_{j+1}|right)=n^{-1/2},sum_{j=1}^n|u_j|left(|u_j|-mu_1^{-2},|u_{j+1}|right)=n^{-1/2},sum_{j=1}^nxi_j,
$$
where $xi_j=|u_j|left(|u_j|-mu_1^{-2},|u_{j+1}|right)$. However, since the $xi_j$ are dependent, I do not know how to proceed.
probability-distributions central-limit-theorem
asked Jan 2 at 15:05
AlmostSureUserAlmostSureUser
326418
$endgroup$
add a comment |
$begingroup$
Suppose that $u_{j}$ is a sequence of iid standard Gaussian random variable, i.e.
$$
u_jstackrel{d}{=}text{N}(0,1).
$$
Call $mu_r=mathbb{E}[|u_j|^r]$. I need to find the asymptotic distribution of the random variable
$$
sqrt{n},left(frac{1}{n}sum_{j=1}^{n}u_j^2-frac{mu_1^{-2}}{n}sum_{j=1}^{n}|u_j|,|u_{j+1}|right)=n^{-1/2},sum_{j=1}^n|u_j|left(|u_j|-mu_1^{-2},|u_{j+1}|right)=n^{-1/2},sum_{j=1}^nxi_j,
$$
where $xi_j=|u_j|left(|u_j|-mu_1^{-2},|u_{j+1}|right)$. However, since the $xi_j$ are dependent, I do not know how to proceed.
probability-distributions central-limit-theorem
asked Jan 2 at 15:05
AlmostSureUserAlmostSureUser
326418
$endgroup$
Suppose that $u_{j}$ is a sequence of iid standard Gaussian random variable, i.e.
$$
u_jstackrel{d}{=}text{N}(0,1).
$$
Call $mu_r=mathbb{E}[|u_j|^r]$. I need to find the asymptotic distribution of the random variable
$$
sqrt{n},left(frac{1}{n}sum_{j=1}^{n}u_j^2-frac{mu_1^{-2}}{n}sum_{j=1}^{n}|u_j|,|u_{j+1}|right)=n^{-1/2},sum_{j=1}^n|u_j|left(|u_j|-mu_1^{-2},|u_{j+1}|right)=n^{-1/2},sum_{j=1}^nxi_j,
$$
where $xi_j=|u_j|left(|u_j|-mu_1^{-2},|u_{j+1}|right)$. However, since the $xi_j$ are dependent, I do not know how to proceed.
probability-distributions central-limit-theorem
probability-distributions central-limit-theorem
asked Jan 2 at 15:05
AlmostSureUserAlmostSureUser
326418
asked Jan 2 at 15:05
AlmostSureUserAlmostSureUser
326418
edited Jan 2 at 16:24
AlmostSureUser
asked Jan 2 at 15:05
AlmostSureUserAlmostSureUser
326418
asked Jan 2 at 15:05
AlmostSureUserAlmostSureUser
326418
asked Jan 2 at 15:05
AlmostSureUserAlmostSureUser
326418
326418
add a comment |
add a comment |
1 Answer
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$begingroup$
Since $(xi_j)_{jle N}$ and $(xi_j)_{jge N+k}$ are independent for $kge 2$, the stochastic process $(xi_j)_{jin mathbb{N}}$ is strongly($alpha$-) mixing. More precisely, we can say that the strong mixing coefficient $alpha(k)$ vanishes for $kgeq 2$. Note that $Exi_j =0$ and every $p$-th moment $E|xi_j|^p$ is finite. By $alpha$-mixing central limit theorem, we have
$$
frac{1}{sqrt{n}}sum_{j=1}^n xi_j to_d mathcal{N}(0,sigma^2)
$$
where $sigma^2 = E[xi_j^2] +2 E[xi_j xi_{j+1}]ge 0$.
answered Jan 2 at 17:39
SongSong
18.5k21651
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Since $(xi_j)_{jle N}$ and $(xi_j)_{jge N+k}$ are independent for $kge 2$, the stochastic process $(xi_j)_{jin mathbb{N}}$ is strongly($alpha$-) mixing. More precisely, we can say that the strong mixing coefficient $alpha(k)$ vanishes for $kgeq 2$. Note that $Exi_j =0$ and every $p$-th moment $E|xi_j|^p$ is finite. By $alpha$-mixing central limit theorem, we have
$$
frac{1}{sqrt{n}}sum_{j=1}^n xi_j to_d mathcal{N}(0,sigma^2)
$$
where $sigma^2 = E[xi_j^2] +2 E[xi_j xi_{j+1}]ge 0$.
answered Jan 2 at 17:39
SongSong
18.5k21651
$endgroup$
add a comment |
$begingroup$
Since $(xi_j)_{jle N}$ and $(xi_j)_{jge N+k}$ are independent for $kge 2$, the stochastic process $(xi_j)_{jin mathbb{N}}$ is strongly($alpha$-) mixing. More precisely, we can say that the strong mixing coefficient $alpha(k)$ vanishes for $kgeq 2$. Note that $Exi_j =0$ and every $p$-th moment $E|xi_j|^p$ is finite. By $alpha$-mixing central limit theorem, we have
$$
frac{1}{sqrt{n}}sum_{j=1}^n xi_j to_d mathcal{N}(0,sigma^2)
$$
where $sigma^2 = E[xi_j^2] +2 E[xi_j xi_{j+1}]ge 0$.
answered Jan 2 at 17:39
SongSong
18.5k21651
$endgroup$
add a comment |
$begingroup$
Since $(xi_j)_{jle N}$ and $(xi_j)_{jge N+k}$ are independent for $kge 2$, the stochastic process $(xi_j)_{jin mathbb{N}}$ is strongly($alpha$-) mixing. More precisely, we can say that the strong mixing coefficient $alpha(k)$ vanishes for $kgeq 2$. Note that $Exi_j =0$ and every $p$-th moment $E|xi_j|^p$ is finite. By $alpha$-mixing central limit theorem, we have
$$
frac{1}{sqrt{n}}sum_{j=1}^n xi_j to_d mathcal{N}(0,sigma^2)
$$
where $sigma^2 = E[xi_j^2] +2 E[xi_j xi_{j+1}]ge 0$.
answered Jan 2 at 17:39
SongSong
18.5k21651
$endgroup$
Since $(xi_j)_{jle N}$ and $(xi_j)_{jge N+k}$ are independent for $kge 2$, the stochastic process $(xi_j)_{jin mathbb{N}}$ is strongly($alpha$-) mixing. More precisely, we can say that the strong mixing coefficient $alpha(k)$ vanishes for $kgeq 2$. Note that $Exi_j =0$ and every $p$-th moment $E|xi_j|^p$ is finite. By $alpha$-mixing central limit theorem, we have
$$
frac{1}{sqrt{n}}sum_{j=1}^n xi_j to_d mathcal{N}(0,sigma^2)
$$
where $sigma^2 = E[xi_j^2] +2 E[xi_j xi_{j+1}]ge 0$.
answered Jan 2 at 17:39
SongSong
18.5k21651
answered Jan 2 at 17:39
SongSong
18.5k21651
answered Jan 2 at 17:39
SongSong
18.5k21651
answered Jan 2 at 17:39
SongSong
18.5k21651
18.5k21651
add a comment |
add a comment |
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