Mixing
Inverse Gaussian Distribution and the Central Limit Theorem
$begingroup$
Let the random variables $Y_1,ldots,Y_n$ be independent and identically distributed (i.i.d.) (standard) Inverse Gaussian random variables with parameters $mu$ and $lambda$.
Then, let the random variables $tilde{A}$ and $A$ be given, which are defined as follows:
$A = sqrt n cdot tilde{A} = sqrt n left(frac{1}{n} cdot sum_{i=1}^n left[e^{b_i Y_i}right] - thetaright)$ ($b_i in mathbb{R}$ ($i=1,ldots,n$) are constants (with $b_i neq b_j$ for any $i neq j$) and $theta$ is a parameter).
Goal: Determine what ultimately follows if one uses the (Lindeberg-Lévy) Central Limit Theorem on $e^{b_i Y_i}$.
The CLT informs us that $sqrt n (overline{Y} - E(Y_i)) xrightarrow{D} mathcal{N}(0,Var(Y_i))$, i.e. $sqrt n left(frac{1}{n} cdot sum_{i=1}^n [Y_i] - mu right) xrightarrow{D} mathcal{N}(0,frac{mu^3}{lambda})$.
It seems to me that it follows that $sqrt n left(frac{1}{n} cdot sum_{i=1}^n left[e^{b_i Y_i}right]- e^{b_i mu}right) xrightarrow{D} ldots$.
Question: How to proceed from here (i.e. how to determine the limit distribution of $sqrt n cdot tilde{A}$)?
probability-theory central-limit-theorem probability-limit-theorems
asked Jan 28 at 19:54
AnnaAnna
7619
$endgroup$
add a comment |
$begingroup$
Let the random variables $Y_1,ldots,Y_n$ be independent and identically distributed (i.i.d.) (standard) Inverse Gaussian random variables with parameters $mu$ and $lambda$.
Then, let the random variables $tilde{A}$ and $A$ be given, which are defined as follows:
$A = sqrt n cdot tilde{A} = sqrt n left(frac{1}{n} cdot sum_{i=1}^n left[e^{b_i Y_i}right] - thetaright)$ ($b_i in mathbb{R}$ ($i=1,ldots,n$) are constants (with $b_i neq b_j$ for any $i neq j$) and $theta$ is a parameter).
Goal: Determine what ultimately follows if one uses the (Lindeberg-Lévy) Central Limit Theorem on $e^{b_i Y_i}$.
The CLT informs us that $sqrt n (overline{Y} - E(Y_i)) xrightarrow{D} mathcal{N}(0,Var(Y_i))$, i.e. $sqrt n left(frac{1}{n} cdot sum_{i=1}^n [Y_i] - mu right) xrightarrow{D} mathcal{N}(0,frac{mu^3}{lambda})$.
It seems to me that it follows that $sqrt n left(frac{1}{n} cdot sum_{i=1}^n left[e^{b_i Y_i}right]- e^{b_i mu}right) xrightarrow{D} ldots$.
Question: How to proceed from here (i.e. how to determine the limit distribution of $sqrt n cdot tilde{A}$)?
probability-theory central-limit-theorem probability-limit-theorems
asked Jan 28 at 19:54
AnnaAnna
7619
$endgroup$
add a comment |
$begingroup$
Let the random variables $Y_1,ldots,Y_n$ be independent and identically distributed (i.i.d.) (standard) Inverse Gaussian random variables with parameters $mu$ and $lambda$.
Then, let the random variables $tilde{A}$ and $A$ be given, which are defined as follows:
$A = sqrt n cdot tilde{A} = sqrt n left(frac{1}{n} cdot sum_{i=1}^n left[e^{b_i Y_i}right] - thetaright)$ ($b_i in mathbb{R}$ ($i=1,ldots,n$) are constants (with $b_i neq b_j$ for any $i neq j$) and $theta$ is a parameter).
Goal: Determine what ultimately follows if one uses the (Lindeberg-Lévy) Central Limit Theorem on $e^{b_i Y_i}$.
The CLT informs us that $sqrt n (overline{Y} - E(Y_i)) xrightarrow{D} mathcal{N}(0,Var(Y_i))$, i.e. $sqrt n left(frac{1}{n} cdot sum_{i=1}^n [Y_i] - mu right) xrightarrow{D} mathcal{N}(0,frac{mu^3}{lambda})$.
It seems to me that it follows that $sqrt n left(frac{1}{n} cdot sum_{i=1}^n left[e^{b_i Y_i}right]- e^{b_i mu}right) xrightarrow{D} ldots$.
Question: How to proceed from here (i.e. how to determine the limit distribution of $sqrt n cdot tilde{A}$)?
probability-theory central-limit-theorem probability-limit-theorems
asked Jan 28 at 19:54
AnnaAnna
7619
$endgroup$
Let the random variables $Y_1,ldots,Y_n$ be independent and identically distributed (i.i.d.) (standard) Inverse Gaussian random variables with parameters $mu$ and $lambda$.
Then, let the random variables $tilde{A}$ and $A$ be given, which are defined as follows:
$A = sqrt n cdot tilde{A} = sqrt n left(frac{1}{n} cdot sum_{i=1}^n left[e^{b_i Y_i}right] - thetaright)$ ($b_i in mathbb{R}$ ($i=1,ldots,n$) are constants (with $b_i neq b_j$ for any $i neq j$) and $theta$ is a parameter).
Goal: Determine what ultimately follows if one uses the (Lindeberg-Lévy) Central Limit Theorem on $e^{b_i Y_i}$.
The CLT informs us that $sqrt n (overline{Y} - E(Y_i)) xrightarrow{D} mathcal{N}(0,Var(Y_i))$, i.e. $sqrt n left(frac{1}{n} cdot sum_{i=1}^n [Y_i] - mu right) xrightarrow{D} mathcal{N}(0,frac{mu^3}{lambda})$.
It seems to me that it follows that $sqrt n left(frac{1}{n} cdot sum_{i=1}^n left[e^{b_i Y_i}right]- e^{b_i mu}right) xrightarrow{D} ldots$.
Question: How to proceed from here (i.e. how to determine the limit distribution of $sqrt n cdot tilde{A}$)?
probability-theory central-limit-theorem probability-limit-theorems
probability-theory central-limit-theorem probability-limit-theorems
asked Jan 28 at 19:54
AnnaAnna
7619
asked Jan 28 at 19:54
AnnaAnna
7619
asked Jan 28 at 19:54
AnnaAnna
7619
asked Jan 28 at 19:54
AnnaAnna
7619
asked Jan 28 at 19:54
AnnaAnna
7619
7619
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You may consider the random variable
begin{align}
frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu})
end{align}
where
$s_n^2 = sum_{i=1}^n sigma_i^2$ and $sigma_i$ is the variance of $e^{b_iY_i}$. Then, you can check whether the Lindeberg's condition is satisfied. If it is satisfied, using the Lindeberg CLT
begin{align}
frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu}) xrightarrow{D} mathcal{N}(0,1)
end{align}
as $ntoinfty$.
answered Jan 31 at 15:19
user52227user52227
1,038512
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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votes
$begingroup$
You may consider the random variable
begin{align}
frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu})
end{align}
where
$s_n^2 = sum_{i=1}^n sigma_i^2$ and $sigma_i$ is the variance of $e^{b_iY_i}$. Then, you can check whether the Lindeberg's condition is satisfied. If it is satisfied, using the Lindeberg CLT
begin{align}
frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu}) xrightarrow{D} mathcal{N}(0,1)
end{align}
as $ntoinfty$.
answered Jan 31 at 15:19
user52227user52227
1,038512
$endgroup$
add a comment |
$begingroup$
You may consider the random variable
begin{align}
frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu})
end{align}
where
$s_n^2 = sum_{i=1}^n sigma_i^2$ and $sigma_i$ is the variance of $e^{b_iY_i}$. Then, you can check whether the Lindeberg's condition is satisfied. If it is satisfied, using the Lindeberg CLT
begin{align}
frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu}) xrightarrow{D} mathcal{N}(0,1)
end{align}
as $ntoinfty$.
answered Jan 31 at 15:19
user52227user52227
1,038512
$endgroup$
add a comment |
$begingroup$
You may consider the random variable
begin{align}
frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu})
end{align}
where
$s_n^2 = sum_{i=1}^n sigma_i^2$ and $sigma_i$ is the variance of $e^{b_iY_i}$. Then, you can check whether the Lindeberg's condition is satisfied. If it is satisfied, using the Lindeberg CLT
begin{align}
frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu}) xrightarrow{D} mathcal{N}(0,1)
end{align}
as $ntoinfty$.
answered Jan 31 at 15:19
user52227user52227
1,038512
$endgroup$
You may consider the random variable
begin{align}
frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu})
end{align}
where
$s_n^2 = sum_{i=1}^n sigma_i^2$ and $sigma_i$ is the variance of $e^{b_iY_i}$. Then, you can check whether the Lindeberg's condition is satisfied. If it is satisfied, using the Lindeberg CLT
begin{align}
frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu}) xrightarrow{D} mathcal{N}(0,1)
end{align}
as $ntoinfty$.
answered Jan 31 at 15:19
user52227user52227
1,038512
answered Jan 31 at 15:19
user52227user52227
1,038512
answered Jan 31 at 15:19
user52227user52227
1,038512
answered Jan 31 at 15:19
user52227user52227
1,038512
1,038512
add a comment |
add a comment |
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