Mixing
The asympotic distribution of a sequence of estimators
$begingroup$
Here's the question
Consider a sequence of estimators $X_{i}=X_{1},X_{2},...,X_{N}$ for $i=1,...,N$.
Each estimator is resulting from a sample $j=1,...,n$.
For each estimator $X_{i}$ the asymptotic normality is satisfied, as $n rightarrow infty$ with $N(0,X_{i}^2)$
and $σ^2=X_{i}^2$. Can we say that the asymptotic normality of the sequence $X_{i}$ is satisfied?
asymptotics
asked Jan 14 at 6:05
Konstantinos GkKonstantinos Gk
11
$endgroup$
add a comment |
$begingroup$
Here's the question
Consider a sequence of estimators $X_{i}=X_{1},X_{2},...,X_{N}$ for $i=1,...,N$.
Each estimator is resulting from a sample $j=1,...,n$.
For each estimator $X_{i}$ the asymptotic normality is satisfied, as $n rightarrow infty$ with $N(0,X_{i}^2)$
and $σ^2=X_{i}^2$. Can we say that the asymptotic normality of the sequence $X_{i}$ is satisfied?
asymptotics
asked Jan 14 at 6:05
Konstantinos GkKonstantinos Gk
11
$endgroup$
$begingroup$
"asymptotic normality is satisfactory" means "asymptotic normality is satisfied/verified" ? ALso, $rightarrow N(0,X_{i}^2)$ makes no sense, you meant $rightarrow N(0,sigma_{i}^2)$ ? Also what is $X$ ?
$endgroup$
– leonbloy
Jan 14 at 14:57
$begingroup$
Thank you for the comments. I made some changes to my question.
$endgroup$
– Konstantinos Gk
Jan 22 at 23:40
add a comment |
$begingroup$
Here's the question
Consider a sequence of estimators $X_{i}=X_{1},X_{2},...,X_{N}$ for $i=1,...,N$.
Each estimator is resulting from a sample $j=1,...,n$.
For each estimator $X_{i}$ the asymptotic normality is satisfied, as $n rightarrow infty$ with $N(0,X_{i}^2)$
and $σ^2=X_{i}^2$. Can we say that the asymptotic normality of the sequence $X_{i}$ is satisfied?
asymptotics
asked Jan 14 at 6:05
Konstantinos GkKonstantinos Gk
11
$endgroup$
Here's the question
Consider a sequence of estimators $X_{i}=X_{1},X_{2},...,X_{N}$ for $i=1,...,N$.
Each estimator is resulting from a sample $j=1,...,n$.
For each estimator $X_{i}$ the asymptotic normality is satisfied, as $n rightarrow infty$ with $N(0,X_{i}^2)$
and $σ^2=X_{i}^2$. Can we say that the asymptotic normality of the sequence $X_{i}$ is satisfied?
asymptotics
asymptotics
asked Jan 14 at 6:05
Konstantinos GkKonstantinos Gk
11
asked Jan 14 at 6:05
Konstantinos GkKonstantinos Gk
11
edited Jan 22 at 23:43
Konstantinos Gk
asked Jan 14 at 6:05
Konstantinos GkKonstantinos Gk
11
asked Jan 14 at 6:05
Konstantinos GkKonstantinos Gk
11
asked Jan 14 at 6:05
Konstantinos GkKonstantinos Gk
11
11
$begingroup$
"asymptotic normality is satisfactory" means "asymptotic normality is satisfied/verified" ? ALso, $rightarrow N(0,X_{i}^2)$ makes no sense, you meant $rightarrow N(0,sigma_{i}^2)$ ? Also what is $X$ ?
$endgroup$
– leonbloy
Jan 14 at 14:57
$begingroup$
Thank you for the comments. I made some changes to my question.
$endgroup$
– Konstantinos Gk
Jan 22 at 23:40
add a comment |
$begingroup$
"asymptotic normality is satisfactory" means "asymptotic normality is satisfied/verified" ? ALso, $rightarrow N(0,X_{i}^2)$ makes no sense, you meant $rightarrow N(0,sigma_{i}^2)$ ? Also what is $X$ ?
$endgroup$
– leonbloy
Jan 14 at 14:57
$begingroup$
Thank you for the comments. I made some changes to my question.
$endgroup$
– Konstantinos Gk
Jan 22 at 23:40
$begingroup$
"asymptotic normality is satisfactory" means "asymptotic normality is satisfied/verified" ? ALso, $rightarrow N(0,X_{i}^2)$ makes no sense, you meant $rightarrow N(0,sigma_{i}^2)$ ? Also what is $X$ ?
$endgroup$
– leonbloy
Jan 14 at 14:57
$begingroup$
"asymptotic normality is satisfactory" means "asymptotic normality is satisfied/verified" ? ALso, $rightarrow N(0,X_{i}^2)$ makes no sense, you meant $rightarrow N(0,sigma_{i}^2)$ ? Also what is $X$ ?
$endgroup$
– leonbloy
Jan 14 at 14:57
$begingroup$
Thank you for the comments. I made some changes to my question.
$endgroup$
– Konstantinos Gk
Jan 22 at 23:40
$begingroup$
Thank you for the comments. I made some changes to my question.
$endgroup$
– Konstantinos Gk
Jan 22 at 23:40
add a comment |
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$begingroup$
"asymptotic normality is satisfactory" means "asymptotic normality is satisfied/verified" ? ALso, $rightarrow N(0,X_{i}^2)$ makes no sense, you meant $rightarrow N(0,sigma_{i}^2)$ ? Also what is $X$ ?
$endgroup$
– leonbloy
Jan 14 at 14:57
$begingroup$
Thank you for the comments. I made some changes to my question.
$endgroup$
– Konstantinos Gk
Jan 22 at 23:40