Mixing
Asymptotic Statistics: Convergence in distribution to a tight CDF implies convergence in probability to a...
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I'm taking a course in asymptotic statistics, and I'm having trouble with how to start proving the following statement. I learn best from examples, rather than just reading the theorems, however I can't find an example that seems similar. Any and all help is appreciated!
$$
left.
begin{matrix}
k_n left( Y_n -c right) xrightarrow{D} H \
k_n to infty
end{matrix}
right} implies Y_n xrightarrow{P} c
$$
There is the additional hint: You may assume without loss of generality that $H$ is the cumulative distribution function of a random variable $V$ which is tight meaning that, given any $epsilon > 0$, there exists a finite number $M$ such that $P(|V | leq M) geq 1 − epsilon/2$.
When I look at this, I instinctively think of the Central Limit Theorem with $k_n = sqrt{n}$, $Y_n = bar{X}_n$, $c=mu$ and $H$ the CDF some normally distributed r.v. $V$. I understand the concept of tightness which implies that none of the PMF escapes to infinity, so that makes me think I'm incorrect in assuming $V$ is normal.
statistics asymptotics weak-convergence
asked Jan 24 at 11:57
StephanieStephanie
83
$endgroup$
add a comment |
$begingroup$
I'm taking a course in asymptotic statistics, and I'm having trouble with how to start proving the following statement. I learn best from examples, rather than just reading the theorems, however I can't find an example that seems similar. Any and all help is appreciated!
$$
left.
begin{matrix}
k_n left( Y_n -c right) xrightarrow{D} H \
k_n to infty
end{matrix}
right} implies Y_n xrightarrow{P} c
$$
There is the additional hint: You may assume without loss of generality that $H$ is the cumulative distribution function of a random variable $V$ which is tight meaning that, given any $epsilon > 0$, there exists a finite number $M$ such that $P(|V | leq M) geq 1 − epsilon/2$.
When I look at this, I instinctively think of the Central Limit Theorem with $k_n = sqrt{n}$, $Y_n = bar{X}_n$, $c=mu$ and $H$ the CDF some normally distributed r.v. $V$. I understand the concept of tightness which implies that none of the PMF escapes to infinity, so that makes me think I'm incorrect in assuming $V$ is normal.
statistics asymptotics weak-convergence
asked Jan 24 at 11:57
StephanieStephanie
83
$endgroup$
add a comment |
$begingroup$
I'm taking a course in asymptotic statistics, and I'm having trouble with how to start proving the following statement. I learn best from examples, rather than just reading the theorems, however I can't find an example that seems similar. Any and all help is appreciated!
$$
left.
begin{matrix}
k_n left( Y_n -c right) xrightarrow{D} H \
k_n to infty
end{matrix}
right} implies Y_n xrightarrow{P} c
$$
There is the additional hint: You may assume without loss of generality that $H$ is the cumulative distribution function of a random variable $V$ which is tight meaning that, given any $epsilon > 0$, there exists a finite number $M$ such that $P(|V | leq M) geq 1 − epsilon/2$.
When I look at this, I instinctively think of the Central Limit Theorem with $k_n = sqrt{n}$, $Y_n = bar{X}_n$, $c=mu$ and $H$ the CDF some normally distributed r.v. $V$. I understand the concept of tightness which implies that none of the PMF escapes to infinity, so that makes me think I'm incorrect in assuming $V$ is normal.
statistics asymptotics weak-convergence
asked Jan 24 at 11:57
StephanieStephanie
83
$endgroup$
I'm taking a course in asymptotic statistics, and I'm having trouble with how to start proving the following statement. I learn best from examples, rather than just reading the theorems, however I can't find an example that seems similar. Any and all help is appreciated!
$$
left.
begin{matrix}
k_n left( Y_n -c right) xrightarrow{D} H \
k_n to infty
end{matrix}
right} implies Y_n xrightarrow{P} c
$$
There is the additional hint: You may assume without loss of generality that $H$ is the cumulative distribution function of a random variable $V$ which is tight meaning that, given any $epsilon > 0$, there exists a finite number $M$ such that $P(|V | leq M) geq 1 − epsilon/2$.
When I look at this, I instinctively think of the Central Limit Theorem with $k_n = sqrt{n}$, $Y_n = bar{X}_n$, $c=mu$ and $H$ the CDF some normally distributed r.v. $V$. I understand the concept of tightness which implies that none of the PMF escapes to infinity, so that makes me think I'm incorrect in assuming $V$ is normal.
statistics asymptotics weak-convergence
statistics asymptotics weak-convergence
asked Jan 24 at 11:57
StephanieStephanie
83
asked Jan 24 at 11:57
StephanieStephanie
83
asked Jan 24 at 11:57
StephanieStephanie
83
asked Jan 24 at 11:57
StephanieStephanie
83
asked Jan 24 at 11:57
StephanieStephanie
83
83
add a comment |
add a comment |
1 Answer
1
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oldest
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Let $epsilon >0$. Let $M$ be as given in the hint. Without loss of generality we may assume that $H$ is continuous at $pm M$. If $k_n >frac M {epsilon}$ then $P{|Y_n-c| >epsilon} =P{|k_n(Y_n-c)| >k_n epsilon}leq P{|k_n(Y_n-c)| >M} to P{|V| >M} <epsilon /2$.
answered Jan 24 at 12:11
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
$endgroup$
$begingroup$
Thanks for your help Kavi! The $to$ implication, does that stem from: as $epsilon$ is arbitrary and $M$ is fixed and finite, $epsilonto 0$ as $k_n to infty$? Or is it an implication of the convergence in distribution? Usually that requires some type of $n$ involved in the terms, is that right? Should $epsilon$ be defined as some type of inverse on $n$ to achieve this?
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– Stephanie
Jan 24 at 13:28
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $epsilon >0$. Let $M$ be as given in the hint. Without loss of generality we may assume that $H$ is continuous at $pm M$. If $k_n >frac M {epsilon}$ then $P{|Y_n-c| >epsilon} =P{|k_n(Y_n-c)| >k_n epsilon}leq P{|k_n(Y_n-c)| >M} to P{|V| >M} <epsilon /2$.
answered Jan 24 at 12:11
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
$endgroup$
$begingroup$
Thanks for your help Kavi! The $to$ implication, does that stem from: as $epsilon$ is arbitrary and $M$ is fixed and finite, $epsilonto 0$ as $k_n to infty$? Or is it an implication of the convergence in distribution? Usually that requires some type of $n$ involved in the terms, is that right? Should $epsilon$ be defined as some type of inverse on $n$ to achieve this?
$endgroup$
– Stephanie
Jan 24 at 13:28
add a comment |
$begingroup$
Let $epsilon >0$. Let $M$ be as given in the hint. Without loss of generality we may assume that $H$ is continuous at $pm M$. If $k_n >frac M {epsilon}$ then $P{|Y_n-c| >epsilon} =P{|k_n(Y_n-c)| >k_n epsilon}leq P{|k_n(Y_n-c)| >M} to P{|V| >M} <epsilon /2$.
answered Jan 24 at 12:11
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
$endgroup$
$begingroup$
Thanks for your help Kavi! The $to$ implication, does that stem from: as $epsilon$ is arbitrary and $M$ is fixed and finite, $epsilonto 0$ as $k_n to infty$? Or is it an implication of the convergence in distribution? Usually that requires some type of $n$ involved in the terms, is that right? Should $epsilon$ be defined as some type of inverse on $n$ to achieve this?
$endgroup$
– Stephanie
Jan 24 at 13:28
add a comment |
$begingroup$
Let $epsilon >0$. Let $M$ be as given in the hint. Without loss of generality we may assume that $H$ is continuous at $pm M$. If $k_n >frac M {epsilon}$ then $P{|Y_n-c| >epsilon} =P{|k_n(Y_n-c)| >k_n epsilon}leq P{|k_n(Y_n-c)| >M} to P{|V| >M} <epsilon /2$.
answered Jan 24 at 12:11
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
$endgroup$
Let $epsilon >0$. Let $M$ be as given in the hint. Without loss of generality we may assume that $H$ is continuous at $pm M$. If $k_n >frac M {epsilon}$ then $P{|Y_n-c| >epsilon} =P{|k_n(Y_n-c)| >k_n epsilon}leq P{|k_n(Y_n-c)| >M} to P{|V| >M} <epsilon /2$.
answered Jan 24 at 12:11
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
answered Jan 24 at 12:11
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
answered Jan 24 at 12:11
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
answered Jan 24 at 12:11
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
67.6k53067
$begingroup$
Thanks for your help Kavi! The $to$ implication, does that stem from: as $epsilon$ is arbitrary and $M$ is fixed and finite, $epsilonto 0$ as $k_n to infty$? Or is it an implication of the convergence in distribution? Usually that requires some type of $n$ involved in the terms, is that right? Should $epsilon$ be defined as some type of inverse on $n$ to achieve this?
$endgroup$
– Stephanie
Jan 24 at 13:28
add a comment |
$begingroup$
Thanks for your help Kavi! The $to$ implication, does that stem from: as $epsilon$ is arbitrary and $M$ is fixed and finite, $epsilonto 0$ as $k_n to infty$? Or is it an implication of the convergence in distribution? Usually that requires some type of $n$ involved in the terms, is that right? Should $epsilon$ be defined as some type of inverse on $n$ to achieve this?
$endgroup$
– Stephanie
Jan 24 at 13:28
$begingroup$
Thanks for your help Kavi! The $to$ implication, does that stem from: as $epsilon$ is arbitrary and $M$ is fixed and finite, $epsilonto 0$ as $k_n to infty$? Or is it an implication of the convergence in distribution? Usually that requires some type of $n$ involved in the terms, is that right? Should $epsilon$ be defined as some type of inverse on $n$ to achieve this?
$endgroup$
– Stephanie
Jan 24 at 13:28
$begingroup$
Thanks for your help Kavi! The $to$ implication, does that stem from: as $epsilon$ is arbitrary and $M$ is fixed and finite, $epsilonto 0$ as $k_n to infty$? Or is it an implication of the convergence in distribution? Usually that requires some type of $n$ involved in the terms, is that right? Should $epsilon$ be defined as some type of inverse on $n$ to achieve this?
$endgroup$
– Stephanie
Jan 24 at 13:28
add a comment |
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