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A Trivial Diagnosis

a transformation $g$ such that $g(S^2)$ has asymptotic distribution that depends on $beta_2$

Variance stabilizing transformationCentral Limit Theorem Application on Multivariate Normalconvergence to standard normal distributionDetermine the asymptotic distribution of $bar X_n$, properly centered and $sqrt n$ scaledExplain why a gamma random variable with parameters $(t, lambda)$ has an approximately normal distribution when $t$ is large.Limiting Distribution with Finite 4th MomentFind the asymptotic distribution of the MME and MLE.Multivariate central limit theorem and coefficient of variationApplying central limit theorem to show convergence in distributionMy simulation of the Central Limit Theorem does not converge to correct value

1

1

$begingroup$

Let $X_1, X_2,dots,X_n$ be i.i.d. RVs with $E|X_1|^4 < infty$. Let $var(X_1) = sigma^2$, $beta_2 = mu_4/sigma^4$.

(a) Using CLT for i.i.d. RVs, show that $sqrt{n}(S^2-sigma^2)rightarrow_L N(0, mu_4-sigma^4)$.

(b) Find a transformation $g$ such that $g(S^2)$ has asymptotic distribution that depends on $beta_2$ alone, not on $sigma^2$.

I have completed part (a). But got stuck on finding $g$. I know I have to use the theorem:

$Y_n$ is $AN(mu, sigma^2_n)$, with $sigma^2_nrightarrow 0$ and $mu$ fixed real. $g$ be differentiable at $mu$ with $g'(mu)neq 0$, then $g(Y_n)$ is $AN(g(mu), [g'(mu)]^2sigma^2_n)$

Any help appreciated. Thanks.

($AN(cdot)$ means asymptotically normal distribution.)

statistics central-limit-theorem

share|cite|improve this question

edited Mar 13 at 9:19

Cettt

1,888622

asked Mar 13 at 8:12

Stat_prob_001Stat_prob_001

333113

$endgroup$

add a comment | 

1

1

$begingroup$

Let $X_1, X_2,dots,X_n$ be i.i.d. RVs with $E|X_1|^4 < infty$. Let $var(X_1) = sigma^2$, $beta_2 = mu_4/sigma^4$.

(a) Using CLT for i.i.d. RVs, show that $sqrt{n}(S^2-sigma^2)rightarrow_L N(0, mu_4-sigma^4)$.

(b) Find a transformation $g$ such that $g(S^2)$ has asymptotic distribution that depends on $beta_2$ alone, not on $sigma^2$.

I have completed part (a). But got stuck on finding $g$. I know I have to use the theorem:

$Y_n$ is $AN(mu, sigma^2_n)$, with $sigma^2_nrightarrow 0$ and $mu$ fixed real. $g$ be differentiable at $mu$ with $g'(mu)neq 0$, then $g(Y_n)$ is $AN(g(mu), [g'(mu)]^2sigma^2_n)$

Any help appreciated. Thanks.

($AN(cdot)$ means asymptotically normal distribution.)

statistics central-limit-theorem

share|cite|improve this question

edited Mar 13 at 9:19

Cettt

1,888622

asked Mar 13 at 8:12

Stat_prob_001Stat_prob_001

333113

$endgroup$

add a comment | 

$begingroup$

Let $X_1, X_2,dots,X_n$ be i.i.d. RVs with $E|X_1|^4 < infty$. Let $var(X_1) = sigma^2$, $beta_2 = mu_4/sigma^4$.

(a) Using CLT for i.i.d. RVs, show that $sqrt{n}(S^2-sigma^2)rightarrow_L N(0, mu_4-sigma^4)$.

(b) Find a transformation $g$ such that $g(S^2)$ has asymptotic distribution that depends on $beta_2$ alone, not on $sigma^2$.

I have completed part (a). But got stuck on finding $g$. I know I have to use the theorem:

$Y_n$ is $AN(mu, sigma^2_n)$, with $sigma^2_nrightarrow 0$ and $mu$ fixed real. $g$ be differentiable at $mu$ with $g'(mu)neq 0$, then $g(Y_n)$ is $AN(g(mu), [g'(mu)]^2sigma^2_n)$

Any help appreciated. Thanks.

($AN(cdot)$ means asymptotically normal distribution.)

statistics central-limit-theorem

share|cite|improve this question

edited Mar 13 at 9:19

Cettt

1,888622

asked Mar 13 at 8:12

Stat_prob_001Stat_prob_001

333113

$endgroup$

share|cite|improve this question

edited Mar 13 at 9:19

Cettt

1,888622

asked Mar 13 at 8:12

Stat_prob_001Stat_prob_001

333113

share|cite|improve this question

edited Mar 13 at 9:19

Cettt

1,888622

asked Mar 13 at 8:12

Stat_prob_001Stat_prob_001

333113

edited Mar 13 at 9:19

Cettt

1,888622

edited Mar 13 at 9:19

Cettt

1,888622

asked Mar 13 at 8:12

Stat_prob_001Stat_prob_001

333113

asked Mar 13 at 8:12

Stat_prob_001Stat_prob_001

333113

1 Answer
1

active

oldest

votes

0

$begingroup$

Rearranging the asymptotic distribution in (a) gives:

$$frac{S_n^2}{sigma^2} overset{text{Asymp}}{sim} text{N} Big( 1, frac{beta_2-1}{n} Big).$$

So, taking the parameter $beta_2$ as fixed, this gives you an asymptotically pivotal quantity (where the distribution does not depend on $sigma^2$ but that parameter appears in the quantity). If you are allowed to use the true variance parameter in your function $g$ then you are done, but if not, then it is unlikely there is a solution. Because it gives the scale of the sample variance, the only way you will be able to remove the variance parameter from this quantity (without having the function degenerate down to a form whose distribution no longer depends on $beta_2$) is if you replace the variance parameter with an alternative estimator of the variance.

share|cite|improve this answer

answered Mar 13 at 9:01

BenBen

1,840215

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it correct to write the asymptotic distribution depending on $n$? I understand it is just a rearrangement, but the asymptotic distribution is obtained when $nto infty$, so I don't think we can write "$ldotsoverset{text{Asymp}}{sim} text{N} Big( 1, frac{beta_2-1}{n} Big)$".
  $endgroup$
  – StubbornAtom
  Mar 14 at 18:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a shorthand notation, so it holds only by specifying its underlying strict meaning (similar to a limit statement). The strict meaning would not involve a limiting distribution that depends on $n$. In the present case I have not elaborated on the strict meaning since the only goal is to find the function $g$. The shorthand statement holds so long as you interpret it as an appropriate limiting statement.'
  $endgroup$
  – Ben
  Mar 14 at 22:09
  
  
  

  
  
  
  

add a comment | 

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0

$begingroup$

Rearranging the asymptotic distribution in (a) gives:

$$frac{S_n^2}{sigma^2} overset{text{Asymp}}{sim} text{N} Big( 1, frac{beta_2-1}{n} Big).$$

So, taking the parameter $beta_2$ as fixed, this gives you an asymptotically pivotal quantity (where the distribution does not depend on $sigma^2$ but that parameter appears in the quantity). If you are allowed to use the true variance parameter in your function $g$ then you are done, but if not, then it is unlikely there is a solution. Because it gives the scale of the sample variance, the only way you will be able to remove the variance parameter from this quantity (without having the function degenerate down to a form whose distribution no longer depends on $beta_2$) is if you replace the variance parameter with an alternative estimator of the variance.

share|cite|improve this answer

answered Mar 13 at 9:01

BenBen

1,840215

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it correct to write the asymptotic distribution depending on $n$? I understand it is just a rearrangement, but the asymptotic distribution is obtained when $nto infty$, so I don't think we can write "$ldotsoverset{text{Asymp}}{sim} text{N} Big( 1, frac{beta_2-1}{n} Big)$".
  $endgroup$
  – StubbornAtom
  Mar 14 at 18:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a shorthand notation, so it holds only by specifying its underlying strict meaning (similar to a limit statement). The strict meaning would not involve a limiting distribution that depends on $n$. In the present case I have not elaborated on the strict meaning since the only goal is to find the function $g$. The shorthand statement holds so long as you interpret it as an appropriate limiting statement.'
  $endgroup$
  – Ben
  Mar 14 at 22:09
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Rearranging the asymptotic distribution in (a) gives:

$$frac{S_n^2}{sigma^2} overset{text{Asymp}}{sim} text{N} Big( 1, frac{beta_2-1}{n} Big).$$

So, taking the parameter $beta_2$ as fixed, this gives you an asymptotically pivotal quantity (where the distribution does not depend on $sigma^2$ but that parameter appears in the quantity). If you are allowed to use the true variance parameter in your function $g$ then you are done, but if not, then it is unlikely there is a solution. Because it gives the scale of the sample variance, the only way you will be able to remove the variance parameter from this quantity (without having the function degenerate down to a form whose distribution no longer depends on $beta_2$) is if you replace the variance parameter with an alternative estimator of the variance.

share|cite|improve this answer

answered Mar 13 at 9:01

BenBen

1,840215

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it correct to write the asymptotic distribution depending on $n$? I understand it is just a rearrangement, but the asymptotic distribution is obtained when $nto infty$, so I don't think we can write "$ldotsoverset{text{Asymp}}{sim} text{N} Big( 1, frac{beta_2-1}{n} Big)$".
  $endgroup$
  – StubbornAtom
  Mar 14 at 18:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a shorthand notation, so it holds only by specifying its underlying strict meaning (similar to a limit statement). The strict meaning would not involve a limiting distribution that depends on $n$. In the present case I have not elaborated on the strict meaning since the only goal is to find the function $g$. The shorthand statement holds so long as you interpret it as an appropriate limiting statement.'
  $endgroup$
  – Ben
  Mar 14 at 22:09
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Rearranging the asymptotic distribution in (a) gives:

$$frac{S_n^2}{sigma^2} overset{text{Asymp}}{sim} text{N} Big( 1, frac{beta_2-1}{n} Big).$$

So, taking the parameter $beta_2$ as fixed, this gives you an asymptotically pivotal quantity (where the distribution does not depend on $sigma^2$ but that parameter appears in the quantity). If you are allowed to use the true variance parameter in your function $g$ then you are done, but if not, then it is unlikely there is a solution. Because it gives the scale of the sample variance, the only way you will be able to remove the variance parameter from this quantity (without having the function degenerate down to a form whose distribution no longer depends on $beta_2$) is if you replace the variance parameter with an alternative estimator of the variance.

share|cite|improve this answer

answered Mar 13 at 9:01

BenBen

1,840215

$endgroup$

share|cite|improve this answer

answered Mar 13 at 9:01

BenBen

1,840215

answered Mar 13 at 9:01

BenBen

1,840215

answered Mar 13 at 9:01

BenBen

1,840215