Symmetry of the Distribution of Sample Mean [closed]

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I've studied all the proofs and the theorems. The central limit theorem and the laws of mean and variance, but i still don't grasp intuitively why the distribution of Sample Mean tends to a symmetrical distribution ?

Are you able to help me ?
Thanks

probability statistics

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edited Jul 28 at 9:44

asked Jul 28 at 9:41

Koinos

535

closed as unclear what you're asking by Yves Daoust, Jyrki Lahtonen, Shailesh, Lord Shark the Unknown, Claude Leibovici Jul 29 at 7:05

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why is there "symmetry" in the title and not in the body ?
  – Yves Daoust
  Jul 28 at 9:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, i edited thanks
  – Koinos
  Jul 28 at 9:44
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

I've studied all the proofs and the theorems. The central limit theorem and the laws of mean and variance, but i still don't grasp intuitively why the distribution of Sample Mean tends to a symmetrical distribution ?

Are you able to help me ?
Thanks

probability statistics

share|cite|improve this question

edited Jul 28 at 9:44

asked Jul 28 at 9:41

Koinos

535

closed as unclear what you're asking by Yves Daoust, Jyrki Lahtonen, Shailesh, Lord Shark the Unknown, Claude Leibovici Jul 29 at 7:05

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why is there "symmetry" in the title and not in the body ?
  – Yves Daoust
  Jul 28 at 9:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, i edited thanks
  – Koinos
  Jul 28 at 9:44
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

I've studied all the proofs and the theorems. The central limit theorem and the laws of mean and variance, but i still don't grasp intuitively why the distribution of Sample Mean tends to a symmetrical distribution ?

Are you able to help me ?
Thanks

probability statistics

share|cite|improve this question

edited Jul 28 at 9:44

asked Jul 28 at 9:41

Koinos

535

I've studied all the proofs and the theorems. The central limit theorem and the laws of mean and variance, but i still don't grasp intuitively why the distribution of Sample Mean tends to a symmetrical distribution ?

Are you able to help me ?
Thanks

probability statistics

share|cite|improve this question

edited Jul 28 at 9:44

asked Jul 28 at 9:41

Koinos

535

share|cite|improve this question

share|cite|improve this question

edited Jul 28 at 9:44

edited Jul 28 at 9:44

edited Jul 28 at 9:44

asked Jul 28 at 9:41

Koinos

535

asked Jul 28 at 9:41

Koinos

535

asked Jul 28 at 9:41

Koinos

535

535

closed as unclear what you're asking by Yves Daoust, Jyrki Lahtonen, Shailesh, Lord Shark the Unknown, Claude Leibovici Jul 29 at 7:05

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

closed as unclear what you're asking by Yves Daoust, Jyrki Lahtonen, Shailesh, Lord Shark the Unknown, Claude Leibovici Jul 29 at 7:05

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why is there "symmetry" in the title and not in the body ?
  – Yves Daoust
  Jul 28 at 9:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, i edited thanks
  – Koinos
  Jul 28 at 9:44
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why is there "symmetry" in the title and not in the body ?
  – Yves Daoust
  Jul 28 at 9:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, i edited thanks
  – Koinos
  Jul 28 at 9:44
  

  
  
  
  

Why is there "symmetry" in the title and not in the body ?
– Yves Daoust
Jul 28 at 9:42

Why is there "symmetry" in the title and not in the body ?
– Yves Daoust
Jul 28 at 9:42

Yeah, i edited thanks
– Koinos
Jul 28 at 9:44

Yeah, i edited thanks
– Koinos
Jul 28 at 9:44

add a comment | 

1 Answer
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The pdf of the sum of two random variables is the result of the convolution of the two initial pdf's, which can be seen as a low-pass filtering of the first distribution with the second, mirrored (in other terms, a sliding weighted average). The mirroring comes from the fact that you accumulate for constant sums ($s=x+yimplies y=s-x$).

The result is a smoother and more symmetric function. Smoother because you are averaging with positive-only coefficients, and more symmetric because of the mirroring (high functions values get a lower weight, and conversely).

The limit pdf is a Gaussian because this function enjoys a special property: the result of a Gaussian convolved with a Gaussian is another Gaussian.

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answered Jul 28 at 9:56

Yves Daoust

110k665203

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

The pdf of the sum of two random variables is the result of the convolution of the two initial pdf's, which can be seen as a low-pass filtering of the first distribution with the second, mirrored (in other terms, a sliding weighted average). The mirroring comes from the fact that you accumulate for constant sums ($s=x+yimplies y=s-x$).

The result is a smoother and more symmetric function. Smoother because you are averaging with positive-only coefficients, and more symmetric because of the mirroring (high functions values get a lower weight, and conversely).

The limit pdf is a Gaussian because this function enjoys a special property: the result of a Gaussian convolved with a Gaussian is another Gaussian.

share|cite|improve this answer

answered Jul 28 at 9:56

Yves Daoust

110k665203

add a comment | 

up vote
2
down vote

The pdf of the sum of two random variables is the result of the convolution of the two initial pdf's, which can be seen as a low-pass filtering of the first distribution with the second, mirrored (in other terms, a sliding weighted average). The mirroring comes from the fact that you accumulate for constant sums ($s=x+yimplies y=s-x$).

The result is a smoother and more symmetric function. Smoother because you are averaging with positive-only coefficients, and more symmetric because of the mirroring (high functions values get a lower weight, and conversely).

The limit pdf is a Gaussian because this function enjoys a special property: the result of a Gaussian convolved with a Gaussian is another Gaussian.

share|cite|improve this answer

answered Jul 28 at 9:56

Yves Daoust

110k665203

add a comment | 

up vote
2
down vote

up vote
2
down vote

The pdf of the sum of two random variables is the result of the convolution of the two initial pdf's, which can be seen as a low-pass filtering of the first distribution with the second, mirrored (in other terms, a sliding weighted average). The mirroring comes from the fact that you accumulate for constant sums ($s=x+yimplies y=s-x$).

The result is a smoother and more symmetric function. Smoother because you are averaging with positive-only coefficients, and more symmetric because of the mirroring (high functions values get a lower weight, and conversely).

The limit pdf is a Gaussian because this function enjoys a special property: the result of a Gaussian convolved with a Gaussian is another Gaussian.

share|cite|improve this answer

answered Jul 28 at 9:56

Yves Daoust

110k665203

The pdf of the sum of two random variables is the result of the convolution of the two initial pdf's, which can be seen as a low-pass filtering of the first distribution with the second, mirrored (in other terms, a sliding weighted average). The mirroring comes from the fact that you accumulate for constant sums ($s=x+yimplies y=s-x$).

The result is a smoother and more symmetric function. Smoother because you are averaging with positive-only coefficients, and more symmetric because of the mirroring (high functions values get a lower weight, and conversely).

The limit pdf is a Gaussian because this function enjoys a special property: the result of a Gaussian convolved with a Gaussian is another Gaussian.

share|cite|improve this answer

answered Jul 28 at 9:56

Yves Daoust

110k665203

share|cite|improve this answer

share|cite|improve this answer

answered Jul 28 at 9:56

Yves Daoust

110k665203

answered Jul 28 at 9:56

Yves Daoust

110k665203

answered Jul 28 at 9:56

Yves Daoust

110k665203

110k665203

add a comment | 

add a comment |