convolution the standard normal distribution

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Quote: If you convolution the standard normal distribution with "+ 1, -1 = 50/50%" then again there will be a standard normal distribution.

Answer: The standard normal distribution is a centered normal (Gaussian) distribution with a normalized variance. Convolution corresponds to the distribution of the sum of independent random variables. Dispersions of a random variable add up when added, so, given that the variance of the -1.1 -distribution is not zero (equal to 1), the convolution variance will be more than one (equal to 2), that is, convolution will no longer be the standard normal distribution. Correctly?

probability-theory

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asked Jul 24 at 6:09

evs

1

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Two beers and a shot... I'm not that drunk. What is the question?
  – Sean Roberson
  Jul 24 at 6:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  maybe difficulties with translation!
  – evs
  Jul 24 at 6:30
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

Quote: If you convolution the standard normal distribution with "+ 1, -1 = 50/50%" then again there will be a standard normal distribution.

Answer: The standard normal distribution is a centered normal (Gaussian) distribution with a normalized variance. Convolution corresponds to the distribution of the sum of independent random variables. Dispersions of a random variable add up when added, so, given that the variance of the -1.1 -distribution is not zero (equal to 1), the convolution variance will be more than one (equal to 2), that is, convolution will no longer be the standard normal distribution. Correctly?

probability-theory

share|cite|improve this question

asked Jul 24 at 6:09

evs

1

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Two beers and a shot... I'm not that drunk. What is the question?
  – Sean Roberson
  Jul 24 at 6:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  maybe difficulties with translation!
  – evs
  Jul 24 at 6:30
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Quote: If you convolution the standard normal distribution with "+ 1, -1 = 50/50%" then again there will be a standard normal distribution.

Answer: The standard normal distribution is a centered normal (Gaussian) distribution with a normalized variance. Convolution corresponds to the distribution of the sum of independent random variables. Dispersions of a random variable add up when added, so, given that the variance of the -1.1 -distribution is not zero (equal to 1), the convolution variance will be more than one (equal to 2), that is, convolution will no longer be the standard normal distribution. Correctly?

probability-theory

share|cite|improve this question

asked Jul 24 at 6:09

evs

1

Quote: If you convolution the standard normal distribution with "+ 1, -1 = 50/50%" then again there will be a standard normal distribution.

Answer: The standard normal distribution is a centered normal (Gaussian) distribution with a normalized variance. Convolution corresponds to the distribution of the sum of independent random variables. Dispersions of a random variable add up when added, so, given that the variance of the -1.1 -distribution is not zero (equal to 1), the convolution variance will be more than one (equal to 2), that is, convolution will no longer be the standard normal distribution. Correctly?

probability-theory

share|cite|improve this question

asked Jul 24 at 6:09

evs

1

share|cite|improve this question

share|cite|improve this question

asked Jul 24 at 6:09

evs

1

asked Jul 24 at 6:09

evs

1

asked Jul 24 at 6:09

evs

1

1

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Two beers and a shot... I'm not that drunk. What is the question?
  – Sean Roberson
  Jul 24 at 6:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  maybe difficulties with translation!
  – evs
  Jul 24 at 6:30
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Two beers and a shot... I'm not that drunk. What is the question?
  – Sean Roberson
  Jul 24 at 6:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  maybe difficulties with translation!
  – evs
  Jul 24 at 6:30
  

  
  
  
  

2

2

Two beers and a shot... I'm not that drunk. What is the question?
– Sean Roberson
Jul 24 at 6:16

Two beers and a shot... I'm not that drunk. What is the question?
– Sean Roberson
Jul 24 at 6:16

maybe difficulties with translation!
– evs
Jul 24 at 6:30

maybe difficulties with translation!
– evs
Jul 24 at 6:30

add a comment | 

1 Answer
1

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up vote
2
down vote

Yes, your answer is correct, the statement you quote is false.

Some additional comments:

• The verb corresponding to the noun “convolution” is “convolve”.


• “Dispersions of a random variable add up when added” – here you implicitly used the independence.


• The resulting distribution is not only not a standard normal distribution, but not even a normal distribution. It's the sum of two Gaussians, which can't be written as a single Gaussian.

share|cite|improve this answer

edited Jul 24 at 9:27

answered Jul 24 at 7:05

joriki

164k10180328

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "Yes, that's correct." is slightly odd in view of the (correct) "additional comments" that come after, since the last of these (correctly) contradicts the belief that the sum is normal which the question seems to (incorrectly) endorse.
  – Did
  Jul 24 at 8:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did: The question seemed heavily in need of interpretation. I interpreted it to mean that the OP saw this quote, wanted to check it and disproved in the part labeled "Answer". I've clarified my answer :-).
  – joriki
  Jul 24 at 9:27
  

  
  
  
  

add a comment | 

Your Answer

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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

Yes, your answer is correct, the statement you quote is false.

Some additional comments:

• The verb corresponding to the noun “convolution” is “convolve”.


• “Dispersions of a random variable add up when added” – here you implicitly used the independence.


• The resulting distribution is not only not a standard normal distribution, but not even a normal distribution. It's the sum of two Gaussians, which can't be written as a single Gaussian.

share|cite|improve this answer

edited Jul 24 at 9:27

answered Jul 24 at 7:05

joriki

164k10180328

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "Yes, that's correct." is slightly odd in view of the (correct) "additional comments" that come after, since the last of these (correctly) contradicts the belief that the sum is normal which the question seems to (incorrectly) endorse.
  – Did
  Jul 24 at 8:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did: The question seemed heavily in need of interpretation. I interpreted it to mean that the OP saw this quote, wanted to check it and disproved in the part labeled "Answer". I've clarified my answer :-).
  – joriki
  Jul 24 at 9:27
  

  
  
  
  

add a comment | 

up vote
2
down vote

Yes, your answer is correct, the statement you quote is false.

Some additional comments:

• The verb corresponding to the noun “convolution” is “convolve”.


• “Dispersions of a random variable add up when added” – here you implicitly used the independence.


• The resulting distribution is not only not a standard normal distribution, but not even a normal distribution. It's the sum of two Gaussians, which can't be written as a single Gaussian.

share|cite|improve this answer

edited Jul 24 at 9:27

answered Jul 24 at 7:05

joriki

164k10180328

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "Yes, that's correct." is slightly odd in view of the (correct) "additional comments" that come after, since the last of these (correctly) contradicts the belief that the sum is normal which the question seems to (incorrectly) endorse.
  – Did
  Jul 24 at 8:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did: The question seemed heavily in need of interpretation. I interpreted it to mean that the OP saw this quote, wanted to check it and disproved in the part labeled "Answer". I've clarified my answer :-).
  – joriki
  Jul 24 at 9:27
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

Yes, your answer is correct, the statement you quote is false.

Some additional comments:

• The verb corresponding to the noun “convolution” is “convolve”.


• “Dispersions of a random variable add up when added” – here you implicitly used the independence.


• The resulting distribution is not only not a standard normal distribution, but not even a normal distribution. It's the sum of two Gaussians, which can't be written as a single Gaussian.

share|cite|improve this answer

edited Jul 24 at 9:27

answered Jul 24 at 7:05

joriki

164k10180328

Yes, your answer is correct, the statement you quote is false.

Some additional comments:

• The verb corresponding to the noun “convolution” is “convolve”.


• “Dispersions of a random variable add up when added” – here you implicitly used the independence.


• The resulting distribution is not only not a standard normal distribution, but not even a normal distribution. It's the sum of two Gaussians, which can't be written as a single Gaussian.

share|cite|improve this answer

edited Jul 24 at 9:27

answered Jul 24 at 7:05

joriki

164k10180328

share|cite|improve this answer

share|cite|improve this answer

edited Jul 24 at 9:27

edited Jul 24 at 9:27

edited Jul 24 at 9:27

answered Jul 24 at 7:05

joriki

164k10180328

answered Jul 24 at 7:05

joriki

164k10180328

answered Jul 24 at 7:05

joriki

164k10180328

164k10180328

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "Yes, that's correct." is slightly odd in view of the (correct) "additional comments" that come after, since the last of these (correctly) contradicts the belief that the sum is normal which the question seems to (incorrectly) endorse.
  – Did
  Jul 24 at 8:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did: The question seemed heavily in need of interpretation. I interpreted it to mean that the OP saw this quote, wanted to check it and disproved in the part labeled "Answer". I've clarified my answer :-).
  – joriki
  Jul 24 at 9:27
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "Yes, that's correct." is slightly odd in view of the (correct) "additional comments" that come after, since the last of these (correctly) contradicts the belief that the sum is normal which the question seems to (incorrectly) endorse.
  – Did
  Jul 24 at 8:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did: The question seemed heavily in need of interpretation. I interpreted it to mean that the OP saw this quote, wanted to check it and disproved in the part labeled "Answer". I've clarified my answer :-).
  – joriki
  Jul 24 at 9:27
  

  
  
  
  

"Yes, that's correct." is slightly odd in view of the (correct) "additional comments" that come after, since the last of these (correctly) contradicts the belief that the sum is normal which the question seems to (incorrectly) endorse.
– Did
Jul 24 at 8:19

"Yes, that's correct." is slightly odd in view of the (correct) "additional comments" that come after, since the last of these (correctly) contradicts the belief that the sum is normal which the question seems to (incorrectly) endorse.
– Did
Jul 24 at 8:19

@Did: The question seemed heavily in need of interpretation. I interpreted it to mean that the OP saw this quote, wanted to check it and disproved in the part labeled "Answer". I've clarified my answer :-).
– joriki
Jul 24 at 9:27

@Did: The question seemed heavily in need of interpretation. I interpreted it to mean that the OP saw this quote, wanted to check it and disproved in the part labeled "Answer". I've clarified my answer :-).
– joriki
Jul 24 at 9:27

add a comment | 

 

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