Can I increase the standard deviation of a series without increasing its mean?

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I have a series of values. Can I modify them in a consistent manner such that it's mean remain the same but standard deviation increase by 1? 1 is just arbitrary in this example

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asked Jul 27 at 7:29

Boon Heng Tan

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I have a series of values. Can I modify them in a consistent manner such that it's mean remain the same but standard deviation increase by 1? 1 is just arbitrary in this example

average

share|cite|improve this question

asked Jul 27 at 7:29

Boon Heng Tan

1

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

I have a series of values. Can I modify them in a consistent manner such that it's mean remain the same but standard deviation increase by 1? 1 is just arbitrary in this example

average

share|cite|improve this question

asked Jul 27 at 7:29

Boon Heng Tan

1

I have a series of values. Can I modify them in a consistent manner such that it's mean remain the same but standard deviation increase by 1? 1 is just arbitrary in this example

average

share|cite|improve this question

asked Jul 27 at 7:29

Boon Heng Tan

1

share|cite|improve this question

share|cite|improve this question

asked Jul 27 at 7:29

Boon Heng Tan

1

asked Jul 27 at 7:29

Boon Heng Tan

1

asked Jul 27 at 7:29

Boon Heng Tan

1

1

add a comment | 

add a comment | 

2 Answers
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If $mu$ denotes the mean and $sigma$ the standard deviation then replace value $x$ by value: $$(1+frac1sigma)(x-mu)+mu$$

The new mean will be $mu$ and the new standard deviation will be: $(1+frac1sigma)sigma=sigma+1$.

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answered Jul 27 at 7:40

drhab

86.1k541118

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In effect this would be adding $fracx-musigma$ to each $x$
  – Henry
  Jul 27 at 8:18
  

  
  
  
  

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up vote
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You can insert mean $pm c$ which won't change the mean but will change the variance by a calculatable amount in terms of $c$ and the original population size.

Note this only works on probability distributions on finite sets.

share|cite|improve this answer

edited Jul 27 at 14:47

answered Jul 27 at 7:37

coffeemath

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

If $mu$ denotes the mean and $sigma$ the standard deviation then replace value $x$ by value: $$(1+frac1sigma)(x-mu)+mu$$

The new mean will be $mu$ and the new standard deviation will be: $(1+frac1sigma)sigma=sigma+1$.

share|cite|improve this answer

answered Jul 27 at 7:40

drhab

86.1k541118

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In effect this would be adding $fracx-musigma$ to each $x$
  – Henry
  Jul 27 at 8:18
  

  
  
  
  

add a comment | 

up vote
1
down vote

If $mu$ denotes the mean and $sigma$ the standard deviation then replace value $x$ by value: $$(1+frac1sigma)(x-mu)+mu$$

The new mean will be $mu$ and the new standard deviation will be: $(1+frac1sigma)sigma=sigma+1$.

share|cite|improve this answer

answered Jul 27 at 7:40

drhab

86.1k541118

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In effect this would be adding $fracx-musigma$ to each $x$
  – Henry
  Jul 27 at 8:18
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

If $mu$ denotes the mean and $sigma$ the standard deviation then replace value $x$ by value: $$(1+frac1sigma)(x-mu)+mu$$

The new mean will be $mu$ and the new standard deviation will be: $(1+frac1sigma)sigma=sigma+1$.

share|cite|improve this answer

answered Jul 27 at 7:40

drhab

86.1k541118

If $mu$ denotes the mean and $sigma$ the standard deviation then replace value $x$ by value: $$(1+frac1sigma)(x-mu)+mu$$

The new mean will be $mu$ and the new standard deviation will be: $(1+frac1sigma)sigma=sigma+1$.

share|cite|improve this answer

answered Jul 27 at 7:40

drhab

86.1k541118

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 7:40

drhab

86.1k541118

answered Jul 27 at 7:40

drhab

86.1k541118

answered Jul 27 at 7:40

drhab

86.1k541118

86.1k541118

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In effect this would be adding $fracx-musigma$ to each $x$
  – Henry
  Jul 27 at 8:18
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In effect this would be adding $fracx-musigma$ to each $x$
  – Henry
  Jul 27 at 8:18
  

  
  
  
  

1

1

In effect this would be adding $fracx-musigma$ to each $x$
– Henry
Jul 27 at 8:18

In effect this would be adding $fracx-musigma$ to each $x$
– Henry
Jul 27 at 8:18

add a comment | 

up vote
0
down vote

You can insert mean $pm c$ which won't change the mean but will change the variance by a calculatable amount in terms of $c$ and the original population size.

Note this only works on probability distributions on finite sets.

share|cite|improve this answer

edited Jul 27 at 14:47

answered Jul 27 at 7:37

coffeemath

1,2491212

add a comment | 

up vote
0
down vote

You can insert mean $pm c$ which won't change the mean but will change the variance by a calculatable amount in terms of $c$ and the original population size.

Note this only works on probability distributions on finite sets.

share|cite|improve this answer

edited Jul 27 at 14:47

answered Jul 27 at 7:37

coffeemath

1,2491212

add a comment | 

up vote
0
down vote

up vote
0
down vote

You can insert mean $pm c$ which won't change the mean but will change the variance by a calculatable amount in terms of $c$ and the original population size.

Note this only works on probability distributions on finite sets.

share|cite|improve this answer

edited Jul 27 at 14:47

answered Jul 27 at 7:37

coffeemath

1,2491212

You can insert mean $pm c$ which won't change the mean but will change the variance by a calculatable amount in terms of $c$ and the original population size.

Note this only works on probability distributions on finite sets.

share|cite|improve this answer

edited Jul 27 at 14:47

answered Jul 27 at 7:37

coffeemath

1,2491212

share|cite|improve this answer

share|cite|improve this answer

edited Jul 27 at 14:47

edited Jul 27 at 14:47

edited Jul 27 at 14:47

answered Jul 27 at 7:37

coffeemath

1,2491212

answered Jul 27 at 7:37

coffeemath

1,2491212

answered Jul 27 at 7:37

coffeemath

1,2491212

1,2491212

add a comment | 

add a comment | 

 

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