Answers.Body: <p>I've found the answer, the L2 is Standard Deviation, and the L1 is Mean Deviation. Standard deviation describes the variation better and the values are always different on different sets of X while the Mean Deviation gives the same values sometimes.</p>
  
  <blockquote>
    <p>*Footnote: Why square the differences? If we just add up the differences from the mean ... the negatives cancel the positives:</p>
    
    <p>standard deviation why a       (4 + 4 − 4 − 4) / 4  = 0 So that won't work.
    How about we use absolute values?</p>
    
    <p>standard deviation why a       (|4| + |4| + |−4| + |−4|) / 4  =  (4 + 4 + 4 +
    4) / 4 = 4 That looks good (and is the Mean Deviation), but what about
    this case:</p>
    
    <p>standard deviation why b       (|7| + |1| + |−6| + |−2|) / 4  =  (7 + 1 + 6 +
    2) / 4 = 4 Oh No! It also gives a value of 4, Even though the differences
    are more spread out.</p>
    
    <p>So let us try squaring each difference (and taking the square root at
    the end):</p>
    
    <p>standard deviation why a      √( (42 + 42 + 42 + 42) / 4 ) = √( 644) = 4
    standard deviation why b      √( (72 + 12 + 62 + 22) / 4 ) = √( 904) = 4.74...</p>
  </blockquote>
  
  <p>Reference:<br/>
  <a href="https://www.mathsisfun.com/data/standard-deviation.html" rel="nofollow noreferrer">https://www.mathsisfun.com/data/standard-deviation.html</a></p>
  
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Answers.CreationDate: 2020-02-04T11:33:24.413Z
Answers.Text: You can use MathJax to format your answer.
Answers.UserDisplayName: nbro
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Body: <p>Standard deviation and variance are in statistics but the formula for variance is somehow related to the L1 and L2.</p>
  
  <p>Mathematically (L2 in machine learning sense),
  <span class="math-container">$$Variance = \dfrac{(X_1-Mean)^2+..+(X_n-Mean)^2}{N}$$</span>
  and,
  <span class="math-container">$$Standard\ Deviation= \sqrt(Variance)$$</span></p>
  
  <p>Why shouldn't it be (L1 in machine learning sense):
  <span class="math-container">$$Variance = \dfrac{|X_1-Mean|+..+|X_n-Mean|}{N}$$</span>
  and,
  <span class="math-container">$$Standard\ Deviation= Variance$$</span></p>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2020-02-04T11:43:54.737Z
Comments.Text: It's a little bit unclear what you're asking. Do L1 and L2 in your answer refer to the L1 and respectively L2 norms? Please, edit your post to clarify what you're asking and don't just write some formulas.
Comments.UserDisplayName: nbro
Tags: <machine-learning><math><statistics>
Title: Why is Standard Deviation based on L2 Variance and not L1 Variance
UserDisplayName: Dan D.
ThreadText: # Why is Standard Deviation based on L2 Variance and not L1 Variance
  
  (Asked by: username_0 on 2020-02-04)
  
  Standard deviation and variance are in statistics but the formula for variance is somehow related to the L1 and L2.
  
  Mathematically (L2 in machine learning sense),
  $$Variance = \dfrac{(X\_1-Mean)^2+..+(X\_n-Mean)^2}{N}$$
  and,
  $$Standard\ Deviation= \sqrt(Variance)$$
  
  Why shouldn't it be (L1 in machine learning sense):
  $$Variance = \dfrac{|X\_1-Mean|+..+|X\_n-Mean|}{N}$$
  and,
  $$Standard\ Deviation= Variance$$
  
  **Tags:** machine-learning, math, statistics
  
  > **Comments:**
  > **username_1**: It's a little bit unclear what you're asking. Do L1 and L2 in your answer refer to the L1 and respectively L2 norms? Please, edit your post to clarify what you're asking and don't just write some formulas.  
  
  ---
  
  ## 1 answers
  
  ### Answer by username_0 on 2020-02-04. Score: 4 | Accepted answer
  
  I've found the answer, the L2 is Standard Deviation, and the L1 is Mean Deviation. Standard deviation describes the variation better and the values are always different on different sets of X while the Mean Deviation gives the same values sometimes.
  
  > \*Footnote: Why square the differences? If we just add up the differences from the mean ... the negatives cancel the positives:
  >
  > standard deviation why a (4 + 4 − 4 − 4) / 4 = 0 So that won't work.
  > How about we use absolute values?
  >
  > standard deviation why a (|4| + |4| + |−4| + |−4|) / 4 = (4 + 4 + 4 +
  > 4) / 4 = 4 That looks good (and is the Mean Deviation), but what about
  > this case:
  >
  > standard deviation why b (|7| + |1| + |−6| + |−2|) / 4 = (7 + 1 + 6 +
  > 2) / 4 = 4 Oh No! It also gives a value of 4, Even though the differences
  > are more spread out.
  >
  > So let us try squaring each difference (and taking the square root at
  > the end):
  >
  > standard deviation why a √( (42 + 42 + 42 + 42) / 4 ) = √( 644) = 4
  > standard deviation why b √( (72 + 12 + 62 + 22) / 4 ) = √( 904) = 4.74...
  
  Reference:  
  <https://www.mathsisfun.com/data/standard-deviation.html>
  
  > **Comments:**
  > **username_1**: You can use MathJax to format your answer.  
  
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