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, $$Standard\ Deviation= Variance$$
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
