Why is standard deviation usually defined as sqrt(var(x)) and not simply as abs(X-u) while the latter is simpler to understand and both give same values?
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5Both do not give the same values. $\sqrt{\sum (x_i - \mu)^2} \ne \sum|x_i - \mu|$ – Greenparker Apr 04 '16 at 16:42
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2It's the difference between an L1 vs an L2 norm. The sqrt(var(x)) is L2 and abs(X-u) is an L1. – Mike Hunter Apr 04 '16 at 16:47
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People who came up with the quadratic deviation are not the same people who would think that absolute value is simpler to understand. – Aksakal Apr 04 '16 at 17:35
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They do not give the same distance.
If you are trying to get from point $a$ to point $b$ in a city with regular, parallel streets, then
$$ \sum_i \left| a_i - b_i \right| $$
is the total driving distance along the streets from $a$ to $b$. Because we are forced to drive either directly to the east/west or directly to the north/south, this is less efficient (more total driving) than the distance a bird would take, flying from $a$ to $b$ directly. The bird distance is what
$$ \sqrt{\sum_i (a_i - b_i)^2} $$
measures (for a small city, ignoring the curvature of the earth).
The driving distance is always greater than or equal to the bird distance, with the two being equal if and only if the bird is flying directly east/west or north/south.
Matthew Drury
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2(+1) Great explanation of Jensen's inequality using such an intuitive analogy! – Apr 04 '16 at 17:08
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They do not give the same result at all, because the square root operator is not linear.
Quantuple
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2I don't understand this, the expectation *is* a linear operator. It's the square root that is not. – Matthew Drury Apr 04 '16 at 16:52
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You are of course totally write. Wrote too quickly and my ideas got mixed up. Your answer nails it anyway – Quantuple Apr 04 '16 at 17:00
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1It's worth an edit to your answer, I make the same kind of mistake all the time! – Matthew Drury Apr 04 '16 at 17:01
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The main reason we use $||X||_2$ is because of its nice mathematical properties: it is differnetiable, which allows for more of Real Analysis to be applied.
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Worth noting that the differetiabilty in some sense makes $| X |^2$, in the OP's words, "simpler to understand". – Matthew Drury Apr 04 '16 at 16:56
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By this argument we could just as well be using $||X||_p$ for any $1 \lt p \lt \infty$. That suggests this answer might be part of the reason but it's certainly not the whole reason. – whuber Apr 04 '16 at 17:44
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@whuber couldn't agree more ... I wasnt asserting that differentiability was the whole story...however it shows why THE most obvious choice is not chosen... – Apr 04 '16 at 18:38