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mathman

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Then the variance V is DEFINED by V=E((X-A)

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Mute

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If you want to find the typical deviation from the average, you can't calculate the average of x -<x>, because that average is zero. You need a measure where the deviations from the average don't cancel out. Taking the average of |x-<x>| works, but that's not nice because the absolute value function isn't differentiable at zero. (x-<x>)^2 also works, and is differentiable. However, if x has units, you can't compare <(x-<x>)^2> directly to x, because the units don't match. So, you need take the square root of to get something with the same units of x that you can treat as a deviation from the mean.

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chiro

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One other thing to keep in mind is that the definition provided for obtaining the variance is used directly in statistical theory like the normal distribution: ie the std deviation that is calculated with the variance that uses difference in squares is used in the normal pdf.

Also you'll find that this definition is useful when dealing with other properties of random variables.

Another thing to keep in mind is that you could picture the variance as a length in an n dimensional euclidean space where n is the number of elements in the random variable.

For example if we have a three dimensional vector where X(1), X(2), AND X(3) represent the difference between the average and the element of the random variable, then the "length" of this vector is basically found using the pythagorean theorem where length = SQRT(X(1)^2 + X(2)^2 + X(3)^2). This definition makes sense when you interpret this geometrically as the length of a vector in an n dimensional euclidean space.

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