What is the Difference between Square & Absolute Deviation?”

[icystrike]

I went for a lecture and the lecturer said that the square of the difference between the x sub i and the mean is the take precaution of the negative value. This has been bugging me , i was wondering why don't they just take absolute because there is a difference between :
\sqrt{\frac{\sum(x-\mu)^2}{f}} and\frac{\sum \left|(x-\mu)\right|}{f}

 

[statdad]

Yes, there is a difference, as

<br /> \sqrt{\sum(x-\mu)^2} \ne \sum |x - \mu |<br />

There is actually quite a history about whether a measure based on

<br /> \sqrt{\frac{\sum (x-\mu)^2 }{f}}<br />

or

<br /> \sqrt{\frac{\sum |x-\mu|}{f}}<br />

should be used. Basically, the measure based on the sum of squared deviations won out because, statistically, when it is assumed that the data are drawn from a normal distribution (equivalently, when it is assumed the random noise is Gaussian).

 

[icystrike]

Basically, the measure based on the sum of squared deviations won out because, statistically, when it is assumed that the data are drawn from a normal distribution (equivalently, when it is assumed the random noise is Gaussian).

Thanks for your help
Random noise, i got to check this out !

 

[boboYO]

heh, i remember my stats lecturer said that too.
an analogy can be drawn with why we take the squares of the sides (pythagoras) to work out the hypotenuse and not the absolute value.

 

[daviddoria]

The squared distance is also used because it is continuous, where the absolute distance function has a discontinuity. This is a big problem in optimization.

 

[statdad]

The squared distance is also used because it is continuous, where the absolute distance function has a discontinuity. This is a big problem in optimization.

Not really the case in statistics - the median, median deviation, and other procedures use the absolute value.

 

[HallsofIvy]

The squared distance is also used because it is continuous, where the absolute distance function has a discontinuity. This is a big problem in optimization.

The absolute distance function does not have a derivative at a point. There is no discontinuity.

 

[HallsofIvy]

Yes, there is a difference, as

<br /> \sqrt{\sum(x-\mu)^2} \ne \sum |x - \mu |<br />

There is actually quite a history about whether a measure based on

<br /> \sqrt{\frac{\sum (x-\mu)^2 }{f}}<br />

or

<br /> \sqrt{\frac{\sum |x-\mu|}{f}}<br />

When you sum the absolute values, you should not have a square root.

should be used. Basically, the measure based on the sum of squared deviations won out because, statistically, when it is assumed that the data are drawn from a normal distribution (equivalently, when it is assumed the random noise is Gaussian).

There is a third used occasionally:
\frac{max |x-\mu|}{f}

The end of your last sentence seems to be missing!

 

[statdad]

Halls, i wish i had your proof-reading skills. Thanks for catching my missed comment.

You are also correct that the absolute value expression has no derivative, but again, for statistics, I'd add that really isn't a problem.

Why did I miss the unneeded square root? Let me know when you figure it out, because I can't.