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The two most popular types of loss functions are

1) squared error: $L(y,f(x))=(y-f(x))^2$ --> best estimate is the $E(Y|x) $
2) absolute error: $L(y,f(x))=|y-f(x)|$ --> best estimate is the $median(Y|x)$

I have two questions.

1) Why do people use the squared error method? The absolute error method makes much more intuitive sense. You get the difference between the actual and the estimate. Plain and simple. If you square the difference, then won't you get "warped" values depending on the size of the difference?

2) This also got me thinking about what is "expected value." Expected value is defined as the mean. However, the best estimate under the absolute error loss function is the median. So is the "expected value" the median? I would very much appreciate it if someone can help me clarify my thinking. Thanks.

irwenqiang
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    If you google your own question ("Why do people use the squared error method"), you will be directed [here](http://stats.stackexchange.com/questions/118/why-square-the-difference-instead-of-taking-the-absolute-value-in-standard-devia) – imran.fanaswala Jul 14 '14 at 14:20
  • I don't know why $argmin_{f(x)}E_{Y|x}(Y-f(x))^2 = E(Y|x)$. – irwenqiang Jul 14 '14 at 14:30
  • 1) What do you mean by 'warped' there? 2) The term 'expected value' is already defined to be the mean. It doesn't matter what the best estimate is under a loss function you like. – Glen_b Jul 14 '14 at 14:54
  • see [Modes, Medians and Means: A Unifying Perspective](http://www.johnmyleswhite.com/notebook/2013/03/22/modes-medians-and-means-an-unifying-perspective/) for a explanation. – irwenqiang Jul 14 '14 at 15:05

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