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Here is "mean squared error" function: C = $\frac{1}{2n}$ * $\sum(length(y - a)^2)$

As I understand, this is like paraboloid in multidimensional space. So, I guess, there is only one extremum: global mimum. But subconsciously I'm not 100% sure of it. Can you explain whether I'm right or wrong?

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    $\begingroup$ It is a convex function, and so any minimum is a global minimum. $\endgroup$ – Anna SdTC Mar 14 '17 at 21:30
  • $\begingroup$ @AnnaSdTC, why do not you write a normal answer? $\endgroup$ – Dmitry Nalyvaiko Mar 14 '17 at 21:51
  • $\begingroup$ What is a normal answer? You asked if there is a global minimum, I say yes, any minimum is a global minimum. $\endgroup$ – Anna SdTC Mar 15 '17 at 1:34
  • $\begingroup$ > You asked if there is a global minimum;;; I asked: Is there LOCAL MINIMUMS in MSE function?;;; > any minimum is a global minimum;;; "any" means that there any multiple minimums, but there are SINGLE minimum $\endgroup$ – Dmitry Nalyvaiko Mar 15 '17 at 10:51
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    $\begingroup$ No, it means that, if you find a minimum (i.e., a point with lower value of the function than any point in its neighborhood), then it is a global minimum. It also means that any point for which all the partial derivatives are zero is a global minimum. $\endgroup$ – Anna SdTC Mar 15 '17 at 17:38
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Yes, you are right. MSE(Mean Square Error) could only have a convex plane which bent only downward. So, There couldn't exist local minima.

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