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Mean Squared Error vs Loss

  1. Sep 13, 2012 #1
    Hi Guys,

    I am just starting readings on machine learning and came across ways that the error can be used to learn the target function. The way I understand it,

    Error: [itex] e = f(\vec{x}) - y* [/itex]
    Loss: [itex] L(\vec{x}) = \frac{( f(\vec{x}) - y* )^2}{2} [/itex]
    Empirical Risk: [itex] R(f) = \sum_{i=o}^{m} \frac{( f(\vec{x}) - y* )^2}{2m} [/itex]

    where y* is the desired function, [itex] \vec{x} [/itex] is the sample vector (example) and m is the number of examples in your sample space.

    I don't understand why the factor of 2 is present in the expression for loss. The only condition my instructor placed on loss was that it had to non-negative, hence the exponent 2. But the division by two only seems to make the loss less than it really is.

    I also came across the expression for mean squared error, and it is essentially the loss without the factor of 2. If anyone could shed light on why the factor of 2 is there, I would be grateful
  2. jcsd
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