Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted