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i am trying to understand the error metrics for images. i had no statistics class or anything and i'm having a bit of problem with even the simplest things.

i have started reading this http://unitstep.net/downloads/school/image-qa-presentation/

just for starters, it says that mean squared error is (page 3)

sum_over_i_j( |ImageA(i,j)-ImageB(i,j)|^2 ) / amount_of_pixels

i understand that this is trying to be something like euclidean distance between two points, but why is there no square root then?

then when you go to the next page of the presentation it tells you that this is a special case of minkowski distance and there IS the square root included in case of mean squared error!

and then, shouldn't we calculate distance of two points by summing the powers of the differences of their dimension components? (`[ (x1-x2)^2 + (y1-y2)^2 + ...] ) and only then square root the sum? then do it for all the pixels and average it.

while on the page 4 it says that we power, sum and square root the pixels first and then do this for all the channels (dimensions) separately?

i don't understand :(

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First of all, the mean squared error should not have a square root. As the name implies, it is the mean of the squared difference. What you are referring to is the root mean squared error. Second, the this is a pixel-wise comparison between to images. Therefore the number of pixels is the dimension, and the sum is performed on it. There is even a comment on that page about color images.

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  • $\begingroup$ thank you. then if i set ɣ=2 in the minkowski metric, i get root mean square error? then why does it say that "mean square error is a specific form of a Minkowski metric". when it is actually not rooted $\endgroup$ – user1916182 Jun 15 '14 at 22:51
  • $\begingroup$ The square root makes no difference in terms of comparing errors. You cam use either. $\endgroup$ – ThP Jun 16 '14 at 6:45
  • $\begingroup$ hmm. ok, my brain is able to settle with this idea. however, i still find it strange that they say "this is just a special case of minkowski metric, where ɣ=2" while in reality, it is not, because it lacks the square root. it doesn't matter that it makes no difference when comparing errors. it is then not a special case of minkowski metric by definition... $\endgroup$ – user1916182 Jun 16 '14 at 11:22

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