I Gradient descent, hessian(E(W^T X)) = cov(X),Why mean=/=0?

  • I
  • Thread starter Thread starter NotASmurf
  • Start date Start date
  • Tags Tags
    Gradient
NotASmurf
Messages
150
Reaction score
2
Backpropagation algorithm E(X,W) = 0.5(target - W^T X)^2 as error, the paper I'm reading notes that covarience matrix of inputs is equal to the hessian, it uses that to develop its learning weight update rule V(k+1) = V(k) + D*V(k), slightly modified (not relevant for my question) version of normal backpropagation feedforward gradient descent, but he uses a mean of zero for the inputs, just like in all other neural networks, but doesn't shifting the mean not affect the covariance matrix, so the eigenvectors ,entropy H=|cov(x)|, second derivative is shouldn't change. It's not even a conjugate prior, its not like I am encoding some terrible prior beliefs if we view it as a distribution mapper. Why does the mean being zero matter here and ostensibly in all ANN's, any help appreciated.
 
Physics news on Phys.org
Sounds like the author is performing "whitening" Given some data set, take the eigenbasis divide by the eigenvalues and you get a normalized. If the data is multivariate Gaussian, then the data now as a mean of zero with a covariance matrix equal to the identity. It's a pretty common practice for image processing, unless there's a lot of white noise.

edit: Actually misread the post, ignore me :)
 
First, thanks for responding.
"If the data is multivariate Gaussian" it is a multivariate gaussian normal yes.

"data now as a mean of zero with a covariance matrix equal to the identity." but he also performs a gaussian integral on det(lambda*I - cov(X)), if cov(X) = I then it would have no eigenvalues, which need to be unique for this algorithm to work, and B) lambda*I - cov(X) is used where cov(X) inverse should be used, i know that A-lambda*I has no inverse, I don't know the logic behind him using lambda*I-A,

Also he says he's using something called "standard fresnel representation for the determinate of a symmetric matrix R" but "fresnel representation determinant" doesn't turn up anything that looks "standard" at all, the paper is over 25 years old to be fair. You have any idea what that is? Cos he seems to just be doing some gaussian multivariate optimization.
 
Without reading the paper, it's hard to make comments. However, there is a relationship between Fresnel Integral and Multivariate Gaussian, but I'm not well versed enough to say anything meaningful about it. I simply recall in grad school, my friend research random matrix and thesis was over such relationship.
 
"Fresnel integral" Thank you! that's just what i was looking for, the distrubutions derived for the eigenvalues from random matrices looks less arbitrary now. Last thing,

"data now as a mean of zero with a covariance matrix equal to the identity." are you saying that if mean is zero cov(X) is a diagonal with all dimensions having same magnitude?
 
Namaste & G'day Postulate: A strongly-knit team wins on average over a less knit one Fundamentals: - Two teams face off with 4 players each - A polo team consists of players that each have assigned to them a measure of their ability (called a "Handicap" - 10 is highest, -2 lowest) I attempted to measure close-knitness of a team in terms of standard deviation (SD) of handicaps of the players. Failure: It turns out that, more often than, a team with a higher SD wins. In my language, that...
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Back
Top