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In the course, in lecture "Reducing Loss: Gradient Descent" is "Convex problems have only one minimum; that is, only one place where the slope is exactly 0.
by graycat 9y ago
In the course, in lecture
"Reducing Loss: Gradient Descent"
is
"Convex problems have only one minimum; that is, only one place where the slope is exactly 0. That minimum is where the loss function converges."
The first sentence is flatly wrong: E.g., for positive integer n and the set of real numbers R, function f: R^n --> R where for all x in R^n f(x) = 0, f is convex, concave, and linear, and for all x in R^n x is a minimum and a maximum of f.
Can there be uncountably infinitely many alternative minima for the Google ML problems? Yes, e.g., just enter one of the independent variables twice.
The second sentence is nonsense.
Grotesque, outrageous incompetence!!!!
It has long been known that minimizing a convex function, even a differentiable convex function, with just gradient descent can be just horribly inefficient. A LOT is known about how to do much better than just gradient descent. E.g., there is Newton iteration (right, that Newton, hundreds of years ago) and quasi-Newton. And there's more.
Why so inefficient? Well, draw a picture like Google did except use just two independent variables instead of just the one in the Google picture. Then see that the resulting, convex "bowl" can be like a long, narrow boat with a very gentle slope in one direction and a very steep slope in an orthogonal direction. Yes the cross section of the bowl can be, first cut, an ellipse with one short axis and one long one. Sure, the axes are eigenvectors, etc. and the ellipse is part of a local quadratic approximation. Well, gradient descent keeps going back and forth nearly parallel to the short axis of the ellipse and making nearly no progress on the long axis. People have known this and known good things to do about it for, uh, at least half a century.
For the Google ML problems, might (1) tweak Newton iteration to improve the rate of convergence to a minimum or (2) at each iteration don't get a gradient descent but get a supporting hyperplane of the epigraph of the convex function, as the iterations proceed, accumulate these hyperplanes, notice that they lead to an approximation of the full epigraph, and use linear programming or some tweak of that to minimize the hyperplane approximation to the convex function -- there is much more that can be said here. By the way, the convex function for that ML problem is quite special, e.g., quadratic.
A lot has long been known and well polished in regression and classification, e.g., with TeX markup:
N.\ R.\ Draper and
H.\ Smith,
{\it Applied Regression Analysis,\/}
John Wiley and Sons,
New York,
1968.\ \
Leo Breiman,
Jerome H.\ Friedman,
Richard A.\ Olshen,
Charles J.\ Stone,
{\it Classification and Regression Trees,\/}
ISBN 0-534-98054-6,
Wadsworth \& Brooks/Cole,
Pacific Grove, California,
1984.\ \
C.\ Radhakrishna Rao,
{\it Linear Statistical Inference and
Its Applications:\ \
Second Edition,\/}
ISBN 0-471-70823-2,
John Wiley and Sons,
New York,
1967.\ \
A good start on convexity is
Wendell H.\ Fleming,
{\it Functions of Several Variables,\/}
Addison-Wesley,
Reading, Massachusetts,
1965.\ \
Total, fun, dessert ice cream on convexity is Jensen's inequality; right away can use it to prove a lot of classic inequalities.
Gee, look on the upside!!!! From this sample, the claims of machine learning (ML) revolutionizing the economy are nonsense!!!! And for startups, don't much have to worry about serious competition from Google!!!!
- stared 9y agohttp://p.migdal.pl/2017/04/30/teaching-deep-learning.html http://p.migdal.pl/2017/04/30/teaching-deep-learning.html -> "What mathematicians thing I do" (Full disclaimer - I did theoretical physics, so understand both sides. :))
- telchar 9y agoYou should recheck your definitions on convexity. >function f: R^n --> R where for all x in R^n f(x) = 0 This hyperplane is not convex. A convex curve by definition can not be equal to its tangent at any point. Edit: I should specify, I mean a convex curve cannot be completely equal to any of its tangents, obviously it will equal each tangent at a single point.
- tzs 9y agoIt sounds like you are describing curves that are strictly convex. Curves that are convex, but not strictly convex, can intersect their tangents at more than one point, or even at every point. I'm going by the definition of convex function given in Rudin's "Principles of Mathematical Analysis", Apostol's "Calculus", Wikipedia, and MathWorld.
- graycat 9y agoYou don't want to consider just "tangents" and, instead, consider what I defined as supporting hyperplanes of the epigraph and subgradients of the function. If the gradient exists, that is, if the function is differentiable, then the subgradient really is a tangent. Otherwise can have many different subgradients supporting at one point on the curve and its epigraph. It's simple: A cube has supporting planes at each point that is an edge or corner, but those points do not have tangents.