maxminminmax's comments

Yes, but that’s not the relevant datum, because of selection effects. The relevant question is how well employed is the person who had a choice to do a Ph.D. or not compared to the counterfactual person who made an opposite choice.

As an example, an Ivy graduate makes more than state school graduate on average, but there was a study showing that those offered Ivy admission but deciding to go to a state school made just as much (that study setup has its own selection bias issues, but hopefully those gives an idea of what I mean).

*then dg/du = x and dg - xdu =0

It’s ironic to see the link to that post which implicitly assumes that functions are defined on R^n, under a post about Legendre transform, whose point is that functions are defined on state spaces of systems, and only become represented by functions on R^n once we parameterize the state spaces by state variables. So the value of f at a given point doesn’t depend on what your favorite letter (aka state variable) is, but the value of f’ certainly does. And Legendre transform, as is actually explained, albeit cryptically, in Goldstein comes from the fact that we have 2d state space - phase space of 1d system with config variable y, velocity variable x, and momentum variable u, - on which we have have non-linearly related variables x and u.

In Legendre transform, what we have (the y variable is a red herring, and I will ignore it; everything happens "pointwise in y"), is curve in u-x plane, which we lift to u-x-z space in two ways -- that is, we find functions f and g defined for the points on that curve such that: 1) if the curve is parametrized by x, so that f is a function of x, then df/dx=u 2)if the curve is parametrized by u, so that g is a function of u, then dg/dx = u. (Why do we want this? Presumably because when x is velocity and f is energy, u is momentum, and we want g to have same property going back. And yes, there are conditions when one can parametrize a curve by one of the coordinates, either locally, or globally; one such is that u is monotone increasing function of x - that corresponds to convexity of f.) Of course now "derivatives of f and g are inverse" is tautological.

If we already know f(x), but don't know neither u nor g we could set u = df/dx and try to compute g. Or we could do it the way Goldstein does it: dg/du=x, so dg=xdu (this is an ODE), integrating it "by parts" g=int x du = xu - int u dx = xu - f.

(In advanced speak, u-x curve is a Lagrangian in the u-x plane, which is symplectic as every sum of vector space and its dual is; the functions f and g correspond to lifts of this Lagrangian to Legendrians based on choice of "canonical" 1-forms udx and xdu, respectively,so that df - udx=0 and g-xdu=0.)

This is one of the most strawman (to put it mildly) things I have ever read.

>but probabilistically the strategy is optimal.

For what value function? It is basically never the case that my value function is "all choices other than the optimal are equally bad" -- which is what this rule is based on.

As a personal opinion, this drives me up the wall. There is a great problem here, and there is a whole area (several of them, actually!) of applied math dedicated to it (Statistical Decision Theory, Reinforcement Learning, you name it). Instead we get this toy version -- which at best is an oversimplified intro to he subject, and at worst an excuse to bamboozle with math-fairy-dust -- brought out as some kind of rule "to live by". Your algorithm is bad, and you should feel bad.

https://www.amazon.com/Nuts-Bolts-Proofs-Introduction-Mathem... https://www.amazon.com/Art-Craft-Problem-Solving/dp/04717890...

105B/39.51M=2657. Can I just take the money?