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No, the whole idea of the lexicographic preference is that there's no "compensation" here. If bundle y has less of the first good than x does, then it's worse than x -- no matter how much more of the second good y has than x has. That's what the definition of the lexicographic preferences says. If you take away even a tiny amount of the first good, no amount of the second good will compensate for that.

I hope!

You're welcome! I'm glad it was helpful.

I'm glad it was helpful. Thank you for letting me know.

Thanks for the kind words. I'm glad you're finding it helpful.

I'm glad it was helpful.

Great, I'm glad it was helpful.

The problem here is that your point "really close to 0.5, but on the right-hand side of 0.5" is still to the right of 0.5 -- so there are still points (in fact, a continuum of points) between that point and 0.5. It doesn't matter how close "really close" is, that little continuum is still going to be there. So a move to the left will have no effect on the image set *until* the move has traversed that little continuum -- i.e., the leftward move won't cause the image set to blow up "instantly."

I've just posted a link and an explanation.

I've just posted a link and an explanation.

As I mentioned right at the beginning, it's in an earlier lecture. Note that lectures 36(A&B) are titled "Solution Function and Value Function." It's also identified in this lecture (38A) at the 7:38 mark and subsequently.

@@ArizonaMathCamp thank you! i'll check out the previous videos.

For the pre-image of the origin, (0,0): Ax = (0,0). So 2x_1 - x_2 = 0 and also -2x_1 + x_2 = 0. Both equations yield x_2 = 2x_1. For the pre-image of the point (1,-1): Ax = (1,-1). So 2x_1 - x_2 = 1 and -2x_1 + x_2 = -1. Both equations yield x_2 = 2x_1 - 1.

I'm glad it was so helpful, and thanks for the great comment. (Maybe I wouldn't go quite as far as "a great thing for humanity", but I really appreciate the accolade.)

Glad it was helpful!

You're welcome! Glad it was helpful.

That's good to hear! I'm glad they were helpful, and thanks for the positive feedback!

Yes, notation can be tricky. There are lots of things for which different people use different notation. Matrix transposes is just one of them. There are a number of places in my lectures here where I indicate that some people use notation that's different from the notation I'm using.

@@ArizonaMathCamp Thanks for replying. I'm looking forward to checking them out.

The lectures are all on mathematics -- the mathematics you need for studying economics (any economics).

Well, I won't argue with you! Thanks for the positive feedback.

The proof requires that we can do this for *every* x' that's larger than x -- i.e., if x'>x then it has to be the case that (x',0) is preferred to (x,1). That can't happen for every x'>x if the preference is continuous: continuity says the weak upper-contour set of (x,1) is closed, so if a sequence of points (x(n),0) converges to (x,0) and every point in the sequence is even weakly preferred to (x,1), then (x,0) must be as well -- but we know that in fact (x,1) is strictly preferred to (x,0). And of course the representation theorem tells us that if the preference is continuous then there *will* be a utility function for it, and in that case we can't do the construction of the function f(.) in the proof.

Well, I think your own hard work is the real key. But if my videos are helping, that's great! Thanks for the positive feedback!

That's the right idea, but it's not quite as simple as that. An important key to dealing with minimization and with >= constraints is to note that (a) minimizing a function f is identical to maximizing -f, and (b) multiplying both sides of a >= constraint by -1 turns it into a <= constraint. This will enable you to apply the KT Conditions as given, or alternatively to convert them to conditions for minimization and/or >= inequalities. (How has your year gone so far?)

@@ArizonaMathCamp Hey, good to hear from you! Apologies for my delay I am in the peak final exam season for mine, six-year pilgrimage through Econ/Maths. Frustratingly, the KKT conditions are still haunting me. The optimisation course I’m currently using for example does not even note the condition of our Lagrange derivative j being ≤ 0 if xj = 0. It also handles minimisation problems by just subtracting the sum of constraints…I.e. it doesn't multiply both sides by, -1. I just find the KKT conditions are taught in so many different ways!!

The textbook for this current course is built off is - Sundaram, L.R. A First Course in Optimisation Theory. I don't find it that great at times. Do you have a single source of truth for Lagrangians the KKT conditions? With good practice questions for all the edge cases? (obviously your material is great, but doesn't cover constraint qualifications for example).

@@cormackjackson9442 Well, you've already cleared the first hurdle, by recognizing that there is not just one CQ. A CQ is a condition *sufficient* to ensure that the first-order (necessary) conditions will hold at a solution (a maximum). A number of different CQs have been identified, each one sufficient for this. Most treatments give just one CQ, which obscures this multiplicity. I like Sundaram's book a lot, but it has just one, I believe. The only book I can recall (it's not the only one there is, I just don't recall any of the others) that covers multiple CQs is a book by Peter B. Morgan titled "An Explanation of Constrained Optimization." The book provides good intuition, and discusses multiple CQs, with examples.

Thanks for the greetings! I'm glad the video was helpful.

I'm glad it was helpful. Thanks for the positive feedback.

Glad you found it helpful, and many thanks for the positive feedback!

I'm glad it was helpful. Thanks for letting me know.

Thanks for the positive feedback!

Glad it was helpful! Thanks for the positive feedback.

Writing on a glass screen, in the normal way, with camera on the other side, and then flipping the resulting video via software. Watch this: ruclips.net/video/Adm2raocQZ4/видео.html Matt Anderson uses a mirror in order to do a live presentation, but I flip it in post-processing. If an ad comes up at the beginning, click "Skip" in the lower right corner.

@@ArizonaMathCamp Thanks for the explanation. More importantly, thank you for the content. It assisted me with passing my discrete mathematics course. It is much appreciated.

Very glad to hear you passed and that I might've help a bit. Thanks for the positive feedback.

Thanks for the kind words. I can't really give you an answer to your question -- I don't know what math your computer science course includes. Good luck on your exams!

@@ArizonaMathCamp thank you!!

Yes, that's correct. The antecedent in each definition is: for every set V [in the target space] such that f(x-bar) is "in" V [i.e., an element of V for a function; a subset of V for a correspondence]. And the consequent in each case is: there exists a set U [in the domain] ... etc.

I'm not clear what you mean when you say you're going to "replace the part f(x) is a subset of V in the definition of UHC by this implication." In particular, I don't see what you mean by replacing the subset statement with an implication.

That way is OK. The key fact is that if λ is 0 or 1 then the two sides of the inequality are the same, so they're equal. So you do have to use λ ∈ (0,1) for strict concavity; but it doesn't matter which you use for concavity.

@@ArizonaMathCamp Thank you for the videos and for taking the time to reply. It's much appreciated!

You're a little bit confused about open and closed sets. See my Lecture 10. The empty set *is* closed as well as open -- that is not a contradiction. Its complement (the entire set of which the empty set is a subset) is also both open and closed. In other (non-Euclidean) metrics, other sets as well can be both open and closed. In your second example, you've assumed that there is an element in the empty set (which is false), from which you derive that the empty set is infinite, which is also false, thereby showing that the original assumption must have been false -- indeed, a proof by contradiction, or indirect proof.

@@ArizonaMathCamp Thank you! I think I'll need to carefully work through some examples to clear up my confusion. Thanks for the lectures, they're very helpful!

Yes, carefully working through examples is absolutely the best way to gain a solid understanding -- especially if you have someone who can tell you whether or not you're on the right track. Note, by the way, that it's important to recognize that open and closed subsets don't work the same as open and closed doors, windows, etc, where it's binary: a door is either open or closed and never both, but a subset can be neither open nor closed, or both open and closed.

I'm glad this was helpful. Thanks for the positive feedback.

You're welcome. I'm glad it was helpful.

I'm glad this was helpful. Thanks for the positive feedback.