5.13. Example of violating NDCQ.

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From the course by National Research University Higher School of Economics

Mathematics for economists

1 rating

National Research University Higher School of Economics

Mathematics for economists

1 rating

This course is an important part of the undergraduate stage in education for future economists. It's also useful for graduate students who would like to gain knowledge and skills in an important part of math. It gives students skills for implementation of the mathematical knowledge and expertise to the problems of economics. Its prerequisites are both the knowledge of the single variable calculus and the foundations of linear algebra including operations on matrices and the general theory of systems of simultaneous equations. Some knowledge of vector spaces would be beneficial for a student. The course covers several variable calculus, both constrained and unconstrained optimization. The course is aimed at teaching students to master comparative statics problems, optimization problems using the acquired mathematical tools. Home assignments will be provided on a weekly basis. The objective of the course is to acquire the students’ knowledge in the field of mathematics and to make them ready to analyze simulated as well as real economic situations. Students learn how to use and apply mathematics by working with concrete examples and exercises. Moreover this course is aimed at showing what constitutes a solid proof. The ability to present proofs can be trained and improved and in that respect the course is helpful. It will be shown that math is not reduced just to “cookbook recipes”. On the contrary the deep knowledge of math concepts helps to understand real life situations.

From the lesson

Constrained optimization for n-dim space. Bordered Hessian.

Week 5 of the Course is devoted to the extension of the constrained optimization problem to the n-dimensional space. This week students will grasp how to apply bordered Hessian concept to classification of critical points arising in different constrained optimization problems.

5.1. Example of the solution of the constrained optimization.6:41

5.2. Weierstrass theorem. Compact sets.7:39

5.3. Bordered Hessian.9:43

5.4. Constrained optimization in general case (part 1).5:31

5.5. Constrained optimization in general case (part 2).6:36

5.6. Application of the bordered Hessian in the constrained optimization.7:51

5.7. Generalization of the constrained optimization problem for the n variables case.8:21

5.8. Example of constrained optimization for the case of more than two variables (part 1).7:11

5.9. Example of constrained optimization for the case of more than two variables (part 2).9:03

5.10. Example of constrained optimization problem on non-compact set.5:31

5.11. Theorem for determining definiteness (positive or negative) or indefiniteness of the bordered matrix.5:44

5.12. Example of application bordered Hessian technique for the constrained optimization problem.8:09

5.13. Example of violating NDCQ.10:33

Meet the Instructors

Kirill Bukin
Associate Professor, Candidate of sciences (phys.-math.)
Faculty of Economic Sciences, Department of Theoretical Economics

The structure of this metric suggests that

the signs or the leading principal minors we need to evaluate,

do not depend on the value of this parameter M itself.

That means we can choose the most comfortable value for M we wish.

The most comfortable value is one,

because one being raised to any power is still one.

So, let M be one,

then our bordered Hessian matrix

becomes a matrix whose entries are just numbers.

We need to evaluate firstly, determinant.

We'll be using the techniques drawn from linear algebra.

Of course when we perform elementary operations on rows of this matrix,

the determinant doesn't change.

For instance, we can take the first row

and subtract the first row from the third and the fourth,

then we get a simpler,

a matrix in terms of determinant calculations.

So this determinant of course,

equals minus.

Once again, we can simplify this matrix by

adding the first row to the second and the third.

We keep the minus sign from here,

and we get two by two,

four minus one, three, minus three negative.

Now, that was the leading principal minor of the biggest size.

Now, we need to find a smaller ordered principal, leading principal minor,

by deleting the final row and the rightmost column and calculate,

let it be delta three,

three means the order of this determinant,

negative two. Now we

need to make a statement based on the size of these leading principal minors.

We deal with the problem where n is three and m is one,

both being odd numbers.

Then we compare the signs of two consecutive leading principal minors, negative negative.

So they keep those signs leading principal minors.

That tells us that the quadratic form in our case,

this is the second order differential of the Lagrangian function is positive definite.

So, that means that the critical point is a point of minimum.

Now, let us proceed with the question,

what happens if NDCQ condition is violated?

I'll show you the importance of checking NDCQ on the basis of such an example.

So, let's consider a problem,

the function of two variables but it actually depends on x

alone is maximized subject to constraint.

What if I forget to check NDCQ condition and proceed with forming Lagrangian immediately?

Setting the system, all three first-order conditions,

and the constraint itself.

How to solve the system?

Usually we start analyzing the system from the simplest of all equations,

looks like the second equation is the simplest.

So, either lambda is zero or y is zero or both,

if lambda is zero,

we have a contradiction with the first equation,

because negative one cannot be zero.

But if y is zero we

substitute the zero value to the third equation and we get also x is zero.

We still have a contradiction,

we get minus one equals zero.

So then, quick answer to this problem would be no solutions,

but this is not true because we can try to solve it again using geometrical approach.

So, we can consider the constraint set in xy-plane.

How to draw that?

So, we can try to figure out.

So, we have two branches of the curve and this is how constraint is set.

Now, when minus x is maximized,

the maximum value will be zero value,

because for any positive x value this is a negative number.

So, the maximum value will be zero.

So, actually this problem has a unique solution.

The solution will be at the origin,

this is the maximum point.

How can we explain this failure?

The thing is that the NDCQ condition is not met at

the origin because both derivatives of the constraint function turns zero at the origin.

What would I recommend in this case?

I mean that well,

clearly it's always desirable to start solving a problem with checking NDCQ.

We can re-consider the solution by checking NDCQ,

and we see that at the origin the NDCQ condition is not met.

This simply tells us that along with the critical points of

the Lagrangian function we may find by solving the first-order conditions.

We also need to check the points where NDCQ is not met.

Exactly what happens here,

so whenever we have

the NDCQ condition violated at some point,

we need to consider the behavior or the objective function on the constraint sets,

in the neighborhood of this particular point.

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