Starr’s ’69 paper considered Walrasian equilibria in exchange economies with non-convex preferences i.e., upper contour sets of utility functions are non-convex. Suppose ${n}$ agents and ${m}$ goods with ${n \geq m}$. Starr identified a price vector ${p^*}$ and a feasible allocation with the property that at most ${m}$ agents did not receiving a utility maximizing bundle at the price vector ${p^*}$.

A poetic interlude. Arrow and Hahn’s book has a chapter that describes Starr’s work and closes with a couple of lines of Milton:

A gulf profound as that Serbonian Bog
Betwixt Damiata and Mount Casius old,
Where Armies whole have sunk.

Milton uses the word concave a couple of times in Paradise Lost to refer to the vault of heaven. Indeed the OED lists this as one of the poetic uses of concavity.

Now, back to brass tacks. Suppose ${u_i}$ is agent ${i}$‘s utility function. Replace the upper contour sets associated with ${u_i}$ for each ${i}$ by its convex hull. Let ${u^*_i}$ be the concave utility function associated with the convex hulls. Let ${p^*}$ be the Walrasian equilibrium prices wrt ${\{u^*_i\}_{i=1}^n}$. Let ${x^*_i}$ be the allocation to agent ${i}$ in the associated Walrasian equilibrium.

For each agent ${i}$ let

$\displaystyle S^i = \arg \max \{u_i(x): p^* \cdot x \leq p^*\cdot e^i\}$

where ${e^i}$ is agent ${i}$‘s endowment. Denote by ${w}$ the vector of total endowments and let ${S^{n+1} = \{-w\}}$.

Let ${z^* = \sum_{i=1}^nx^*_i - w = 0}$ be the excess demand with respect to ${p^*}$ and ${\{u^*_i\}_{i=1}^n}$. Notice that ${z^*}$ is in the convex hull of the Minkowski sum of ${\{S^1, \ldots, S^n, S^{n+1}\}}$. By the Shapley-Folkman-Starr lemma we can find ${x_i \in conv(S^i)}$ for ${i = 1, \ldots, n}$, such that ${|\{i: x_i \in S^i\}| \geq n - m}$ and ${0 = z^* = \sum_{i=1}^nx_i - w}$.

When one recalls, that Walrasian equilibria can also be determined by maximizing a suitable weighted (the Negishi weights) sum of utilities over the set of feasible allocations, Starr’s result can be interpreted as a statement about approximating an optimization problem. I believe this was first articulated by Aubin and Elkeland (see their ’76 paper in Math of OR). As an illustration, consider the following problem :

$\displaystyle \max \sum_{j=1}^nf_j(y_j)$

subject to

$\displaystyle Ay = b$

$\displaystyle y \geq 0$

Call this problem ${P}$. Here ${A}$ is an ${m \times n}$ matrix with ${n > m}$.

For each ${j}$ let ${f^*_j(\cdot)}$ be the smallest concave function such that ${f^*_j(t) \geq f_j(t)}$ for all ${t \geq 0}$ (probably quasi-concave will do). Instead of solving problem ${P}$, solve problem ${P^*}$ instead:

$\displaystyle \max \sum_{j=1}^nf^*_j(y_j)$

subject to

$\displaystyle Ay = b$

$\displaystyle y \geq 0$

The obvious question to be answered is how good an approximation is the solution to ${P^*}$ to problem ${P}$. To answer it, let ${e_j = \sup_t [f_j^*(t) - f_j(t)]}$ (where I leave you, the reader, to fill in the blanks about the appropriate domain). Each ${e_j}$ measures how close ${f_j^*}$ is to ${f_j}$. Sort the ${e_j}$‘s in decreasing orders. If ${y^*}$ is an optimal solution to ${P^*}$, then following the idea in Starr’s ’69 paper we get:

$\displaystyle \sum_{j=1}^nf_j(y^*_j) \geq \sum_{j=1}^nf^*_j(y^*_j)- \sum_{j=1}^me_j$

Here is the question from Ross’ book that I posted last week

Question 1 We have two coins, a red one and a green one. When flipped, one lands heads with probability ${P_1}$ and the other with probability ${P_2}$. Assume that ${P_1>P_2}$. We do not know which coin is the ${P_1}$ coin. We initially attach probability ${p}$ to the red coin being the ${P_1}$ coin. We receive one dollar for each heads and our objective is to maximize the total expected discounted return with discount factor ${\beta}$. Find the optimal policy.

This is a dynamic programming problem where the state is the belief that the red coin is ${P_1}$. Every period we choose a coin to toss, get a reward and updated our state given the outcome. Before I give my solution let me explain why we can’t immediately invoke uncle Gittins.

In the classical bandit problem there are ${n}$ arms and each arm ${i}$ provides a reward from an unknown distribution ${\theta_i\in\Delta([0,1])}$. Bandit problems are used to model tradeoffs between exploitation and exploration: Every period we either exploit an arm about whose distribution we already have a good idea or explore another arm. The ${\theta_i}$ are randomized independently according to distributions ${\mu_i\in \Delta(\Delta([0,1]))}$, and what we are interested in is the expected discounted reward. The optimization problem has a remarkable solution: choose in every period the arm with the largest Gittins index. Then update your belief about that arm using Bayes’ rule. The Gittins index is a function which attaches a number ${G(\mu)}$ (the index) to every belief ${\mu}$ about an arm. What is important is that the index of an arm ${i}$ depends only on ${\mu_i}$ — our current belief about the distribution of the arm — not on our beliefs about the distribution of the other arms.

The independence assumption means that we only learn about the distribution of the arm we are using. This assumption is not satisfied in the red coin green coin problem: If we toss the red coin and get heads then the probability that the green coin is ${P_1}$ decreases. Googling multi-armed bandit’ with dependent arms’ I got some papers which I haven’t looked at carefully but my superficial impression is that they would not help here.

Here is my solution. Call the problem I started with the difficult problem’ and consider a variant which I call the easy problem’. Let ${r=p/(p+\sqrt{p(1-p)}}$ so that ${r^2/(1-r)^2=p/1-p}$. In the easy problem there are again two coins but this time the red coin is ${P_1}$ with probability ${r}$ and ${P_2}$ with probability ${1-r}$ and, independently, the green coin is ${P_1}$ with probability ${(1-r)}$ and ${P_2}$ with probability ${r}$. The easy problem is easy because it is a bandit problem. We have to keep track of beliefs ${p_r}$ and ${p_g}$ about the red coin and the green coin (${p_r}$ is the probability that the red coin is ${P_1}$), starting with ${p_r={r}}$ and ${p_g=(1-r)}$, and when we toss the red coin we update ${p_r}$ but keep ${p_g}$ fixed. It is easy to see that the Gittins index of an arm is a monotone function of the belief that the arm is ${P_1}$ so the optimal strategy is to play red when ${p_r\ge p_g}$ and green when ${p_g\ge p_r}$. In particular, the optimal action in the first period is red when ${p\ge 1/2}$ and green when ${p\le 1/2}$.

Now here comes the trick. Consider a general strategy ${g}$ that assigns to every finite sequence of past actions and outcomes an action (red or green). Denote by ${V_d(g)}$ and ${V_e(g)}$ the rewards that ${g}$ gives in the difficult and easy problems respectively. I claim that

$\displaystyle \begin{array}{rcl} &V_e(g)=r(1-r) \cdot P_1/(1-\beta)+ \\ &r(1-r) \cdot P_2/(1-\beta) + (r^2+(1-r)^2) V_d(g).\end{array}$

Why is that ? in the easy problem there is a probability ${r(1-r)}$ that both coins are ${P_1}$. If this happens then every ${g}$ gives payoff ${P_1/(1-\beta)}$. There is a probability ${r(1-r)}$ that both coins are ${P_2}$. If this happens then every ${g}$ gives payoff ${P_2/(1-\beta)}$. And there is a probability ${r^2+(1-r)^2}$ that the coins are different, and, because of the choice of ${r}$, conditionally on this event the probability of ${G}$ being ${P_1}$ is ${p}$. Therefore, in this case ${g}$ gives whatever ${g}$ gives in the difficult problem.

So, the payoff in the easy problem is a linear function of the payoff in the difficult problem. Therefore the optimal strategy in the difficult problem is the same as the optimal strategy in the easy problem. In particular, we just proved that, for every ${p}$, the optimal action in the first period is red when ${p\ge 1/2}$ and green with ${p\le 1/2}$. Now back to the dynamic programming formulation, from standard arguments it follows that the optimal strategy is to keep doing it forever, i.e., at every period to toss the coin that is more likely to be the ${P_1}$ coin given the current information.

See why I said my solution is tricky and specific ? it relies on the fact that there are only two arms (the fact that the arms are coins is not important). Here is a problem whose solution I don’t know:

Question 2 Let ${0 \le P_1 < P_2 < ... < P_n \le 1}$. We are given ${n}$ coins, one of each parameter, all ${n!}$ possibilities equally likely. Each period we have to toss a coin and we get payoff ${1}$ for Heads. What is the optimal strategy ?

It states that the Minkowski sum of a large number of sets is approximately convex. The clearest statement  as well as the nicest proof  I am familiar with is due to J. W. S. Cassels. Cassels is a distinguished number theorist who for many years taught the mathematical economics course in the Tripos. The lecture notes  are available in a slender book now published by Cambridge University Press.

This central limit like quality of the lemma is well beyond the capacity of a hewer of wood like myself. I prefer the more prosaic version.

Let ${\{S^j\}_{j=1}^n}$ be a collection of sets in ${\Re ^m}$ with ${n > m}$. Denote by ${S}$ the Minkowski sum of the collection ${\{S^i\}_{i=1}^n}$. Then, every ${x \in conv(S)}$ can be expressed as ${\sum_{j=1}^nx^j}$ where ${x^j \in conv(S^j)}$ for all ${j = 1,\ldots, n}$ and ${|\{j: x^j \not \in S^j| \leq m}$.

How might this be useful? Let ${A}$ be an ${m \times n}$ 0-1 matrix and ${b \in \Re^m}$ with ${n > m}$. Consider the problem

$\displaystyle \max \{cx: Ax = b, x_j \in \{0,1\}\ \forall \,\, j = 1, \ldots, n\}.$

Let ${x^*}$ be a solution to the linear relaxation of this problem. Then, the lemma yields the existence of a 0-1 vector ${x}$ such that ${cx \geq cx^* = z}$ and ${||Ax - b||_{\infty} \leq m}$. One can get a bound in terms of Euclidean distance as well.

How does one do this? Denote each column ${j}$ of the ${A}$ matrix by ${a^j}$ and let ${d^j = (c_j, a^j)}$. Let ${S^j = \{d^j, 0\}}$. Because ${z = cx^*}$ and ${b = Ax^*}$ it follows that ${(z,b) \in conv(S)}$. Thus, by the Lemma,

$\displaystyle (z, b) = \sum_{j=1}^n(c_j, a^j)y_j$

where each ${y_j \in [0,1]}$ and ${|\{j: y_j \in (0,1) \}| \leq m }$. In words, ${y}$ has at most ${m}$ fractional components. Now construct a 0-1 vector ${y^*}$ from ${y}$ as follows. If ${y_j \in \{0,1\}}$, set ${y^*_j = y_j}$. If ${y_j }$ is fractional, round ${y^*_j}$ upto 1 with probability ${y_j}$ and down to zero otherwise. Observe that ${||Ay - b||_{\infty} \leq m}$ and the ${E(cy) = cx^*}$. Hence, there must exist a 0-1 vector ${x}$ with the claimed properties.

The error bound of ${m}$ is to large for many applications. This is a consequence of the generality of the lemma. It makes no use of any structure encoded in the ${A}$ matrix. For example, suppose $x^*$ were an extreme point and $A$ a totally unimodular matrix. Then, the number of fractional components of $x^*$ are zero. The rounding methods of Kiralyi, Lau and Singh as well as of Kumar, Marathe, Parthasarthy and Srinivasan exploit the structure of the matrix. In fact both use an idea that one can find in Cassel’s paper. I’ll follow the treatment in Kumar et. al.

As before we start with ${x^*}$. For convenience suppose ${0 < x^*_j < 1}$ for all ${j = 1, \ldots, n}$. As ${A}$ as has more columns then rows, there must be a non-zero vector ${r}$ in the kernel of ${A}$, i.e., ${Ar = 0}$. Consider ${x + \alpha r}$ and ${x -\beta r}$. For ${\alpha > 0}$ and ${\beta > 0}$ sufficiently small, ${x_j + \alpha r_j, x_j - \beta r_j \in [0,1]}$ for all ${j}$. Increase ${\alpha}$ and ${\beta}$ until the first time at least one component of ${x +\alpha r}$ and ${x- \beta r}$ is in ${\{0,1\}}$. Next select the vector ${x + \alpha r}$ with probability ${\frac{\beta}{\alpha + \beta}}$ or the vector ${x- \beta r}$ with probability ${\frac{\alpha}{\alpha + \beta}}$. Call the vector selected ${x^1}$.

Now ${Ax^1 = b}$. Furthermore, ${x^1}$ has at least one more integer component than ${x^*}$. Let ${J = \{j: x^1_j \in (0,1)\}}$. Let ${A^J}$ be the matrix consisting only of the columns in ${J}$ and ${x^1(J)}$ consist only of the components of ${x^1}$ in ${J}$. Consider the system ${A^Jx^1(J) = b - \sum_{j \not \in J}x^1_j}$. As long as ${A^J}$ has more columns then rows we can repeat the same argument as above. This iterative procedure gives us the same rounding result as the Lemma. However, one can do better, because it may be that even when the number of columns of the matrix is less than the number of rows, the system may be under-determined and therefore the null space is non-empty.

In a sequel, I’ll describe an optimization version of the Lemma that was implicit in Starr’s 1969 Econometrica paper on equilibria in economies with non-convexities.

Economists, I told my class, are the most empathetic and tolerant of people. Empathetic, as they learnt from game theory, because they strive to see the world through the eyes of others. Tolerant, because they never question anyone’s preferences. If I had the  talent I’d have broken into song with a version of Why Can’t a Woman be More Like a Man’ :

Psychologists are irrational, that’s all there is to that!
Their heads are full of cotton, hay, and rags!
They’re nothing but exasperating, irritating,
vacillating, calculating, agitating,
Maddening and infuriating lags!

Why can’t a psychologist be more like an economist?

Back to earth with preference orderings. Avoided  the word rational to describe the restrictions placed on preference orderings, used consistency’ instead. More neutral and conveys the idea that inconsistency makes prediction hard rather that suggesting a Wooster like IQ. Emphasized that utility functions were simply a succinct representation of consistent preferences and had no meaning beyond that.

In a bow to tradition went over the equi-marginal principle, a holdover from the days when economics students were ignorant of multivariable calculus. Won’t do that again. Should be banished from the textbooks.

Now for some meat: the income and substitution (I&S) effect. Had been warned this was tricky. No shirt Sherlock,’ my students might say. One has to be careful about the set up.

Suppose price vector $p$ and income $I$. Before I actually purchase anything, I contemplate what I might purchase to maximize my utility. Call that $x$.
Again, before I purchase $x$, the price of good 1 rises. Again, I contemplate what I might consume. Call it $z$. The textbook discussion of the income and substitution effect is about the difference between $x$ and $z$.

As described, the agent has not purchased $x$ or $z$. Why this petty foggery? Suppose I actually purchase $x$ before the price increase. If the price of good 1 goes up, I can resell it. This is both a change in price and income, something not covered by the I&S effect.

The issue is resale of good 1. Thus, an example of an I&S effect using housing should distinguish between owning vs. renting. To be safe one might want to stick to consumables. To observe the income effect, we would need a consumable that sucks up a largish’ fraction of income. A possibility is low income consumer who spends a large fraction on food.

Here is the bonus question from the final exam in my dynamic optimization class of last semester. It is based on problem 8 Chapter II in Ross’ book Introduction to stochastic dynamic programming. It appears there as guess the optimal policy’ without asking for proof. The question seems very natural, but I couldn’t find any information about it (nor apparently the students). I have a solution but it is tricky and too specific to this problem. I will describe my solution next week but perhaps somebody can show me a better solution or a reference ?

We have two coins, a red one and a green one. When flipped, one lands heads with probability P1 and the other with probability P2. We do not know which coin is the P1 coin. We initially attach probability p to the red coin being the P1 coin. We receive one dollar for each heads and our objective is to maximize the total expected discounted return with discount factor β. Describe the optimal policy, including a proof of optimality.

Bertrand, Cournot and Hotelling was on the menu. In addition to covering the mechanics of computing equilibria, spent time trying to motivate each model. Cournot is, I think, the hardest. The short version of the story I gave was this. Firms choose their capacities/quantities, then go to a middleman (should have said platform, so much sexier these days) who auctions of their joint output via a uniform price auction. Wholesale electricity markets come close to this story, which allows one to use Cournot to convey the idea of supply (demand) reduction and the scandal of the California power markets.

Hotelling is easier to motivate, and is a useful vehicle to illustrate why they should always break’ a model to learn something about it. In the standard Hotelling setup, no reservation price is specified for the buyers. Now, allow the two firms to merge and act like a monopolist. The monopolist’s profit function is unbounded! However, you can still write down a first order condition and solve it. Thus, it is also a useful reminder of the dangers of blindly differentiating and setting to zero.

Contrasted Cournot with Hotelling, for example, the effect on consumer surplus when a merger results in a cost reduction for the merged firm. Also provided an opportunity to remind the class about monopoly and evaluating consumer surplus.

Concluded the module on imperfect competition by applying what  had been discussed to Amazon vs. Apple vs the publishers. Another opportunity to walk down memory lane with double marginalization and then add a wrinkle involving competition in the downstream market.

Nicolas Copernicus’s de Revolutionibus, in which he advocated his theory that Earth revolved ar ound the sun, was first printed just before Copernicus’ death in 1543. It therefore fell on one Andreas Osiander to write the introduction. Here is a passage from the introduction:

[An astronomer] will adopt whatever suppositions enable [celestial] motions to be computed correctly from the principles of geometry for the future as well as for the past…. These hypotheses need not be true nor even probable. On the contrary, if they provide a calculus consistent with the observations, that alone is enough.

In other words, the purpose of the astronomer’s study is to capture the observed phenomena — to provide an analytic framework by which we can explain and predict what we see when we look at the sky. It turns out that it is more convenient to capture the phenomena by assuming that Earth revolved around the sun than by assuming, as the Greek astronomers did, a geo-centric epicyclical planet motion. Therefore let’s calculate the right time for Easter by making this assumption. As astronomers, we shouldn’t care whether this is actually true.

Whether or not Copernicus would have endorsed this approach is disputable. What is certain is that his book was at least initially accepted by the Catholic Church whose astronomers have used Copernicus’ model to develop the Gregorian Calendar. (Notice I said the word model btw, which is probably anachronistic but, I think, appropriately captures Osiander’s view). The person who caused the scandal was Galileo Galilei, who famously declared that if earth behaves as if it moves around the sun then, well, it moves around the sun. Yet it moves. It’s not a model, it’s reality. Physicists’ subject matter is the nature, not models about nature.

What about economists ? Econ theorists at least don’t usually claim that the components of their modeling of economic agents (think utilities, beliefs, discount factors, ambiguity aversions) correspond to any real elements of the physical world or of the cognitive process that the agent performs. When we say that Adam’s utility from apple is log(c) we don’t mean that Adam knows anything about logs. We mean — wait for it — that he behaves as if this is his utility, or, as Osiander would have put it, this utility provides a calculus consistent with the observations, and that alone is enough.

The contrast between theoretical economists’ as if’ approach and physicists’ and yet it moves’ approach is not as sharp as I would like it to be. First, from the physics side, modern interpretations of quantum physics view it, and by extension the entire physics enterprise, as nothing more than a computational tool to produce predictions. On the other hand, from the economics side, while I think it is still customary to pay lip service to the as if’ orthodoxy at least in decision theory classes, I don’t often hear it in seminars. And when neuro-economists claim to localize the decision making process in the brain they seem to view the components of the model as more than just mathematical constructions.

Yep, I am advertising another paper. Stay tuned :)

Uber posts a price ${p}$ per ride and keeps a commission ${\alpha}$ on the price. Suppose Uber is the only ride matching service in town. If ${D(p)}$ is the demand function for rides at per ride price ${p}$ and ${S(w)}$ is the supply curve for drivers at wage ${w}$ per ride, Uber must choose ${\alpha}$ and ${p}$ to solve the following:

$\displaystyle \max_{\alpha, p} \alpha p D(p)$

subject to

$\displaystyle D(p) \leq S((1-\alpha)p)$

The last constraint comes from the assumption that Uber is committed to ensuring that every rider seeking a ride at the posted price gets one.

Suppose, Uber did not link the payment to driver to the price charged to rider in this particular way. Then, Uber would solve

$\displaystyle \max_{p,w} pD(p) - wS(w)$

subject to

$\displaystyle D(p) \leq S(w)$

The first optimization problem is clearly more restrictive than the second. Hence, the claim that Uber is not profit maximizing. Which raises the obvious puzzle, why is Uber using a revenue sharing scheme?

Sydney Afriat arrived in Purdue in the late 60′s with a Bentley in tow. Mort Kamien described him as having walked out of the pages of an Ian Flemming novel. Why he brought the Bentley was a puzzle, as there were no qualified mechanics as far as the eye could see. In Indiana, that is a long way. Afriat would take his Bentley on long drives only to be interrupted by mechanical difficulties that necessitated the Bentley being towed to wait for parts or specialized help.

I came upon Afriat when I learnt about the problem of rationalizability.  One has a model of choice and a collection of observations about what an agent selected. Can one rationalize the observed choices by the given model of choice? In Afriat’s seminal paper on the subject, the observations consisted of price-quantity pairs for a vector of goods and a budget. The goal was to determine if the observed choices were consistent with an agent maximizing a concave utility function subject to the budget constraint. Afriat’s paper has prompted many other papers asking the same question for different models of choice. There is an aspect of these papers, including Afriat’s, that I find puzzling.

To illustrate, consider rationalizing expected utility (Eran Shmaya suggested that expected consumption’ might be more accurate). Let ${S = \{1,2 \ldots, n\}}$ be the set of possible states. We are given a sequence of observations ${\{x^{i},p^{i}\}_{i=1}^{m}}$ and a single budget ${b}$. Here ${x^i_j}$ represents consumption in state ${j}$ and ${p^i_j}$ is the unit price of consumption in state ${j}$ in observation ${i}$. We want to know if there is a probability distribution over states, ${v=(v_{1},...,v_{n})}$, such that each ${(x^i, p^i)}$ maximizes expected utility. In other words, ${(x^i, p^i)}$ solves

$\displaystyle \max \sum_{j=1}^{n}v_{j}x^i_{j}$

subject to

$\displaystyle \sum_{j=1}^{n}p^i_{j}x_{j}\leq b$

$\displaystyle x^i_{j}\geq 0\,\,\forall j \in S$

The solution to the above program is obvious. Identify the variable with the largest objective coefficient to constraint ratio and make it as large as possible. It is immediate that a collection of observations ${\{x^{i},p^{i}\}_{i=1}^{m}}$ can be rationalized by a suitable set ${\{v_{j}\} _{j=1}^{n}}$ of non-zero and nonnegative ${v_{j}}$‘s if the following system has a feasible solution:

$\displaystyle \frac{v_{r}}{p^i_r}\geq \frac{v_{j}}{p^i_{j}} \,\,\forall j, \,\, x^i_r> 0$

$\displaystyle \sum_{j \in S}v_{j}=1$

$\displaystyle v_{j}\geq 0\,\,\forall j \in S$

This completes the task as formulated by Afriat. A system of inequalities has been identified, that if feasible means the given observations can be rationalized. How hard is this to do in other cases? As long as the model of choice involves optimization and the optimization problem is well behaved in that first order conditions, say, suffice to characterize optimality, its a homework exercise. One can do this all day, thanks to Afriat; concave, additively separable concave, etc. etc.

Interestingly, no rationalizability paper stops at the point of identifying the inequalities. Even Afriat’s paper goes a step farther and proceeds to characterize’ when the observations can be rationalized. But, feasibility of the inequalities themselves is just such a characterization. What more is needed?

Perhaps, the characterization involving inequalities lacks interpretation’. Or, if the given system for a set of observations was infeasible, we may be interested in the obstacle to feasibility. Afriat’s paper gave a characterization in terms of the strong axiom of revealed preference, i.e., an absence of cycles of certain kinds. But that is precisely the Farkas alternative to the system of inequalities identified in Afriat. The absence of cycles condition follows from the fact that the initial set of inequalities is associated with the problem of finding a shortest path (see the chapter on rationalizability in my mechanism design book). Let me illustrate with the example above. It is equivalent to finding a non-negative and non trivial solution to

$\displaystyle \frac{v_{r}}{v_j}\geq \frac{p^i_{r}}{p^i_{j}} \,\,\forall j, \,\, x^i_r> 0$

Take logs:

$\displaystyle \ln{v_r} - \ln{v_j} \geq \ln{\frac{p^i_{r}}{p^i_{j}}} \,\,\forall j, \,\, x^i_r> 0$

This is exactly the dual to the problem of finding a shortest path in a suitable network (I believe that Afriat has a paper, that I’ve not found, which focuses on systems of the form $b_{rs} >$ $x_s - x_r$ ).The cycle characterization would involve products of terms like ${\frac{p^i_{r}}{p^i_{j}}}$ being less than 1 (or greater than 1 depending on convention). So, what would this add?

Completed what I wanted about monopoly and launched into imperfect competition and introduced the nash equilibrium. I follow the set up in the chapter of pricing from my pricing book with Lakshman Krishnamurthi. The novelty, if any, is to start with Bertrand competition, add capacity and then differentiation. I do this to highlight the different forces at play so that they are not obscured by the algebra of identifying reaction functions and finding where they cross. We’ll get to those later on. Midterm Day, 12. I am, as Enoch Powell once remarked in another, unattractive context,

…. filled with foreboding. Like the Roman, I seem to see “the River Tiber foaming with much blood”.

Much of my energy has been taken up with designing homework problems and a midterm exam on monopoly. Interesting questions are hard to come by. Those lying around make me want to make gnaw my feet off. I started with the assumption, which I may live to regret, that my students are capable of the mechanical and have good memories. The goal, instead, is to get them to put what they have learnt in class to use. Here is an example. Two upstream suppliers, A and B, who each supply an input to a Retailer. The Retailer is characterized by a production function that tells you how much output it generates from the inputs supplied by A and B as well as a demand curve for the final product. Fix the price set by B to, w, say. Now compute the price that A should charge to maximize profit. Its double marginalization with a twist. Suppose the inputs are substitutes for each other. If B raises its price above w what effect will that have on A’s profits? There are two effects. The retailers costs will go up of course, so reducing its output. However, A will retain a larger share of the smaller output. Which will be bigger? Its a question that requires them to put various pieces together. I’ve had them work up to it by solving the various pieces under different guises in different problems. Am I expecting too much? I’ll find out after the midterm. Yet, I cannot see anyway around having questions like this. What is the point of the mathematics we require them to use and know if we don’t ask them to apply it when blah-blah alone is insufficient?

I now have a modest store of such problems, but coming up with them has been devilish hard. Working out the solutions is painful, because it involves actual algebra and one cannot afford errors (on that account I’m behind the curve). To compound matters, one is unable to recycle the exam and homework problems given various sharing sites. I begin to regret not making them do just algebra.