## Treating Markets Mechanically – An Example

April 27, 2011

The aim of this post is to provide the transition from time-independence to time-dependence within a simple economic model for further reference.

For that purpose we consider a single consumer-worker. This agent obeys a time constraint on labour L and free time F

$L + F = 1.$

We introduce a utility function U as

$U = C^\alpha F^{1-\alpha}$

for a given $0< \alpha<1$. There is a budget constraint W given as price p times consumption C equals wage rate w  times labour L.

$p C = w L.$

The agent now maximizes $U$ such that $p C + w F = w$. Let us solve that. Lagrange equations yield

$-\frac{\partial U}{\partial C} = \lambda \frac{\partial W}{\partial C}$

and

$-\frac{\partial U}{\partial F} = \lambda \frac{\partial W}{\partial F}$

with constraint

$W = p C + w F -w = 0.$

Thus

$\frac{p}{w}=\frac{\alpha C^{\alpha-1} F^{1-\alpha}}{(1-\alpha) C^\alpha F^{-\alpha}}=\frac{\alpha F}{(1-\alpha) C}.$

Solving that for C and plugging it into the budget constraint yields

$\frac{\alpha}{(1-\alpha)}w F+w F -w =0$.

Solving this for F and again using the budget constraint shows that

$F=1-\alpha$

and

$C=\alpha\frac{w}{p}$

solves the maximization problem. So far there is no time evolution. To introduce such a dynamics we mimic mechanics and set $C=C(p,\dot{p})$. Demand is a function of price and its derivative. For economists the $\dot{p}$ comes from nowhere. Especially since it is not obvious at all how to define the derivative of a price evolution. For now it has to suffice that eventually we shall understand the derivative in a distributional sense and until then we treat it as a formal parameter.

The time-dependent utility function for the consumer-worker

$U = C^\alpha F^{1-\alpha}r^t$

for a discount rate $0. The agent now maximizes $\int_0^T U d t$ under the constraint $p C + w F = w$.

We make the following assumption due to S. Smale (for excess demand):

$\dot{p}=C \textnormal{ and }\dot{w}=L.$

Euler-Lagrange equations yield

$\frac{d}{d t}\frac{\partial U}{\partial C} = \lambda \frac{\partial W}{\partial C}$

and

$\frac{d}{d t}\frac{\partial U}{\partial L} = \lambda \frac{\partial W}{\partial L}$

with constraint

$W = p C - w L = 0.$

For the Lagrange multipliers we get

$\lambda =-p^{-1}\alpha r^t C^{\alpha-2}F^{-\alpha}((\alpha-1)(C\dot{F}-F\dot{C})-C F \log r)$

and

$\lambda =w^{-1}(\alpha-1) r^t C^{\alpha-1}F^{-1-\alpha}(\alpha(C\dot{F}-F\dot{C})-C F \log r).$

Equating, plugging in the constraint and dividing by $C F$ yields

$\frac{\dot{C}}{C}-\frac{\dot{F}}{F}= \frac{(\alpha w - p C)\log r}{\alpha(1-\alpha)w}$

First we discuss the case $r=1$. Then

$\frac{\dot{C}}{C}=\frac{\dot{F}}{F}$

and thus (consider $\frac{d}{d t}\ln C$) there is a positive, constant K such that C = K F and we get because of the budget constraint

$F = \frac{w}{p K + w}, C = \frac{w K }{p K + w}.$

The constant $K(p,w,\alpha)$ is unique and maximizes $\int_0^T C^\alpha F^{1-\alpha} ds = \int_0^T \frac{w K^\alpha}{p K + w} ds$. In equilibrium we have $p = p^*$ and $w = w^*$. Maximizing  K yields $\frac{d}{d K}\frac{w^* K^\alpha}{p^* K + w^*}= 0$ and thus $K=\frac{\alpha w^*}{(1-\alpha) p^*}$. Now

$F = 1-\alpha$

and

$C = \alpha \frac{w^*}{p^*}.$

The case $r<1$. In equilibrium $\dot{C}=\dot{F}=0$ we immediately obtain $C=\alpha \frac{w^*}{p^*}$. Plugging this into the budget constraint yields $F=1-\alpha$.

Interestingly enough, we get an equilibrium equal to the solution of the time-independent model. How justified is S. Smale’s assumption $C=\dot{p}$? Economists often use linear demand theory and set $C=T-p$. Both approaches seem to be incompatible and both have a draw back. When you scale prices (e.g. by introducing a new currency) demand should stay the same. This is not the case in both settings. One needs currency dependent constants that scale accordingly to fix that. One possibility to avoid that is $C=\frac{\dot{p}}{p}$. As usual, more options do not improve clarity and calculating the whole model in the general case, i.e. $C=C(p,\dot{p})$ is not totally conclusive either. For a solution of the Euler-Lagrange equations one obtains under moderate assumptions on the partial derivatives that

$\alpha (1-\alpha)w\frac{\partial C}{\partial \dot{p}}\left(\frac{\dot{C}}{C}-\frac{\dot{F}}{F}\right) = (\alpha w - p C)\left(\frac{\partial C}{\partial \dot{p}}\log r + \frac{d}{d t}\frac{\partial C}{\partial \dot{p}}-\frac{\partial C}{\partial p}\right).$

Linear demand has $\frac{\partial C}{\partial \dot{p}}=0$ and thus $\alpha w - p C=0$. The budget constraint implies $F=1-\alpha$ which is a constant. We thus can safely exclude linear demand from our considerations. The above equation cannot distinguish between Smale’s assumption and $C=\frac{\dot{p}}{p}$. However, hidden in the technical assumptions, there seems to be some advantage in Smale’s approach. It remains to clarify the price-scaling issue.

## Polymath5 – A collection of results concerning the completely multiplicative case

March 23, 2011

In this post I have collected results of the fifth Polymath project on the Erdös discrepancy problem. With hundreds and maybe even thousands of posts I have decided to restrict this to the (maybe) simplest non-trivial special case, namely completely multiplicative functions with values in ${\{\pm 1\}}$. That has some justification since there are results reducing the problem to the completely multiplicative case (albeit with values in ${S_1}$). Furthermore my focus is on elementary examples related to the problem and how far one can get with textbook methods. It that sense this can be seen as a beginner’s guide with the quoted results being nothing more than exercises. I have arranged the results within the following sections

• Statement of the problem
• Examples related to the problem
• Necessary conditions for bounded discrepancy
• Discrepancy results

What fascinates me about the problem is that it can be stated without much technicalities and that a solution seems to be within reach due to recent results in number theory.

Discrepancy Problem. Find or prove the non-existence of a function ${f:\mathbb{N}\rightarrow\{\pm 1\}}$ with the following properties:

• ${f}$ is completely multiplicative, i.e. ${f(m n)=f(m) f(n) }$ for all ${m,n\in\mathbb{N}}$.
• ${f}$ has bounded partial sums (bounded discrepancy), i.e. there is a ${C>0}$ such that ${\left|\sum_{n\leq x}f(n)\right|\leq C}$ for all ${x\in\mathbb{R}}$.

Let me fix some notation before I present examples related to the above question. In the following ${f}$ denotes a completely multiplicative function with values in ${\{\pm 1\}}$, ${p,q}$ denote primes and ${s=\sigma + i t}$ is a complex number with real part ${\sigma}$ and imaginary part ${t}$.

Examples related to the problem.

• ${f(n)=1}$ for all ${n\in\mathbb{N}}$ is completely multiplicative and has unbounded discrepancy ${\sum_{n\leq x}1=\lfloor x\rfloor}$.
• Liouville function ${\lambda(p):=-1}$ for all primes ${p}$.
• Dirichlet characters of modulus ${k}$ are completely multiplicative with values in ${\{0\}\cup\{e^{2 \pi i n / k}:1\leq n \leq k\}}$. Since Dirichlet characters assume the value ${0}$ they are strictly speaking not in the focus of this post. However, they can be used to construct ${\{\pm 1\}}$-valued examples (cf. the next example) and they play a central role in the discussions.
• The functions $\displaystyle \mu_p(q):=\left\{ {(\frac{q}{p}) \textnormal{ if } q\not= p} \atop -1 {\textnormal{ if } q=p}\right.$ and $\displaystyle \lambda_p(q):=\left\{ {(\frac{q}{p}) \textnormal{ if } q\not= p} \atop 1 {\textnormal{ if } q=p}\right.$with ${(\frac{q}{p})}$ being the quadratic Dirichlet character mod ${p}$ (Legendre symbol) and ${p}$ an odd prime.
• To generalize the last example fix ${m \in \mathbb{N}}$ and a set ${S\subseteq \{1,\ldots, m-1\}}$. Define ${\lambda_{m,S}(n):=1}$ if the last non-zero digit of ${n}$ in its ${m}$-ary representation is in ${S}$. In all other cases ${\lambda_{m,A}(n):=-1}$. Immediately from the definition we find $\displaystyle \lambda_{m,S}(n+l)=\lambda_{m,S}(n)$ if ${\textnormal{ord}_m l>\textnormal{ord}_m n}$. To not obtain unbounded discrepancy it is necessary that ${\sum_{i=1}^{m/d-1}\lambda_{m,S}(i)=0}$ for ${1. Moreover, we find ${\sum_{i=1}^{n}\lambda_{m,S}(i)=\sum_{j=0}^l \sum_{i=1}^{d_j}\lambda_{m,S}(i) }$ which implies logarithmic growth. Since ${\lambda_{m,S}}$ is completely multiplicative we have ${\lambda_{m,S}(n)=1 }$ if n is a quadratic residue mod ${m}$, implying ${\lambda_{p,S}=\lambda_p}$ for ${p}$ an odd prime. We construct some ${\lambda_{m,S}}$ for small composite ${m}$:
• For ${m=9}$ we have for the quadratic residues ${\{1,4,7\} \subseteq S}$ and thus depending on whether we include ${3}$ either ${\lambda_{9,\{1,3,4,7\}}=\lambda_3}$ or ${\lambda_{9,\{1,4,6,7\}}=\mu_3}$.
• For ${m=15}$ we get ${\{1,4,6,9,10\} \subseteq S}$. Since ${\sum_{i=1}^4 \lambda_{15,S}(i)=0}$ for all ${S}$ we cannot include ${2}$ and ${3}$. Because ${10}$ is included this forces ${\lambda_{15,S}(5)=-1}$ for all ${S}$. Now using complete multiplicativity and ${\sum_{i=1}^{14} \lambda_{15,S}(i)=0}$ we get the four cases: ${S_{15,\{1,4,6,9,10,11,14\}}}$, ${S_{15,\{1,4,6,7,9,10,11\}}}$, ${S_{15,\{1,4,6,9,10,13,14\}}}$ and ${S_{15,\{1,4,6,7,9,10,13\}}}$.
• For ${m=21}$ we have that ${9^2\equiv 18 \mod 21}$. Thus ${1=\lambda_{21,S}(18)=\lambda_{21,S}(2)}$. Since ${3|21}$ we have ${\sum_{i=1}^2\lambda_{21,S}(i)=0}$, a contradiction.

Necessary conditions for bounded discrepancy.

• If ${f}$ has bounded partial sums, then the Dirichlet series ${\sum_{n\leq x} \frac{f(n)}{n^s}}$ converges for ${\sigma>0}$. Proof idea: Integration by parts.
• If ${f}$ has bounded partial sums, then ${\sum_{p\leq x} \frac{f(p)}{p}=A+O\left(\frac{1}{\log{x}}\right)}$. Proof idea: One first establishes ${\sum_{p\leq x} \frac{f(p)\log{p}}{p}=O(1)}$ like in the proof of Dirichlet’s theorem on primes in arithmetic progressions. This estimate is then used in the textbook proof on the growth of ${\sum_ {p\leq x} \frac{1}{p}}$.
• If ${f}$ has bounded partial sums, then for infinitely many primes ${p}$ one has ${f(p)=1}$ and for infinitely many primes ${q}$ one has ${f(q)=-1}$. Proof idea: Otherwise one would get a contradiction to the standard result on the growth of ${\sum_{p\leq x} \frac{1}{p}}$.
• With not so elementary methods one can prove results ‘in the spirit of’: Let ${f:\mathbb{N}\rightarrow S_1}$ have bounded partial sums and let ${\chi}$ be a non-principal Dirichlet character. Then ${\sum_{p}\frac{|1-f(p)\overline{\chi(p)}|}{p}=\infty}$. Here more serious number theory kicks in and this is probably where most are currently heading to.

Discrepancy results.

• Computer experiments (no link found) show that there are ${f}$ with ${\left|\sum_{n\leq x}f(n)\right|\leq 2}$ for all ${x\leq 246}$. No such sequence exists for ${x=247}$. There is no traditional proof of this, not even for ${x=\infty}$.
• Liouville’s function does not have bounded partial sums since its Dirichlet series equals ${\frac{\zeta(2 s)}{\zeta(s)}}$ and thus has a singularity at ${s=\frac{1}{2}}$.Alternatively one can proceed as in the proof of Dirichlet’s theorem on primes in arithmetic progressions and apply Dirichlet’s hyperbola method to ${A(x):= \sum_{n\leq x}\frac{1*\lambda(n)}{\sqrt{n}}}$ ultimately showing that under the assumption of bounded partial sums ${A(x)}$ grows like ${\sqrt{x}}$. This is a contradiction since ${A(x)}$ grows logarithmically. Showing that the discrepancy of the Liouville function is ${O(n^{1/2+\varepsilon})}$ for all ${\varepsilon>0}$ is equivalent to solving the Riemann hypothesis.
• For primitive Dirichlet characters ${\chi}$ mod ${k}$ one has ${\max_{n\leq x}\left|\sum_{i\leq n}\chi(i) \right|\geq \frac{\sqrt{k}}{2\pi}}$. Proof idea: Apply partial summation to the associated Gauss sum ${G(1,\chi):=\sum_{1\leq m\leq k} \chi(m) e^{2 \pi i m /k}}$ and use ${|G(1,\chi)|^2=k}$.
• We have ${\mu_3(n) = \lambda_{9,\{1,4,6,7\}}(n)}$ and  $\displaystyle \left|\sum_{i=1}^n\mu_3(i)\right|\leq\log_9(8n+1)$ with equality holding only for ${n=\frac{9^m-1}{8}}$ and ${m\in\mathbb{N}}$.
• ${\mu_3}$ is optimal in the following sense: $\displaystyle \limsup_{x\rightarrow\infty}\frac{\left|\sum_{n\leq x}\mu_p(n)\right|}{\log{x}}\geq \frac{1}{2 \log{3}}$ with equality holding only for ${p=3}$. Proof idea: One constructs a sequence ${x_i}$ such that ${\lim_{x_i\rightarrow\infty}\frac{\left|\sum_{n\leq x_i}\mu_p(n)\right|}{\log{x_i}}\geq \frac{M_p-m_p}{2\log{p}}}$ with ${M_p:=\max_{1\leq n< p}\sum_{1\leq i\leq n} (\frac{i}{p})}$ and ${m_p:=\min_{1\leq n< p}\sum_{1\leq i\leq n} (\frac{i}{p})}$. With the lower bound for the discrepancy of primitive Dirichlet characters one can eliminate all but finitely many cases. These can then be checked by computer. A similar argument shows that the sums of ${\lambda_p}$ grow faster than ${\mu_3}$.
• Computer experiments so far did not find a ${\lambda_{m, S}}$ that grows slower than ${\mu_3}$. For a given ${\lambda_{m,S}}$ let ${M_{m,S}:=\max_{1\leq n< m}\left|\sum_{1\leq i\leq n} \lambda_{m,S}(i)\right|}$. Then ${m=39}$ with ${S=}$ $\displaystyle \{1,3,4,9,10,12,13,14,16,19,22,25,27,29,30,31,34,35,36\}$ is the largest ${m}$ such that there is an ${S}$ with ${M_{m,S}=2}$. Proof idea: Do a computer search among all completely multiplicative discrepancy 2 functions with domain ${\{1,\ldots,m-1\}}$ (${m}$ must be less than 247). A ${\lambda_{m, S}}$ beating ${\mu_3}$ thus has at least ${M_{m,S}=3}$ and thus ${\frac{1}{\log 9}>\frac{3}{\log m}}$ implying ${m>729}$.

## A recent example of physical notions not being functions of each other

March 22, 2010

Let me quote John Baez from his recent issue “This Week’s Finds in Mathematical Physics (Week 294)”:

The point of these examples is that most linear resistors let us treat current as a function of voltage or voltage as a function of current, since R is neither zero nor infinite. But in the these two limiting cases – the short circuit and the open circuit – that’s not true. To fit these cases neatly in a unified framework, we shouldn’t think of the relation between current and voltage as defining a function. It’s just a relation!

That is another example of basic notions not being a function of each other.

Maybe that makes my last comment on the relation between ‘price of a good’ and ‘demand for a good’ not being a function of each other more accessible. Between these economic notions there is just a (commutation) relation. They are not functions of each other. I will certainly elaborate on this …

## General Equilibrium in a Nutshell

February 17, 2010

My goal today is to tell the story of textbook general equilibrium theory and relate it to some of the things I have done so far in this blog.

Let me first quote from Stephen Smale, Mathematical Problems for the next Century, 1998:

The following problem is not one of pure mathematics, but lies on the interface of economics and mathematics. It has been solved only in quite limited situations.

Extend the mathematical model of general equilibrium theory to include price adjustments.

What is the mathematical model of general equilibrium theory? Usually there are ${n}$ goods and prices ${p_i}$ attached to each good for ${1\leq i\leq n}$. Furthermore there is an excess demand function ${z:\mathbb{R}^n_+\rightarrow\mathbb{R}^n}$ (considered as the difference of demand and supply) from the set of prices to the set of goods. Excess demand satisfies three axioms:

1. Homogeneity of degree zero: ${z(\lambda p) = z(p)}$ for all ${0<\lambda\in\mathbb{R}}$ and all ${p\in \mathbb{R}^n_+}$.
2. Walras’ law: ${\sum_{i=1}^n p_i z_i(p)=0}$.
3. Positive demand for a free good: ${z_i(p)>0}$ if ${p_i=0}$.

For this vector field Hopf’s theorem ensures the existence of an equilibrium price vector ${p^*}$ such that ${z(p^*)=0}$. This is often refered to as “supply equals demand”. The problem now is to find a dynamical model explaining the time evolution of prices maybe even as actions of agents acting in the market. This model should preferable be compatible with the existing equilibrium theory.

How does this relate to what I have said so far in this blog? The essential question:

Is (excess) demand indeed a function of price?

To my knowledge there is hardly any economic evidence/experiment to settle this question. “What else could it be?” is certainly not a valid viewpoint. While there are reasons to believe this in the stationary setting, as soon as time hits the scene a functional relation needs justification. Agents may learn or exhibit other hard to explain behavior.

If we do not assume excess demand to be a function of price, how far do we get? Quite far, actually. The reason is axiom 1. Its content in everyday speech: demand does not depend on price-scaling. Since price-scalings form a group we are in a position to find a representation, for example, as linear operators on a vector space. Doing this we can derive a kind of uncertainty principle for demand and price. They cannot be measured simultaneously with arbitrary precision. This is certainly necessary if we wanted to define a functional relation between them. Moreover, as an invariant of the group action we get a two parameter family of operators. One of these parameters is readily identified as “endowment”, the other represents a willingness-to-pay/willingness-to-accept discrepancy. Goods are considered more valuable if we own them compared to if we want them. We get what economists call the endowment effect for free.

All this could be done without the other axioms. As soon as this problem is tackled I comment on these too. Not today however …

## An Uncertainty Principle for Markets

December 9, 2009

Today our goal is to derive an exact formulation of an uncertainty principle in markets. To that purpose we have established in earlier posts a commutation relation between demand ${d_i}$ and price ${p_i}$ of a good ${i}$ in a market. I state it again:

Prices ${p_i}$ and demands ${d_j}$ interact according to

$\displaystyle [p_i,d_j]=i \mu_i p_i \delta_{i,j} \ \ \ \ \ (1)$

for a fixed real ${\mu_i\in\mathbb{R}}$.

What I didn’t tell you so far is how measurement of market observables is supposed to work. Let me just close this gap. Measurement of an observable, e.g. the price of good ${i}$, in a market in state ${\xi}$ (e.g. in this case selling a small quantity of good ${i}$) will result in a jump of the market into a new state ${\zeta}$ being an eigenvector of the observable. The outcome of the measurement will be a real number ${\zeta_i}$ (e.g. the price), the eigenvalue of the observable corresponding to ${\zeta}$ with probability

$\displaystyle \textnormal{prob }(\zeta_i)=\frac{\left\langle \xi|\zeta \right\rangle \left\langle \zeta|\xi\right\rangle}{\|\xi\|^2}.$

For an observable ${a}$ on ${X}$ one can show that its mean value at state ${\xi\in X}$ is given as

$\displaystyle \overline{a_\xi}=\frac{\left\langle a \xi|\xi\right\rangle}{\|\xi\|^2}.$

The dispersion of an observable ${a}$ on ${X}$ is given as

$\displaystyle \overline{\left(\triangle a \right)^2_\xi}= \frac{\left\langle\left(a-\overline{a_\xi} \text{id}_X \right)^2 \xi|\xi\right\rangle}{\|\xi\|^2 }.$

Now we are in the shape to state the uncertainty principle in markets. In essence it claims that prices and demands of a good cannot be measured with arbitrary precision. Moreover, an explicit lower bound on the maximal simultaneous precision is given. Its proof is essentially a straight forward application of Cauchy-Schwarz inequality.

Proposition. For a market in state ${\xi}$ the dispersions of ${p_i}$ and ${d_i}$ satisfy

$\displaystyle \overline{\left(\triangle p_i \right)^2_\xi} \, \overline{\left(\triangle d_i \right)^2_\xi} \geq \frac{\mu_i^2}{4} \| \sqrt{p_i} \xi\|^4.$

In the asymmetric case ${\mu_i\neq0}$, the right-hand side is strictly larger than zero.

Proof. Since dispersion and mean do not depend on the norm of a state we can, without loss of generality, assume that ${\|\xi\|=1}$ and obtain

$\displaystyle \overline{\left(\triangle p_i \right)^2_\xi} \, \overline{\left(\triangle d_i \right)^2_\xi} = \left\langle\left(p_i - \overline{p_i} \text{id}_X \right)^2\xi|\xi\right\rangle \left\langle\left(d_i - \overline{d_i} \text{id}_X \right)^2\xi|\xi\right\rangle.$

Now Cauchy-Schwarz inequality implies

$\displaystyle \begin{array}{rcl} \overline{\left(\triangle p_i \right)^2_\xi}\, \overline{\left(\triangle d_i \right)^2_\xi} & \geq & \left\langle\left(p_i - \overline{p_i} \text{id}_X \right) \left(d_i - \overline{d_i} \text{id}_X \right) \xi|\xi\right\rangle \\ & & \qquad \times \left\langle\left(d_i - \overline{d_i} \text{id}_X \right) \left(p_i - \overline{p_i} \text{id}_X \right) \xi|\xi\right\rangle. \end{array}$

Since ${ab = \frac{1}{2}[a,b]_+ + \frac{1}{2i}i[a,b]}$ with ${[a,b]_+=ab+ba}$ we obtain

$\displaystyle \begin{array}{rcl} \overline{\left(\triangle p_i \right)^2_\xi} \, \overline{\left(\triangle d_i \right)^2_\xi} & \geq & \left\langle\frac{1}{2} [d_i - \overline{d_i} \text{id}_X, p_i - \overline{p_i} \text{id}_X]_+\xi|\xi\right\rangle^2 \\ & & \qquad +\left\langle\frac{1}{2i} [d_i - \overline{d_i} \text{id}_X, p_i - \overline{p_i} \text{id}_X]\xi|\xi\right\rangle^2 \end{array}$

and since the first term is positive

$\displaystyle \begin{array}{rcl} \overline{\left(\triangle p_i \right)^2_\xi} \, \overline{\left(\triangle d_i \right)^2_\xi} & \geq & \left\langle\frac{1}{2i} [d_i - \overline{d_i} \text{id}_X, p_i - \overline{p_i} \text{id}_X]\xi|\xi\right\rangle^2 \\ & \geq & \left\langle\frac{1}{2i} [d_i,p_i]\xi|\xi\right\rangle^2. \end{array}$

Now (1) and the fact that positive observables have a square root yields the final inequality

$\displaystyle \begin{array}{rcl} \overline{\left(\triangle p_i \right)^2_\xi} \, \overline{\left(\triangle d_i \right)^2_\xi} & \geq & \frac{\mu_i^2}{4} \|\sqrt{p_i} \xi\|^4. \end{array}$

Since ${\mu_i \|\sqrt{p_i} \xi\|^4}$ can only be zero if ${\mu_i}$ is zero the proposition is proved.

Admittedly, that was a bit dry, but it does the job and that is sometimes all that is necessary in mathematics. Now the pace increases and we are heading with giant leaps towards the time evolution equations for markets …

## Mechanics and Markets

November 25, 2009

When we talk about markets we often use terms like equilibrium or even market force. We choose this terminology for a reason. The analogy to the well established theories of mechanics and quantum mechanics is intended and the pictures we have in mind are a pendulum or even a simple spring. Their restoring forces seem to model the market forces and therefore we frequently observe argumentations very similar to:

if prices increase, then demand decreases and vice versa finally, because of some process still to be described, the market settles down in an equilibrium (called Walrasian price equilibrium).

As a start, that sounds convincing. There just remains one big question. Is that a good picture? Or, even more to the point:

Are there any justifications for the existence of market forces?

Rather than answering this question (regular readers know my standpoint anyway) I would like to justify why this question is actually reasonable and should be asked and answered. In physics this question is answered to the positive, in economics the situation is a little blurry to say the least. I continue by comparing mechanics with economics in catchwords. Thereby pointing out similarities, but also discrepancies and, in a way, recalling ‘the story so far’.

Basic notions

Let me start with two of the fundamental notions in mechanics, namely position and momentum. In earlier posts we have identified their counterparts in economics as price and demand.

Symmetries

In mechanics the intuition is that momentum is invariant under translation of position. In economics we need demand invariance under price-scaling.

Commutation relations

These symmetries lead to commutation relations of the form ${[A,B]=\text{id}}$ in quantum mechanics and ${[A,B]=A}$ in economics (cf. here). This difference is essential and has a huge impact, albeit not immediately.

Bounded representations

Both commutation relations imply that the symmetry groups do not have representations on a finite-dimensional vector space (cf. here).

Unbounded representations

While there are no bounded representations, we get unbounded representations on the Hilbert space ${L^2(\mathbb{R}^n)}$ of square integrable functions. Momentum and demand operators are differential operators, whereas position and price are (different) multiplication operators (cf. here).

Uncertainty principle

The uncertainty principle of quantum mechanics is well-known. So far I didn’t write about that here in the blog, but in economics the commutation relations imply inequalities which can also be interpreted as some sort of uncertainty principle. I shall come back to this later.

Time evolution

As described in scientific laws to get the time evolution in quantum mechanics one chooses an action, one uses Legendre transform to obtain the energy, one derives the canonical equations and essentially plugs in the above representation to obtain Schrödingers equation governing the time evolution of a quantum system. That surely sounds more complicated than it actually is.

Why can’t we just do that for markets and obtain market equations governing their time evolution? Now, there are a couple of technical difficulties. The most prominent probably is that the Legendre transform of a market action is not invariant under time translation. Hence, in markets there is no conservation of energy. This fact alone makes the usage of a term like market force a little obscure. What is meant by force if there is no energy or at least no energy conservation?

That essentially is the programme for the rest of the year. I shall spell out the maths behind the uncertainty principle for markets and then delve into the technical details of obtaining a time evolution for markets.

Stay tuned …

## Finally Top Ten!

October 31, 2009

Last week I was teaching 8 hours mathematics a day, but that could not prevent me from checking my e-mail. What I found was a link to the following top ten list:

Top Ten List

It contains a joint work with Christian Schwarz on 10th place which is not overly remarkable. The paper is about a fairly general proof of the non-existence of market equilibria.

The twist is that the list also contains a ‘handful’ of papers proving the existence of said market equilibria.