## 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 somewhat surprising observation

April 5, 2011

In this post, Kevin O’Bryant has considered “least nonzero digit”-type examples $\lambda_{m, S}$. Encouraged by his remark that there are only very few possibilities, i.e. feasible sets S, for each modulus m I did a complete enumeration up to modulus 200. The goal was (and still is) to find a slowly growing example possibly beating $\mu_{3}$ the current record holder (please read my last post for more information).

Much to my surprise there are certain “magic” numbers with comparatively many feasible sets S. What follows is a table (modulus m: number of sets S) for those m with more than 4 sets:

(39: 24)
(87: 560)
(95: 3.360)
(111: 5.280)
(119: 42.772)
(135: 5.940)
(145: 11.440)
(159: 64.064)
(183: 13.728)

All other m < 200 have 4 or less feasible sets.

For example, there is no set S for m=105. This can be seen as follows: Logarithmic discrepancy of $\lambda_{105}$ implies $\sum_{i=1}^{d-1} \lambda_{105}(i)=0$ for $d|105$. Since $3,5,7|105$ complete multiplicativity implies $\lambda_{105}(2)=\lambda_{105}(3)=\lambda_{105}(5)=-1$. Now $45^2\equiv 30 \mod 105$ and again since $\lambda_{105}$ is completely multiplicative $1=\lambda_{105}(30)=\lambda_{105}(2)\lambda_{105}(3)\lambda_{105}(5)=-1$ produces a contradiction.

What makes these number “special”? Why is there this huge jump from 4 to “many”? Is this just one more “random” number theoretic artifact or can we understand this by considering how the m-examples are constructed from the d|m-examples?

I do not know the answers, but let me close with something I do know. In my last post I gave a proof that any $\lambda_{m,S}$ that grows slower than $\mu_3$ must have $m>729$. Subsequent computer searches have increased that bound to 1000.