## Thursday, July 27, 2017

### Macro ensembles and factors of production

I was inspired by Dietrich Vollrath's latest blog post to work out the generalization of the macro ensemble version of the information equilibrium condition  to more than one factor of production. However, as it was my lunch break, I didn't have time to LaTeX up all the steps so I'm just going to post the starting place and the result (for now).

We have two ensembles of information equilibrium relationships $A_{i} \rightleftarrows B$ and $A_{j} \rightleftarrows C$ (with two factors of production $B$ and $C$), and we generalize the partition function analogously to multiple thermodynamic potentials (see also here):

$$Z = \sum_{i j} e^{-k_{i}^{(1)} \log B/B_{0} -k_{j}^{(2)} \log C/C_{0}}$$

Playing the same game as worked out in , except with partial derivatives, you obtain:

\begin{align} \frac{\partial \langle A \rangle}{\partial B} = & \; \langle k^{(1)} \rangle \frac{\langle A \rangle}{B}\\ \frac{\partial \langle A \rangle}{\partial C} = & \; \langle k^{(2)} \rangle \frac{\langle A \rangle}{C} \end{align}

This is the same as before, except now the values of $k$ can change. If the $\langle k \rangle$ change slowly (i.e. treated as almost constant), the solution can be approximated by a Cobb-Douglas production function:

$$\langle A \rangle = a \; B^{\langle k^{(1)} \rangle} C^{\langle k^{(2)} \rangle}$$

And now you can read Vollrath's piece keeping in mind that using an ensemble of information equilibrium relationships implies $\beta$ (e.g. $\langle k^{(1)} \rangle$) can change and we aren't required to maintain $\langle k^{(1)} \rangle + \langle k^{(2)} \rangle = 1$.

...

Update 28 July 2017

I'm sure it was obvious to readers, but this generalizes to any number of factors of production using the partition function

$$Z = \sum_{i_{n}} \exp \left( - \sum_{n} k_{i_{n}}^{(n)} \log B^{(n)}/B_{0}^{(n)} \right)$$
where instead of $B$ and $C$ (or $D$), we'd have $B^{(1)}$ and $B^{(2)}$ (or $B^{(3)}$). You'd obtain:

$$\frac{\partial \langle A \rangle}{\partial B^{(n)}} = \; \langle k^{(n)} \rangle \frac{\langle A \rangle}{B^{(n)}}$$