## 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 [1] 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 [1], 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:

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

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