## Wednesday, January 24, 2018

### Varying $\langle k \rangle$

As I mention in my (newly revised) recent paper, we can use the ensemble approach to arrive at an almost identical information equilibrium condition for an ensemble of markets:

$$\frac{d \langle A \rangle}{dB} = \langle k \rangle \; \frac{\langle A \rangle}{B}$$

If $\langle k \rangle$ is slowly varying enough to treat as a constant $\bar{k}$, then we obtain the same solutions we have for a single market. But what if $\langle k \rangle$ has a small dependence on $B$

$$\langle k \rangle \approx \bar{k} + \beta \frac{B}{B_{0}}$$

Where $\beta \ll 1$? The result is actually pretty straightforward (the differential equation is still exactly solvable [1])

$$\frac{\langle A \rangle }{A_{0}} \approx \left( \frac{B}{B_{0}} \right)^{\bar{k}} \left( 1 + \beta \frac{B}{B_{0}}\right)$$

...

Footnotes:

[1] The exact solution is

$$A(B) = A_{0} \exp \left( \beta \frac{B}{B_{0}} + \bar{k} \log \frac{B}{B_{0}} \right)$$
but since $\beta$ is small, we can expand the exponential and rewrite it in the more familiar form showing the $\beta$ term as a perturbation.

1. Am I correct in thinking that term 'bracket A bracket' is a collection that includes term B?

Or are 'bracket A bracket' and B two items within a container, with the relationship described by the equation?

1. It is an ensemble average or expectation value with respect to a partition function (defined in the link to the paper in the post):

https://en.wikipedia.org/wiki/Partition_function_(mathematics)#Expectation_values

It does depend on B since B is the Lagrange multiplier in the partition function.