## Thursday, March 19, 2015

### The slowly variying information transfer index approximation

Previously, I had looked at the changing information transfer index as an approximation (see e.g. here). What if we look at the exact result assuming the functional form of $k$

$$k(N, M) = \frac{\log N/c}{\log M/c}$$

So that

$$\frac{dN}{dM} = k(N, M) \; \frac{N}{M} = \frac{\log N/c}{\log M/c} \; \frac{N}{M}$$

Re-arranging and integrating the differential equation

$$\frac{dN}{N \log N/c} = \frac{dM}{M \log M/c}$$

$$\int_{n_{0}}^{N} \frac{dN'}{N' \log N'/c} = \int_{m_{0}}^{M} \frac{dM'}{M' \log M'/c}$$

$$\log \log N/c - \log \log n_{0}/c = \log \log M/c - \log \log m_{0}/c$$

$$\log \frac{\log N/c}{\log n_{0}/c} = \log \frac{\log M/c}{\log m_{0}/c}$$

$$\frac{\log N/c}{\log n_{0}/c} =\frac{\log M/c}{\log m_{0}/c}$$

$$\log N/c = \frac{\log n_{0}/c}{\log m_{0}/c} \log M/c$$

If we define the constant

$$k_{0} \equiv \frac{\log n_{0}/c}{\log m_{0}/c}$$

We have

$$N = c \left( \frac{M}{c}\right)^{k_{0}}$$

And the price level is

$$P = \alpha k_{0} \left( \frac{M}{c}\right)^{k_{0} - 1}$$

where $\alpha$ represents the freedom to define the price level to be $P = 100$ for any given year. This is basically the same result where we take $k$ to be constant, which means the approximation where we take

$$\int_{n_{0}}^{N} \frac{dN'}{N'} \approx k(N,M) \int_{m_{0}}^{M} \frac{dM'}{M'}$$

for slowly varying $k(N,M)$ represents simply moving to a local fit rather than a global fit.

[Assuming my math is right.]