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.]

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