Let's take a general information equilibrium model $P : A \rightleftarrows B$ with price $P$ information transfer index $k$ and log-linearize it. That notation is shorthand for the differential equation:

$$

P \equiv \frac{dA}{dB} = k \; \frac{A}{B}

$$

Define the variables $A \equiv a \; e^{\tilde{a}_{t}}$, $B \equiv b \; e^{\tilde{b}_{t}}$, and $P \equiv p \; e^{\tilde{p}_{t}}$. Substitution into the equation above yields

d\tilde{a}_{t} = k \; d\tilde{b}_{t}

$$

or as a finite difference equation:

$$

\tilde{a}_{t+1} - \tilde{a}_{t} = k \left( \tilde{b}_{t+1} - \tilde{b}_{t} \right)

$$

The general solution to the differential equation gives us the formula for the price $P$

P = ck \left( \frac{B}{B_{ref}} \right)^{k-1}

$$

Using the substitutions above, $B_{ref} \equiv b$, and a little algebra, we can show

$$

\tilde{p}_{t} = \left( k - 1\right) \tilde{b}_{t} + \log k + c_{p}

$$

where $c_{p}$ is a constant (parameter). Therefore ...

__Log-linear information equilibrium relationship__$$

\begin{align}

\tilde{a}_{t+1} & = k \left( \tilde{b}_{t+1} - \tilde{b}_{t} \right) + \tilde{a}_{t}\\

\tilde{p}_{t} & = \left( k - 1\right) \tilde{b}_{t} + \log k + c_{p}

\end{align}

$$

for which we can define the notation $\tilde{p}_{t} : \tilde{a}_{t} \rightleftarrows \tilde{b}_{t}$.

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