Liquidity and interest rates

Consider two cased, fixed term, non-tradeable deposit certificates and loan. And the other case, tradeable deposits and loans.

The non-tradable paper has a fixed market liquidity which is set to the standard value of 1.0.  One means one transaction, essentially.  This is the case where neither the lender nor borrower expects any material changes in the economy, a fixed term non liquid market. At the end of the period, the depositor demands principal and interest, and visa versa for the lender.  The both have to prepare to make the money good.  Hence, the two period planning model. Lender and depositor get one period to adjust their cash in preparation to the two period demand, this is the basic rule of adapted statistics.

Let's define an index, m, which identifies an ordering of markets for which loans and deposits are made with liquidity 1. Set deposits to D, loans to L, then we have:

D(m) is the amount due, principal and interest, over one period in market m. Do the same for loans, L(m). Then we apply the two [period model and take the derivative with respect to a change in markets.

D(m)^2 - L(m)^2 = 1, and its derivative, D(m)D'(m) - L(m)L'(m) = 0.

All this says is that we are dealing with markets having liquidity constant 1, so this differential must be zero.  And we see that D' = L and L'=D. Regardless of the liquidity value, stable markets have constant liquidity. So we get a standard result if the market liquidity is constant and two period planing is in effect.

The form of the deposits and loans

 We are already hyperbolic, lets look:

deposit value = (e^m+1/e^m)/2 and loan value is (e^m-1/e^m)/2
So what is going on here is maintaining liquidity. Over one period we want the deposits to earn 1/2 of the liquidity requirement and loans should retain 1/2 unit of liquidity. So, in one period, we have a half unit earned on deposits and a half unit retained on borrowings. Thus, the required liquidity in an adapted network is met.

Next we deal with liquidity different than one. This is the case when the network has to adapt. Let the adaption liquidity be some Q. Then we have:
D^2-L^2 = Q^2.   So in one period, D has to earn Q/2 and loans reserve Q/2. But our derivative still has to work, and they do according to my wxMaxima machine.

What are rates and balances?

D is the one period value, so:

b(1+r) = (e^m + Q*e^-m)* 1/2 = 1/2 * e^m *(1+Q * e^(-2m))

r= Q* e^(-2m)

b = 1/2 * sqrt(Q/r), with likely math errors 

So there is a fixed relationship between liquidity required and rates for any market m.   For loans, use reverse time like they do with anti-electron or like cash in advance. I will work it later.

So, term period is not known, but there is a frequency relationship between the currency banker at m=1 and the rest of the markets.   And for any given m, Q, the liquidity is a scale factor to the value of deposits and loans. That is, any quantity of deposits and loan values can be simply accumulated as a batch of unitary loans, so scale does not yet come into play. But the number of deposit/loan pairs at any given m, relative to another m, will be constricted by the flow conditions. The probability of a deposit/lona pari across all m have to be one, and that proability is inherent in the loan/deposit ratios at any m.