Answer for Henry #1

Below is my answer to a question posed by Henry on Jason's blog.

"My question is do these terms make sense"

Examining those terms in isolation is misleading. I'm NOT saying the whole expression makes sense (as per Jason's point, I think), but looking at those terms in isolation and examining their sensical-ness (new word) is the wrong approach. For example, what if I told you that x(t) -> inf as t -> 0 and y(t) -> inf as t->0, is it possible that x(t) - y(t) could approach something meaningful as t -> 0? Of course it is, because I'm interested in their difference, not the individual terms.

Allow me to take a stab at explaining the issue in a way that perhaps isn't as general as Jason's approach, but which I nonetheless find to be clearer. Going back to his definition:

γ := rate of government spending, in dollars/year (say)

But instead of making ξ := rate of taxation in dollars/year, let's leave that out of it. Now say

ΔH[n+1,n] = H[n+1] - H[n] = γ*Δt - T (abandoning your <> notation in favor of [] because of technical difficulties)

Now lets say (as is the case in the SIM model if you do the algebra) T = r*H, with r a fraction (a constant fraction to make it simple, say on (0,1), defined as a rate per unit of time τ, which initially happens to be τ = Δt, but of course will not hold if Δt changes), or putting indices on it:

T[n+1,n] = r*H[n]

Now we have:

ΔH[n+1,n] = H[n+1] - H[n] = γ*Δt - T = γ*Δt - r*H[n]

To simplify further, set γ = 0 for now, so we have:

H[n+1] = (1-r)*H[n] = A*H[n] where A = 1-r

This has a well known solution:

H[k] = H0*(A^k), for all k >= 0, and where H0 = H[0] (i.e. H0 is the initial allotment of money in the economy).

And because r is on (0,1) then so is A, so this will be an exponentially decaying function (a sampled exponential).

Now what happens if we change our sample period Δt? Well r doesn't change, nor does τ it's defined over, but

(1) H[n+1] = A*H[n] = (1-r)*H[n] 

No longer holds. Why? Imagine that it did hold, and that say r = A = 0.5. So in one time step Δt, half the money disappears from the economy through taxation. If Δt = τ = 1 year to start out, then in 1 year half the money is gone. Now say Δt is reduced to 1 day... then if equation (1) still held, then in one day half the money is gone. Thus (1) would NOT be describing the same system. We're sampling an underlying continuous time system, so we have to account for that. The real rate of decay (time constant) of the system should be independent of the sample period. The answer is we need to adjust A as Δt changes, because A was always an implicit function of Δt (i.e. A in (1) is A(τ), and so is only valid when Δt = τ). We can write a valid version of (1) if we adjust A for when Δt ≠ τ as follows:
 
(2) A(Δt) = A(τ)^(Δt/τ)

So then using (2) our valid version of (1) for any Δt is as follows:

(3) H[n+1] = A(Δt)*H[n]

Now in this simple case, it turns out that 1-r*Δt/τ ≈ A(Δt), but it's not exact. Also, in the limit as Δt -> 0 we have A(Δt) -> 1, so our difference equation (1) breaks down. We need to replace it with a differential equation:

dh/dt = (ln(A)/τ)*h = a*h, where ln() is natural log

and where a = ln(A)/τ, and H[n] = h(n*Δt). Note that a < 0 in this example because A is on (0,1)

The solution to this equation is:

h(t) = exp(a*t)*H0

That's the underlying continuous time system. Since a < 0, it's a decaying exponential, as before. We can sample it at any Δt we choose:

H[n] = (exp(a*Δt)^n)*H0

or in difference formula form:

H[n+1] = exp(a*Δt)*H[n]

Which suggests an alternative formulation of equation (2) when the continuous-time time constant -1/a is known [1]:


(4) A(Δt) = exp(a*Δt)

I leave as an exercise to the reader the case where γ > 0. :D
  
[Hint, try this]

Or if you want to see it worked out for the SIM example, look at how I calculate B here at the bottom.

What G&L are doing amounts to assuming that (2) below (the first two terms of (1) (also below)) is a  good approximation, and I think they try to restrict alpha1, apha2 and theta to accomplish that. But even with those restrictions, (2) can sometimes be a terrible approximation for (1). Either that or they are assuming the rate of compounding always matches the sample rate. Note that in the more general case where we substitute square matrix A for  a, (1) and (2) still hold, provided we take A raised to n to represent A matrix multiplied by itself n times in (1), and substitute the identity matrix I for 1 in (2).

 

Discrete and continuous time system equivalents, where H[n] = h(n*Ts), etc. Ts being the sample period

NOTES:

[1]  -1/a is defined as the time constant of the decaying exponential. It's the time t  > 0 by which the exponential has decreased from it's value at t=0 (unity) by 63%, or equivalently, decreased down to 37% of 1: since if t = -1/a, then exp(a*t) = exp(-1) ≈ 0.37.