Dynamic Modeling I
Drawn largely from Sage QASS #27, by Huckfeldt,
Kohfeld, and Likens.
Courtney Brown, Ph.D.
Emory University
What Is Dynamic Modeling?
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It is not a statistical technique.
It is used in many fields, like population
biology for modeling species interactions, and
physics for modeling physical and quantum
systems.
It attempts to answer the “why” question by
describing the structure of the system, not
just what influences what.
Statistical Modeling vs. Dynamic Modeling
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Statistical modeling focuses on how certain
variable correlate with other variables. This
shows influence.
Dynamic modeling focuses on the structure of
systems.
For example, there is a correlation between
stepping on the gas pedal of a car and the
speed of the car. But a dynamic model would
specify the physical linkages between the pedal,
the engine, the wheels, and the resulting speed.
Types of Change
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Synchronic change – The system stays the
same, but the inputs and outputs change.
Diachronic change – The system itself
changes, normally requiring a new model.
Some modeling forms can capture dramatic
change within one model, such as
catastrophe theory. This is still synchronic
change as long as the model stays the same.
Basic Gain and Loss Concepts
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ΔMt = Mt+1 – Mt, the “1” is arbitrary, but
customary.
Gains = g(1-Mt), where (1-Mt) are the not yet
mobilized
Losses = fMt, where Mt are the already
mobilized
A naive model would be,
ΔMt = g(1-Mt) – fMt = gains - losses
This can be re-phrased as
Mt+1 – Mt = g(1-Mt) – fMt
Mt+1 = g(1-Mt) – fMt + Mt
Mt+1 = g + Mt(1–g–f), a first-order difference
equation with constant coefficients.
 This can be estimated as
Mt+1 = β0 + β1 Mt, where β0 = g, and β1 = (1-g-f)
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Adding Realism to the Model
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We can make the model more realistic by
adding a limit to the total population that is
available for recruitment. Thus, we have
ΔMt = g(L-Mt) – fMt, where L is the limit.
Now we have,
Mt+1 – Mt = gL - gMt - fMt, or
Mt+1 = gL + Mt(1-g-f), which is isomorphic to
Mt+1 = β0 + β1Mt, where β0 = gL, and β1 = (1g-f)
Equilibrium Value
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We can obtain the equilibrium value from the
reduced form version of the model, Mt+1 = β0
+ β1Mt, which is M*= β0/(1- β1).
We can also deduce the qualitative behavior
of the model as well from this reduced form
version.
But ultimately we want to find the values of
the parameters g, f, and L.
Descriptive Constraints
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0<f<1
0<g<1
0<L<1
If β1 is negative, the trajectory will oscillate.
If β1 is positive, the trajectory will be monotonic.
If (1-f-g) < 0, the trajectory will be oscillatory. This
means that 1-f<g, which says that the retention rate
is less than the recruitment rate.
If (1-f-g) > 0, the trajectory will be monotonic. This
means that 1-f>g, which says that the retention rate
is greater than the recruitment rate.
Estimation
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Since β0 = gL, and β1 = (1-g-f), we have a
problem of trying to obtain g, f, and L from
only β0 and β1. This is a common problem
with these types of models.
You need to impose a value for one of the
unknowns. The question is, “Which one?”
Constraints on Parameter
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Sometimes one can use constraints on the
parameters to produce guidance for resolving
over-determined systems.
This can be tricky, and it often involves a bit
of luck.
The easiest way is to pick some convenient
parameter and assign it some reasonable
value. In our example with the parameter L,
one might simply make L equal to 1, an
ultimate limit.
An Easy Way
If L=1, then using our model
Mt+1 = gL + Mt(1-g-f), where the reduced form
is Mt+1 = β0 + β1Mt, then β0 = g.
 Since β1 = (1-g-f), we can re-arrange this,
substitute β0 = g, and then obtain
f = (1 - β0 - β1). Thus, we have g and f in terms
of β0 and β1.
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An Historical Approach
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Assigning a value of 1 to L will likely make this limit
too large, and we may want to be more realistic.
One approach is to use historical data. We could
find out the highest proportion of people registered
to vote on record and use that for the value of L.
We could also use the highest proportion of people
who ever were mobilized to vote and use that for the
value of L.
But we may want to use a more analytical approach
to assigning a value of a parameter.
A More Difficult Way
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In the example described in Huckfeldt,
Kohfeld, and Likens, the estimated
parameters β0 and β1 are positive (as found
by estimating them using real data).
Since we know that β0 = gL, and thus
β0/g = L, we can also know that 0<L<1. We
already knew this, but let us call this “Result
#1.”
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Now let us substitute β0/g = L into the
inequality 0<L<1, and we have 0< β0/g<1.
Multiply through by g to obtain 0< β0<g. We
know that the inequality signs do not reverse
since g is positive. We will call this “Result
#2.”
Now let us do the same with β1 and f. We
know that β1 = (1-g-f). Thus, f=1-g- β1. Since
0<f<1, we can substitute and say
0 < 1-g- β1< 1, or 0 > -1+g+ β1> -1. Rearrange to yield 1- β1 > g > - β1, or
- β1 < g < 1- β1. Let us call this “Result #3.”
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From Result #2 we know that β0 < g. But
Result #3 tells us - β1 < g < 1- β1. We want to
use all of the model’s constraints, not just
one. Thus, let us use β0 as the lower limit for
g, rather than - β1.
We also recall that the equilibrium value, M*,
is equal β0/(1- β1). This must be a positive
number, which follows from the equilibrium
constraint, 0<M*<1. Thus 0< β0/(1- β1)<1, or
0< β0 <(1- β1).
Thus, we can say that β0< g < (1- β1). Let’s
call this “Result #4.”
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It is now time to assign a number for the
parameter g. From Result #4, we know that
β0< g < (1- β1).
Our solution is to pick the midpoint of this
interval for the value of g. Our reasoning for
picking the midpoint is that we assume a
Normal distribution for the estimated value of
the parameter g. The Normal distribution is
symmetrical. Thus, picking a midpoint in the
interval seems a reasonable choice.
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In the example used by Huckfeldt, Kohfeld,
and Likens, they found the following values
for the parameters by estimating using
ordinary least squares:
β0= 0.14, and β1= 0.62. Thus,
0.14 < g < (1-0.62), or 0.14 < g < 0.38.
Picking the midpoint, g = (0.14 + 0.38)/2, or
g=0.26, which is our final result for this
problem.
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Now we want to substitute g back into the
original equations and solve for g and L
algebraically.
Recall that β0 = gL and β1 = (1-g-f). Thus we
can write 0.14 = (0.26)L, and
0.62 = (1-f-0.26).
Thus, L=0.54, and f=0.12.