Dynamic Modeling I Drawn largely from Sage QASS #27, by Huckfeldt, Kohfeld, and Likens.

1. Dynamic Modeling IDrawn largely from Sage QASS #27, by Huckfeldt, Kohfeld, and Likens. Courtney Brown, Ph.D. Emory University

2. What Is Dynamic Modeling? • 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.

3. Statistical Modeling vs. Dynamic Modeling • 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.

4. Types of Change • 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.

5. Basic Gain and Loss Concepts • 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

6. 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)

7. Adding Realism to the Model • We can make the model more realistic by adding a limit to the total population that is available for recruitment. Thus, we haveΔMt = g(L-Mt) – fMt, where L is the limit. • Now we have,Mt+1 – Mt = gL - gMt - fMt, orMt+1 = gL + Mt(1-g-f), which is isomorphic to Mt+1 = β0 + β1Mt, where β0 = gL, and β1 = (1-g-f)

8. Equilibrium Value • 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.

9. Descriptive Constraints • 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.

10. Estimation • 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?”

11. Constraints on Parameter • 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.

12. 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.

13. An Historical Approach • 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.

14. A More Difficult Way • 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 thusβ0/g = L, we can also know that 0<L<1. We already knew this, but let us call this “Result #1.”

15. 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. Re-arrange to yield 1- β1 > g > - β1, or - β1 < g < 1- β1. Let us call this “Result #3.”

16. 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.”

17. 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.

18. 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.

19. 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.