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

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

2. Nonlinear Difference Equation Models • Most of nature is nonlinear. It makes sense to create nonlinear mathematical models of society. • The most basic forms of nonlinear components for dynamic models include power and interaction terms.

3. Constant Source Gains • Constant source gains can be thought of as gains due to a process that does not mathematically depend on interactions between population groups. For example, exponential growth of a population simply requires that future states grow exponentially from past states. Thus, the rate of growth is dependent on the value of the current state.

4. Thomas Malthus • Exponential growth of a population is a feature of ideas raised by Thomas Malthus. Here we can frame these ideas with the exponential growth model, dY/dt = aY, where growth in the population Y depends on its current value. The solution for this is Yt= Y0eat. Note the exponential character of the solution.

5. A difference equation version of this model is ΔY(t) = cY(t), or Y(t+1) = (1+c)Y(t). Thus, a future state [Y(t+1)] is the result of a constant (1+c) times a past state [Y(t)]. We can think of this as constant growth. • Constant growth processes of change include things like vote mobilization through the use of broadcast media, like television commercials.

6. Interactive Gains: A Mobilization Example • Interactive gains require the interaction between two identifiable groups in the population, such as those who are already mobilized and those who are not yet mobilized. We can write such a model as ΔMt = sMt(1-Mt), where Mt can be thought of as those who are already mobilized and (1-Mt) represents those who are not yet mobilized.

7. Combining Constant Source and Interactive Gains • By combining various gain components in one model, we can enrich the specification of the dynamic process. • Combining constant source and interactive components as described previously, we have, ΔMt = gMt + sMt(1-Mt). We can think of this model as capturing mobilization via radio broadcasts (gMt) and interpersonal interaction [sMt(1-Mt)].

8. Adding Losses • But most processes of change have losses as well as gains. We can model this as|ΔM(t) = (constant source gains) + (interactive gains) – (decay losses). • To make things easy, we can begin with losses being similar to constant source gains, but with a different sign. Thus, we can have losses = -fMt

9. This would produce the following model:ΔMt = gMt + sMt(1-Mt) – fMt • Let us add some realism by giving a limit to the proportion of the population that is available for mobilization. Now we haveΔMt = g(L-Mt) + sMt(L-Mt) – fMt, where L is the mobilization limit. Note that we use this limit in the constant source and interactive components of the model. • But now note that we are double counting, since people can be mobilized through the constant source and the interactive components.

10. To get rid of the double counting, we can subtract the constant source gains from the possible interactive gains. Now we haveΔMt = g(L-Mt) + sMt[(L-Mt) - g(L-Mt)] - fMt, or ΔMt = (sg-s)Mt2 + (sL-f-g-sgL)Mt + gL, or Mt+1 = (sg-s)Mt2 + (1+sL-f-g-sgL)Mt + gL • This is a first-order nonlinear difference equation of degree two.

11. If the double counting check was not included in the model, the model would have beenΔMt = g(L-Mt) + sMt(L-Mt) - fMt, or Mt+1 = -sMt2 + (1-f-g+sL)Mt + gL • But accounting for double counting is more realistic, and thus better. We will use the better version for the rest of this presentation.

12. Parameter Restrictions • We begin with simple descriptive contraints. • 0 ≤ L, g, f ≤ 1 • We also know that 0<s. But s is not necessarily less than 1. (See the top of page 34 in Huckfeldt, Kohfeld, and Likens.) • The possibilities for interaction are: MtMt, yielding no spread (L-Mt)(L-Mt), yielding no spread Mt(L-Mt), yielding the possibility of spread (L-Mt)Mt, yielding the possibility of spread

13. Thus, there are two paths along which interactions between mobilized and nonmobilized individuals in the population can occur and produce interactive gains. This can be described probabilistically in terms of their relative frequency as 2(L-Mt)Mt. It is customary to simply drop the 2 above, and let this be absorbed in the estimated value of the parameter s. This is why the parameter s can be greater than 1. But it still would be nice if we could get an upper limit for s.

14. Additionally, total gains must be less than or equal to all those available for mobilization. Thus we have 0 ≤ g(L-Mt) + sMt[(L-Mt) - g(L-Mt)] ≤ (L-Mt), which is saying 0 ≤ total gains ≤ those available

15. Now, let us take this and try to isolate g. 0 ≤ g(L-Mt) + sMt[(L-Mt) - g(L-Mt)] ≤ (L-Mt) and divide by (L-Mt) to obtain 0 ≤ g + sMt(1 – g) ≤ 1, for Mt ≠ L. Let us call this “Result A.” We can re-arrange this and obtain 0 ≤ g (1 - sMt) ≤ (1 - sMt), or 0 ≤ g ≤ 1. This is the same as before. All this work did not get us any further. 

16. But now let us go back to Result A, and solve the restriction for s instead of g. • Again, Result A is 0 ≤ g + sMt(1 – g) ≤ 1, for Mt ≠ L. Re-arrange this as follows: 0 ≤ g/(1-g) + sMt ≤ 1/(1-g), for g≠1 -g/(1-g) ≤ sMt ≤ 1/(1-g) – g/(1-g) -g/(1-g) ≤ sMt ≤ (1-g)/(1-g) = 1. Divide by Mt and you have -g/[(1-g)Mt] ≤ s ≤ 1/Mt. But 0 ≤ s, so 0 ≤ s ≤ 1/Mt. This give us an upper bound for s based on Mt.

17. 0 ≤ s ≤ 1/Mt. It is good to have an upper bound for s. Note that this is different from the upper bounds for L, g, and f. This upper bound can vary, and depends on the dependent variable. • Such constraints can often be used after estimates are obtained for the parameters. This offers a test of reliability (or believability).

18. Estimation • We estimate the model as Mt+1 = β2Mt2 + β1Mt + β0, and we obtain estimates for β0, β1, and β2. From this we have β0 = gL β1 = 1 + sL – f – g - sgL β2 = sg – s, which is three equations and four unknowns.

19. To solve this problem of having an over-determined system, you can make L=1 as an easy way out. Thus, you would have β0 = g β1 = 1 + s – f – g - sg β2 = sg – s, which is three equations and three unknowns. • But if you have another way to get L, use it. Perhaps you can use historical data.

20. Equilibria • Set all of the Mt equal to M*, and solve for M*. Thus, ΔMt = 0 = (sg-s)M*2 + (sL-f-g-sgL)M* + gL • This is a quadratic equation. We use the quadratic formula to get the roots.

21. The Quadratic Formula • For a quadratic equation set equal to zero, we find the roots with the following formulas: 0 = ax2 + bx + c xroots = [-b ± √(b2 – 4ac)]/2a

22. Solving the Quadratic • Thus, for our model, we note the following: ΔMt = 0 = (sg-s)M*2 + (sL-f-g-sgL)M* + gL a = (sg-s) b = (sL-f-g-sgL) c = gL Substitute these quantities into the quadratic formula, and solve for the roots. This will give you two roots. One will be spurious. But remember 0<M*. From the example used in Huckfeldt, Kohfeld, and Likens (p. 40), M* = 0.54 and -0.66. The answer is 0.54, and the value of -0.66 is a spurious root.

23. Local Stability Analysis • Now that we have a value for the equilibrium of the dependent variable of our model, M*, it would be nice to know if that equilibrium is a stable or unstable equilibrium. • A stable equilibrium attracts. An unstable equilibrium repels. The idea is that Mt may move away from M* if the equilibrium is unstable. But Mt will get closer to M* if the equilibrium is stable.

24. The Neighborhood of the Equilibrium • With local stability analysis, we are concerned about how Mt behaves in the local neighborhood of M*. That is, we want to know what will happen to Mt when it is close to M*. • This is different from a global stability analysis. With global stability, we are interested in whether or not M* attracts Mt regardless of the value of Mt.

25. Setting up the Local Stability Analysis • One way to conduct local stability analysis is to look at small perturbations around M*. • Mt = M* + Xt • Xt is the disturbance to M*. X0 is the initial disturbance. • We want to model the long-term impact of Xt on Mt. • Consider the difference equation, ΔXt = aXt, or equivalently, Xt+1 = (1 + a)Xt. • If we can model Xt, we can know if the perturbations around M* will decay.

26. From our knowledge of first-order linear difference equations with constant coefficients, we know that if (1+a) is between -1 and 1, then we will have convergence to the equilibrium value for Xt. In this instance, note that the equilibrium value for Xt is zero. Thus, if we have convergence for Xt, then the convergence is to zero and the perturbation dies away. • The solution form for the first-order linear difference equation with constant coefficients tells us that Xt = X0(1+a)t. Note what happens when -1<(1+a)<1, or equivalently, when -2 < a < 0.

27. The Taylor Series Expansion • Now we want to expand our equation for ΔMt around M* to see if Mt converges to M*. A Taylor series does this. • Recall our model for ΔMt. ΔMt = (sg-s)Mt2 + (sL-f-g-sgL)Mt + gL • Now we want to calculate the first two terms of the Taylor series for this function around M*. The first two terms of the Taylor series is the equation of the line that is tangent to the model when Mt = M*. These two terms are (1) ΔMt when Mt=M* and (2) (dΔMt/dMt)(Mt-M*) = (d ΔMt/dMt)Xt.

28. The first term equals zero since ΔMt=0 when Mt=M*. Thus, we only have the second term to work with. For the second term, we need to obtain the value of the derivative of ΔMt with respect to Mt when Mt=M*. dΔMt/dMt = 2(sg-s)M* + sL-f-g-sgL • This derivative is actually our value of the parameter a in ΔXt = aXt . How do we know this? • Since Mt = M* + Xt, we can multiply through by the linear operator Δ to obtain ΔMt = ΔM* + ΔXt. Since M* is a constant, ΔMt = ΔXt and dΔMt/dMt = dΔXt/dMt= a.

29. Thus, our Taylor series expansion of our model yields ΔMt = ΔXt = (dΔMt/dMt)Xt. • Substituting for dΔMt/dMt we have ΔXt = [2(sg-s)M* + sL - f - g - sgL]Xt or equivalently, Xt+1 = [1 + 2(sg-s)M* + sL - f - g - sgL] Xt. • Note that 1 + a = [1 + 2(sg-s)M* + sL - f - g - sgL] • Huckfeldt, Kohfeld, and Likens estimate their model using election data for Essex County and find that (1+a) = 0.46 for M*=0.54. This tells us that the equilibrium of 0.54 is stable, since -1<(1+a)<1.