Dynamic Modeling II Drawn largely from Sage QASS

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.

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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.

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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.

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• A difference equation version of this model is
• ?Y(t) cY(t), or Y(t1) (1c)Y(t). Thus, a
  future state Y(t1) is the result of a constant
  (1c) 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.

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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
• ?Mt 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.

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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, ?Mt
  gMt sMt(1-Mt). We can think of this model as
  capturing mobilization via radio broadcasts (gMt)
  and interpersonal interaction sMt(1-Mt).

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Adding Losses

• But most processes of change have losses as well
  as gains. We can model this as?M(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

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• This would produce the following model?Mt 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 have?Mt 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.

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• To get rid of the double counting, we can
  subtract the constant source gains from the
  possible interactive gains. Now we have?Mt
  g(L-Mt) sMt(L-Mt) - g(L-Mt) - fMt, or
• ?Mt (sg-s)Mt2 (sL-f-g-sgL)Mt gL, or
• Mt1 (sg-s)Mt2 (1sL-f-g-sgL)Mt gL
• This is a first-order nonlinear difference
  equation of degree two.

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• If the double counting check was not included in
  the model, the model would have been?Mt
  g(L-Mt) sMt(L-Mt) - fMt, or
• Mt1 -sMt2 (1-f-gsL)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.

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Parameter Restrictions

• We begin with simple descriptive contraints.
• 0 L, g, f 1
• We also know that 0lts. 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

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• 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.

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• 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

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• 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. ?

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• 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.

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• 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).

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Estimation

• We estimate the model as
• Mt1 ß2Mt2 ß1Mt ß0, and we obtain
  estimates for ß0, ß1, and ß2. From this we have
• ß0 gL
• ß1 1 sL f g - sgL
• ß2 sg s,
• which is three equations and four unknowns.

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• To solve this problem of having an
  over-determined system, you can make L1 as an
  easy way out. Thus, you would have
• ß0 g
• ß1 1 s f g - sg
• ß2 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.

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Equilibria

• Set all of the Mt equal to M, and solve for M.
  Thus,
• ?Mt 0 (sg-s)M2 (sL-f-g-sgL)M gL
• This is a quadratic equation. We use the
  quadratic formula to get the roots.

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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 v(b2 4ac)/2a

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Solving the Quadratic

• Thus, for our model, we note the following
• ?Mt 0 (sg-s)M2 (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 0ltM.
• 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.

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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.

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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.

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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, ?Xt aXt, or
  equivalently, Xt1 (1 a)Xt.
• If we can model Xt, we can know if the
  perturbations around M will decay.

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• From our knowledge of first-order linear
  difference equations with constant coefficients,
  we know that if (1a) 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(1a)t. Note what happens
  when -1lt(1a)lt1, or equivalently, when -2 lt a lt 0.

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The Taylor Series Expansion

• Now we want to expand our equation for ?Mt around
  M to see if Mt converges to M. A Taylor series
  does this.
• Recall our model for ?Mt.
• ?Mt (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) ?Mt when
  MtM and (2) (d?Mt/dMt)(Mt-M) (d ?Mt/dMt)Xt.

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• The first term equals zero since ?Mt0 when
  MtM. Thus, we only have the second term to
  work with. For the second term, we need to
  obtain the value of the derivative of ?Mt with
  respect to Mt when MtM.
• d?Mt/dMt 2(sg-s)M sL-f-g-sgL
• This derivative is actually our value of the
  parameter a in ?Xt aXt . How do we know this?
• Since Mt M Xt, we can multiply through by
  the linear operator ? to obtain
• ?Mt ?M ?Xt. Since M is a constant, ?Mt
  ?Xt and d?Mt/dMt d?Xt/dMt a.

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• Thus, our Taylor series expansion of our model
  yields ?Mt ?Xt (d?Mt/dMt)Xt.
• Substituting for d?Mt/dMt we have
• ?Xt 2(sg-s)M sL - f - g - sgLXt
• or equivalently,
• Xt1 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
• (1a) 0.46 for M0.54. This tells us that
  the equilibrium of 0.54 is stable, since
  -1lt(1a)lt1.