Mathematical methods for economic theory

Martin J. Osborne
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9.1 Exercises on first-order difference equations

  1. Solve the following difference equations. In each case, determine whether the solution path is convergent or divergent.
    1. xt = −3xt−1 + 4.
    2. xt = (1/2)xt−1 + 3.

    Solution

    1. xt = 1 + (−3)t(x0 − 1); divergent (and oscillating).
    2. xt = 6 + (1/2)t(x0 − 6); convergent.

  2. Consider a model of price adjustment that differs from the one described in the text only in that the suppliers' price expectations are adaptive, so that
    P*t = P*t−1 + η(pt−1 − P*t−1)
    where 0 < η ≤ 1 and P*t is the expected price in period t, and supply depends on the expected price:
    St = −β + αP*t,
    where α > 0 and β > 0. (Note that the parameters α and β here differ from the parameters α and β in the model presented in the text.) The other components of the model remain Dt = γ − δpt, with γ > 0 and δ > 0, and St = Dt.
    1. Find the difference equation in pt that this model generates (first isolate P*t in the supply function.)
    2. Solve this difference equation for Pt. For what values of η is the time path of pt oscillatory? convergent?

    Solution

    1. After isolating P*t in the supply function, substitute the expression you get into the equation for P*t, and then substitute for St the expression for Dt, to get pt = (1 − η − ηα/δ)pt−1 + η(γ + β)/δ.
    2. pt = (p0 − (γ + β)/(δ + α))(1 − η − ηα/δ)t + (γ + β)/(δ + α). Oscillates if 1 − η − ηα/δ < 0, or η > δ/(δ + α). Converges if |1 − η − ηα/δ| < 1, or 0 < η(δ + α)/δ < 2, or η < 2δ/(δ + α) (given that α > 0, δ > 0, and η > 0).

  3. Consider the difference equation
    yt+1(a + byt) = cyt
    where a, b, and c are positive constants and y0 > 0.
    1. Show that yt > 0 for all t.
    2. Define xt = 1/yt. Show that by using this substitution, the new difference equation is of the type xt = αxt+1 + β. Use this fact to solve the difference equation
      yt+1(2 + 3yt) = 4yt
      with y0 = 1/2. What is the limit of yt as t → ∞?

    Solution

    1. Given that a > 0, b > 0, and c > 0 we have yt > 0 so long as yt−1 > 0. Since y0 > 0 we thus have yt > 0 for all t.
    2. Substituting gives xt+1 = (a/c)xt + b/c. Thus the solution of the specific equation given is, in the xt variable, xt = (1/2)t+1 + 3/2, or yt = [(1/2)t+1 + 3/2]−1. As t → ∞ we have yt → 2/3.