By Walter Enders

Applied Econometric Time sequence, 4th Edition demonstrates smooth options for constructing types able to forecasting, analyzing, and trying out hypotheses referring to financial facts. during this textual content, Dr. Walter Enders commits to utilizing a “learn-by-doing” method of aid readers grasp time-series research successfully and effectively.

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Extra resources for Applied Econometric Time Series

Example text

25; {yt } is a unit root process because a1 + a2 = 1. 6. In the event that the solution ct fails, sequentially try the solutions yt = 2 3 n ct , ct , … , ct . For an nth-order equation, one of these solutions will always be the particular solution. tex V3 - 09/02/2014 12:52pm Page 33 PARTICULAR SOLUTIONS FOR DETERMINISTIC PROCESSES 33 CASE 2 The Exponential Case. Let xt have the exponential form b(d)rt , where b, d, and r are constants. Since r has the natural interpretation as a growth rate, we would expect to encounter this type of forcing process case in a growth context.

At planting time, farmers do not know the price prevailing at harvest time; they base their supply decision on the expected price (p∗t ). The actual quantity produced depends on the planned quantity b + ????p∗t plus a random supply shock ????t . Once the product is harvested, market equilibrium requires that the quantity supplied equals the quantity demanded. Unlike the actual market for wheat, the model does not allow for the possibility of storage. 3 represents the long-run equilibrium price and quantity combination.

For |a| < 1, the infinite sum (1 + aL + a2 L2 + a3 L3 + · · ·)yt = yt ∕(1 − aL). This property of lag operators may not seem intuitive, but it follows directly from properties 2 and 3 above. Proof: Multiply each side by (1 − aL) to form (1 − aL)(1 + aL + a2 L2 + 3 3 a L + · · ·)yt = yt . Multiply the two expressions to obtain (1 − aL + aL − a2 L2 + a2 L2 − a3 L3 + · · ·)yt = yt . Given that |a| < 1, the expression an Ln yt converges to zero as n → ∞. Thus, the two sides of the equation are equal.