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93 Lag Operators A different, more general approach to solving the model is to use lag operators. The lag operator, which we denote by I, is a function that lags variables. That is, the lag operator applied to any variable gives the previous periods value of the variable: Lzt = Zf i. To see the usefulness of lag operators, consider our model without the restriction that m follows a random walk. Equation (6.60) continues to hold. If we proceed analogously to the derivation of (6.62), but without imposing fffrifji = fttf, straightforward algebra yields 1  2A 1  2A Xf = Aixti + £fXt+i) + mt + £f mt+b (6.75) where A is as before. Note that (6.75) simplifies to (6.62) if £f frif+i = mt. The first step is to rewrite this expression using lag operators. Xf i is the lag of Xf: Xf 1 = Ixf. In addition, if we adopt the rule that L lags the date of an expectational variable but not the date of the expectations, xt is the lag of £fXf+i: I£tXf+i = EfXt = X(. Equivalently, using to denote the inverse lag function, £tXf+i = L"Xf. Similarly, Etmt+i = L~mt. Thus we can rewrite (6.75) as Since £tX( i = Xt i and E,m, = m,, we can tliink of all the variables in (6.75) as being expectations as of (. Thus in the analysis that follows, the lag operator should always be interpreted as keeping all variables as expectations as of f. The backshift operator, B, is used to denote the function that lags both the date of the variable and the date of the expectations. Thus, for example, BEt?<t+i = EtiX,. Whether the lag operator or the backshift operator is more useful depends on the application; in the present case it is the lag operator. should not adjust their prices fully in the period of the shock. This in turn implies that it is not optimal for the remaining individuals to adjust their prices fully in the subsequent period. And the knowledge that they will not do this further dampens the initial response of the individuals who adjust their prices in the period of the shock. The end result of these forward and backwardlooking interactions is the gradual adjustment shown in equation (6.65). Thus, as in the model with prices that are predetermined but not fixed, the extent of incomplete price adjustment in the aggregate can be larger than one might expect simply from the knowledge that not all prices are adjusted every period. Indeed, the extent of aggregate price sluggishness is even larger in this case, since it persists even after every price has changed. And again a low value of that is, a high degree of real rigidityis critical to this result. If is 1, then A is 0, and so each pricesetter adjusts his or her price fully to changes in m at the earhest opportunity. If exceeds 1, A is negative, and so p moves by more than m in the period after the shock, and thereafter the adjustment toward the longrun equilibrium is oscillatory.
Xf = A(Lxt + LXt) + mt + ,, (6.76) (7  AI  AIi)Xr = (1 + L)mt. (6.77) Here I is the identity operator (so Iz, = Zf for any z). Thus (7 + L)mt is shorthand for mt + Lmt, and (7  AIAI~i)Xf is shorthand for Xf  AXf i AEfXf+i. Now observe that we can "factor" (7ALAL) as (7 Ali)(/  AI)(A/A), where A is again given by (6.70). Thus we have (7  AI i)(7  ADXf = y + L)mt. (6.78) This formulation of "multiplying" expressions involving the lag operator should be interpreted in the natural way: (7  AI )(7  AI)Xf is shorthand for (7  AI)xt minus A times the inverse lag operator applied to (7  Ai)Xf, and thus equals (Xf  AlXf)  (AI"Xf  AXf). Simple algebra and the definition of can be used to verify that (6.78) and (6.77) are equivalent. As before, to solve the model we need to eliminate the term involving the expectation of the future value of an endogenous variable. In (6.78), ffXf+i appears (imphcitly) on the lefthand side because of the (7  AI") term. It is thus natural to "divide" both sides by (7  \L~). That is, consider applying the operator 7 i AI + AI + aI i to both sides of (6.78). 7 + AI 11 AI 2 + ... times 7  AI 4s simply 7; thus the lefthand side is (7  Al)xt. And 7 + Al"! + AL + times 7 i IMs 7 i (I i A)I t(1 + A)AI2 +0 + A)A2i3 + .15 Thus (6.78) becomes (7  AI)Xf = + (1 + A)I t (1 t A)AI2 + (1 + a)a2i3 +   .] . (6.79) Rewriting this expression without lag operators yields A I  2A Xf = AXfi t [m, I (1 I A)(£fmf+] + + ()]. (6.80) Expression (6.80) characterizes the behavior of newly set prices in terms of the exogenous money supply process. To find the behavior of the Since the operator / + AL + AL + is an inhnite sum, this requires that lim„„(J I AL + AL2 + + A"L")(J + Li)w, exists. This requires that A"L"+4w, (which equals A"£tWt+„+i) converges to zero. For the case where A = Aj (so A < 1) and where m is a random walk, this condition is satisfied.
The Taylor Model and Inflation Inertia The solution of the Taylor model for the case where the process followed by aggregate demand need not be a random walk can be used to discuss one of the models main limitations. As described in Chapter 5, modern Keynesian specifications of the outputinflation tradeoff assume that inflation exhibits inertiathat is, that aggregate demand policies can reduce inflation only at the cost of a period of low output and high unemployment. Such inflation inertia is central to Keynesian accounts of output behavior during many periods of disinflation, such as in the United States in the early 1980s. As "The reason that we cannot assume that (6.82) holds for < 0 is that the law of iterated projections does not apply backward: the expectation today of the expectation at some date in the past of a variable need not equal the expectation today of the variable. For a more thorough introduction to lag operators and their uses, see Sargent (1987a, Chapter 9). aggregate price level and output, we only have to substitute this expression into the expressions for p {pt = (Xf + Xf i)/2) and ( ( = mt  pt). In the special case when m is a random walk, all of the Etmt+, s are equal to mt. In this case, (6.80) simplifies to Xt = AXf i + (l + i) mt. (6.81) It is straightforward to show that expression (6.68), A + AA = A, implies that equation (6.81) reduces to equation (6.65), x, = AXf i (1  A)mf. Thus when w is a random walk, we obtain the same result as before. But we have also solved the model for a general process for m. Although this use of lag operators may seem mysterious, in fact it is no more than a compact way of carrying out perfectly standard manipulations. We could have first derived (6.77) (expressed without using lag operators) by simple algebra. We could then have noted that since (6.77) holds at each date, it must be the case that 1  2A EtXt+k  A£fXf+fc i  A£fXf+fc+i = {Etmt+k + ftmt+k+i) (6.82) for all > 0. Since the left and righthand sides of (6.82) are equal, it must be the case that the lefthand side for = 0 plus A times the lefthand side for / = 1 plus A times the lefthand side for A: = 2 and so on equals the righthand side for / = 0 plus A times the righthand side for = 1 plus A times the righthand side for = 2 and so on. Computing these two expressions yields (6.80). Thus lag operators are not essential; they serve merely to simphfy the notation and to suggest ways of proceeding that might otherwise be missed.!
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