In this section, I explain the solution method of the model. The …rst order conditions of the planner’s problem are10:
1 Ct = t (1 n X j=1
Lj;t) = t(1 )Ki;tOi;tAt1 Li;t
t = Ef t+1[ Ki;t 1Oi;t(AtLi;t)1 + (1 )]g t
Bi;t
= Ef t+1[ Ki;t 1Oi;t1(AtLi;t)1 +
(1 ') Bi;t+1 ]g:
i = 1; 2; :::n The steady state equations are:
1 0By taking …rst order conditions, an interior solution is already assumed. Why can we rule out corner solution? In other words, is it possible that in some periods, certain …rms get zero investment because they are hit by low shocks? The answer is no. The reason is as follows. Assume all …rms start with the same amount of capitals K and O, but …rm A has higher organizational investment-speci…c shock for the next period. Suppose the social planner chooses to concentrate all the new O investment in …rm A and starve other …rms, obviously all the new K investment has to be made in …rm A, too, otherwise too much O makes the marginal productivity of O in …rm A go down so much that it can hardly be optimal. Now think of what happens to other …rms. They get zero new investment, but are still in business with the left-over K and O from last period. But K and O have very di¤erent depreciation rates. Speci…cally, in the model, I assume depreciation for K around 5% per year, but for O about 50%. So in the next period, the marginal productivity of O in other …rms would be much higher than in …rm A, if they don’t receive any new O investment. The situation can be improved if social planner had chosen to invest some O in these low-shock …rms, which means that the investment schedule I assumed in the beginning cannot be optimal. The key thing here is a much higher depreciation rate for O than for K. And this assumption is by no means unrealistic.
(1 n X j=1 Lj) = 1 C(1 )Ki OiLi 1 = Ki 1OiL1i + (1 ) 1 = Ki Oi 1L1i + (1 ') C + n X j=1 Kj) + ' n X j=1 Oj = n X j=1 KjOjL1j A = 1; B = 1: Let
Rki;t = Ki;t 1Oi;t(AtLi;t)1 + (1 )
Ri;to = Ki;t 1Oi;t 1(AtLi;t)1 +
(1 ') Bi;t
And let lower case letters denote log-deviations of variables from the steady state. Log-linearizing the …rst order conditions and constraints around the steady state:
ri;t+1k = K
1O L1
Rk [( 1)ki;t+1+ oi;t+1+ (1 )li;t+1
+(1 a )at+1]
ri;t+1o = K O
1L1
Ro [ ki;t+1+ ( 1)oi;t+1+ (1 )li;t+1
+(1 a )at+1] (1 ')bi;t+1
ct = t
L
1 nLli;t = t+ ki;t+ oi;t+ ( )li;t+ (1 a )at
t = Et( t+1+ ri;t+1k ) t bi;t = Et( t+1+ ri;t+1o )
Cct+ K n X j=1 kj;t+1+ O n X j=1 oj;t+1 = [ K 1O L1 + (1 )K] n X j=1 kj;t+ [ K O 1L1 +(1 ')O] n X j=1 oj;t+ nK O L1 (1 )at +K O L1 (1 ) n X j=1 lj;t+ 'O n X j=1 bj;t
Exogenous shock processes take the form:
at+1= aat+ "t+1; "t+1~iid N (0; 2")
bi;t+1= bbi;t+ i;t+1; i = 1; 2; :::n; i;t+1~iid N (0; 2)
I then solved for the equilibrium law of motion using Schur factorization method proposed by King & Watson (2002).
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