Ragnar Nymoen 29/9 2009. Updated 9/10 2009 Note to IAM Chapter 20.1 (supplement to slide sets to lecture 7) The formal analysis of stabilization policy states that the government and/or central bank aims to minimize the social loss function SL = 2 y 2 + , > 0, (1) is a parameter that measures the degree to which the society (through the mandated decisions makers) values stable in‡ation relative to output stability. 2 2 are the mathematical variances of output and in‡ation that are y and implied by our model of the economy. Since 2y and 2 are derived from the model, we need to consider the solution of the AD-AS model consisting of (yt y) = ( t = t 1 t + (yt ) + zt , and (2) y) + st ; (3) In the lectures we have already worked with the …nal equation for (yt is: (yt y) = (yt 1 y) + (zt zt 1 ) st , where = 1 1+ ; 0< <1 y). It (4) (5) given the assumptions of the model. As stated in IAM p 612, 2y and 2 are interpreted as the unconditional variances of the output and in‡ation, also called the marginal variances, or asymptotical variances. Finding 2 y The most direct way of obtaining an expression for 2y is to use (4), and note that asymptotically the variance should not depend on time, so: V ar [(yt y)] = V ar [(yt 1 y)] = 2 y. We then take the variance on each side of “=” in (4). This gives: 2 y = 2 2 y + 2 V ar[zt zt 1] +( )2 V ar[st ] where we haved used that the variance of a sum of uncorrelated variables is the sum of the variances, and also have used the rule that V ar[aU ] = V ar[ aU ] = a2 V ar[U ], where U is a stochastic variable and a is a constant. (NB: How can 1 yt 1 be uncorrelated with zt 1 ? Probably an inconcistency here but we use that assumption to obtain the formulae in the book). From this we obtain (1 2 ) 2 y = 2 V ar[zt ) = zt 1] )2 V ar[st ]: +( (6) Note that 2 (1 2 = (1 + = 2 1 1+ + ( )2 1 2 ) 2 2 )2 +( so that (6) can be written as 2 y = V ar[zt + 2 V ar[st ] : + ( )2 zt 1] 2 (7) Finally, introduce the notation that V ar[zt ] = 2z and V ar[st ] = the same expression as in equation (11) on page 612 in IAM: 2 y = 2 2 2 z + +( 2 s, 2 2 s )2 and obtain (8) An alternative derivation uses repeated backward substitution in the …nal equation, and lim j ((yt j 1 y) ! 0 j!1 to obtain: (yt y) = 1 X j (zt zt j 1 j) 1 X + j=0 y)] = 2 = 2 2 2 z 2 2 z 1 X 2 j +( ) 2 2 s 2 1 This is the same expression for 1 X 2 j j=0 1 +( 2 ) 2 s 1 2 1 2 2 = j, y): j=0 = st j=0 which we can use to calculate the variance of (yt V ar[(yt j 2 2z ( ) 2s + + ( )2 ] 2 [2 + ( )2 ] 2 2 2 2z s + . 2 +( ) 2 + ( )2 [2 2 2 y as we found in equation (8). 2 2 NOTE added 9/10 2009: Tord Krogh has pointed out that this derivation presumes that zt j is uncorrelated with zt j . This is correct, so the above is only valid if zt follows a random walk so that zt is without any autocorrelation. In that interpretation 2 2z can be replaced by 2 z for example. Random walk for zt does however imply that the demand shocks are (fully) persistent. To obtain the expression that is consistent with intended interpetation that 1 P j the demand shocks are temporaty, express (zt j zt 1 j ), in terms of j=0 the individual zt 1 X j (zt j ’s: zt j 1 j) = zt = zt (1 )zt 2 + 1 (1 )zt 2 + ::: j=0 (1 )[ zt 1 2 + zt 2 + :::]: The variance V ar[ 1 X j (zt zt j 1 j )] = 2 2 z = 2 2 z( )2 + (1 2 2 Z[ +( 2 2 ) +( 2 3 ) + :::] j=0 = replaces 2 2 2 1 z1 2 2 )2 ( + (1 2 2 (1 z y)] = (1 + 2 1 1) ) 2 1 in the expression for V ar[(yt V ar[(yt 1 2 2z )(2 + ) y)] above. This then gives + 2 2 s 2 +( )2 as the expression that is consistent with the assumption that zt and zt j (j 1) 2 1 are uncorrelated. (Again using = 1+1 and (1 2 ) = 2 +( )2 .) I also found the following link with corrections of typos to IAM http://www.econ.ku.dk/okojacob/MAKRO-2-F07/typos-gbc-…rst-and-secondprint.pdf. The above is noted on page 7. Finding To derive of (2) in (3): t = 2 2 we use the …nal equation for 1 (1 + ) t 1 + (1 + ) zt + 3 t, 1 (1 + which we obtain by insertion ) st + (1 + ) Taking the variance on both sides, and collecting terms gives: 2 + ( )2 (1 + )2 2 2 = (1 + and then 2 2 z ) + 2 2 z = + 2s + ( )2 2 1 (1 + 2 ) 2 s (9) which is (12) p 612 in IAM. Both 2y and 2 are functions of the parameters b and h in the Taylor-rule function. This is because, as we have seen = and zt = which means that the variance 1 2 z is 2 z = 2h (1 + 2 b) (gt g) + vt (1 + 2 b) 2 v (1 + 4 2 2 b)