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Ensaios Econômicos - Sistema de Bibliotecas da FGV - Fundação ...

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Using (2.11), with CES utility and recalling that the conditional variance of logged observed<br />

consumption is ! t 2 tP<br />

1<br />

= 11 t + <br />

2 22 j + 2 tP<br />

1<br />

12 j, we can compute the welfare cost of business<br />

j=0<br />

cycle when 12 6= 0. The analogous formula to (2.13) is:<br />

8<br />

>< exp ( (2e 12 + e 22 ) =2) 1 for 6= 1,<br />

P =<br />

>:<br />

exp ((2e 12 + e 22 ) =2) 1 for = 1.<br />

j=0<br />

: (2.22)<br />

Similarly, using (2.14), we can compute the welfare cost of economic-growth variation. The analogous<br />

formula to (2.15) is:<br />

8<br />

><<br />

T =<br />

>:<br />

<br />

(1 (1+1 ) (1 ) exp( (1 ) 11 =2))exp( (1 )e 12 =2)<br />

(1 (1+ 1 ) (1 ) )<br />

<br />

exp<br />

11 +2(1 )e 12<br />

2(1 )<br />

1<br />

(1 )<br />

1 for = 1.<br />

1 for 6= 1,<br />

Finally, using (2.17), we can compute the welfare cost of all macroeconomic uncertainty.<br />

analogous formula to (2.18) is:<br />

8 <br />

><<br />

D =<br />

>:<br />

exp((1 )(2e 12 +e 22 )=2)(1 (1+ 1 ) (1 ) exp( (1 ) 11 =2))<br />

(1 (1+ 1 ) (1 ) )<br />

exp<br />

<br />

11 +(1 )(2e 12 +e 22 )<br />

2(1 )<br />

1<br />

(1 )<br />

1 for = 1.<br />

1 for 6= 1,<br />

(2.23)<br />

The<br />

(2.24)<br />

The key di¤erence between (2.13), (2.15), and (2.18), and their respective counterparts in the<br />

1P<br />

previous section is their <strong>de</strong>pen<strong>de</strong>nce on e 12 = 12 j, the unconditional counterpart of 12 .<br />

j=0<br />

Finally, it is straightforward to compute the marginal welfare costs of business cycles, economicgrowth<br />

variation, and all macroeconomic uncertainty. They are, respectively:<br />

8<br />

>< exp ( (2e 12 + e 22 )) 1 for 6= 1<br />

0 P (0) =<br />

(2.25)<br />

>:<br />

exp (2e 12 + e 22 ) 1 for = 1<br />

8<br />

<br />

<br />

exp(2e 12 ) 1 (1+ 1 ) (1 ) (1 )11<br />

exp<br />

2<br />

<br />

><<br />

0 1 (1+ 1 ) (1 ) (1+)11<br />

1 for 6= 1<br />

exp<br />

2<br />

T (0) =<br />

; and, (2.26)<br />

>:<br />

exp(2e 12 )(1 )<br />

1 exp( 11 )<br />

1 for = 1<br />

8<br />

<br />

<br />

exp((2e 12 +e 22 )) 1 (1+ 1 ) (1 ) (1 )11<br />

exp<br />

2<br />

<br />

><<br />

0 1 (1+ 1 ) (1 ) (1+)11<br />

1 for 6= 1<br />

exp<br />

2<br />

D (0) =<br />

: (2.27)<br />

>:<br />

exp(2e 12 +e 22 )(1 )<br />

1 exp( 11 )<br />

1 for = 1<br />

14

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