Economic Models - Convex Optimization

88 Andrew Hughes Hallet
Substituting Eqs. (13)–(15) back into Eq. (11), we can now get the
stage 1 solution from:
{
min ELg (δ, λ cb ) = 1 (1 − δ)β(φ − η)λ cb + δ(βφ + γ)λ g } 2
1
δ,λ cb 2 α[β(φ − η) + δ(βη + γ)]
{
+ λg 2 (1 − δ)βs(βη + γ)(λ cb − λ g 1 )
} 2
2 [β(φ − η) + δ(βη + γ)]λ g . (19)
2
This part of the problem has first-order conditions:
and
(1 − δ)(φ − η)λ g 2 {(1 − δ)β(φ − η)λcb + δ(βφ + γ)λ g 1 }
−(1 − δ) 2 (βη + γ) 2 α 2 s 2 β(λ g 1 − λcb ) = 0 (20)
{(1 − δ)β(φ − η)λ cb + δ(βφ + γ)λ g 1 }
× (λ g 1 − λcb ) {δ(1 − δ) + (φ − η)} λ g 2
−(1 − δ)(βη + γ)α 2 s 2 β {(βη + γ) − (1 − δ)β} (λ g 1 − λcb ) 2 = 0.
(21)
where = ∂η/∂δ. There are two real-valued solutions which satisfy this
pair of first-order conditions. 19 Both are satisfied when δ = 1 and λ cb = λ g 1 .
This solution describes a fully dependent central bank, which is not appropriate
in the Eurozone case. And, it turns out to be inferior to the second
solution: δ = λ cb = 0. In this solution, the central bank is fully independent
and exclusively concerned with the economy’s inflation performance.
Out of the two possibilities, the solution which yields the lowest welfare
loss, as measured by the government’s (society’s) loss function, can be
identified by comparing Eq. (19) to the expected loss that would be suffered
under the alternative institutional arrangement. Substituting, δ = 1 and
λ cb = λ g 1 in Eq. (19) results in: EL g = (λg 1 )2
2α 2 . (22)
19 Because η is a function of δ, Eq. (21) is quartic in δ. This polynomial has four distinct
roots, of which only two are real-valued. The complete solution may be found in Hughes
Hallett and Weymark (2002).