Mathias Christensen-3 - Aevering 2 Mathias V Christensen...

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∂C 1 ∂T 1 = θ ( Y L 1 ) - ( T 1 ) + ( 1 1 + r ( Y L 2 - T 2 ) + V 1 ) = - θ θ θ θ = 1 1 + ( 1 + r ) σ - 1 ( 1 + φ ) - σ 0 < θ < 1 r > 0 φ > 0 σ r φ φ φ θ σ σ < 1 θ σ = 1 σ > 1 σ C 2 C 1 MRS ( C 2 ; C 1 ) σ σ 1 + r 1 + φ r > φ σ θ dT 1 = dT 2 dC 1 = θ dY L 1 - dT 1 + 1 1 + r ( dY L 2 - dT 2 ) + dV 1 = θ dY L 1 - dT 1 + 1 1 + r ( dY L 2 - dT 1 ) + dV 1 dY L 1 = dY L 2 = dV 1 = 0 dC 1 = θ - dT 1 - 1 1 + r ( dT 1 ) dC 1 dT 1 = - θ ( 1 + 1 1 + r )
1 + 1 1 + r 1 + 1 1 + r θ r = φ dT 1 = dT 2 ∂C 1 ∂T 1 = - 1 1 + ( 1 + r ) σ - 1 ( 1 + φ ) - σ ( 1 + 1 1 + r ) = - 1 1 + ( 1 + r ) σ - 1 ( 1 + r ) - σ ( 1 + 1 1 + r ) = - 1 1 + ( 1 + r ) - 1 ( 1 + 1 1 + r ) = - 1 + r ( 1 + r ) + 1 ( 1 + 1

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