Perturbation and LQ Wouter J. Den Haan London School of Economics - PowerPoint PPT Presentation

Perturbation and LQ Wouter J. Den Haan London School of Economics c � 2011 by Wouter J. Den Haan March 24, 2014

No uncertainty With uncertainty Global method Linear-Quadratic Neoclassical growth model - no uncertainty β t − 1 c 1 − γ ∞ − 1 ∑ t max 1 − γ { c t , k t + 1 } ∞ t = 1 t = 1 s.t. c t + k t + 1 = k α t + ( 1 − δ ) k t k 1 is given � � c − γ = β c − γ α k α − 1 t + 1 + 1 − δ t t + 1

No uncertainty With uncertainty Global method Linear-Quadratic Neoclassical growth model - no uncertainty When we substitute out consumption using the budget constraint we get t + ( 1 − δ ) k t − k t + 1 ) − γ ( k α = � − γ � � � k α α k α − 1 t + 1 + ( 1 − δ ) k t + 1 − k t + 2 t + 1 + 1 − δ β ,

No uncertainty With uncertainty Global method Linear-Quadratic General specification I f ( x , x � , y , y � ) = 0. • x : n x × 1 vector of endogenous & exogenous state variables • y : n y × 1 vector of endogenous choice variable

No uncertainty With uncertainty Global method Linear-Quadratic General specification II Model: f ( x �� , x � , x ) = 0 for a known function f ( · ) . Solution is of the form: x � = h ( x ) Thus, F ( x ) ≡ f ( h ( h ( x )) , h ( x ) , x ) = 0 ∀ x

No uncertainty With uncertainty Global method Linear-Quadratic Neoclassical growth model again f ( k � , k , c � , c ) = � − c − γ + β ( c � ) − γ � � � α ( k � ) α − 1 + 1 − δ − c − k � + k α + ( 1 − δ ) k Solution is of the form: k � = h ( k ) c = g ( k ) Thus, F ( k ) ≡ � − g ( k ) − γ + β g ( h ( k )) − γ � � � α h ( k ) α − 1 + 1 − δ − g ( k ) − h ( k ) + k α + ( 1 − δ ) k

No uncertainty With uncertainty Global method Linear-Quadratic Neoclassical growth model again ( − k α − ( 1 − δ ) k − k � ) − γ f ( k �� , k � , k ) = + β (( k � ) α + ( 1 − δ ) k � − k �� ) − γ � � α ( k � ) α − 1 + 1 − δ , for known values of α , δ , and γ Solution is of the form: k � = h ( k ) Thus F ( k ) ≡ ( − k α − ( 1 − δ ) k − h ( k )) − γ + β ( h ( k ) α + ( 1 − δ ) h ( k ) − h ( h ( k ))) − γ � � α h ( k ) α − 1 + 1 − δ ,

No uncertainty With uncertainty Global method Linear-Quadratic Key condition F ( k ) = 0 ∀ x

No uncertainty With uncertainty Global method Linear-Quadratic Linear, Log-linear, t(x) linear, etc • All first-order solutions are linear in something • Specification in last slide • = ⇒ solution that is linear in the level of k

No uncertainty With uncertainty Global method Linear-Quadratic Linear, Log-linear, t(x) linear, etc • How to get a solution that is linear in ˜ k = ln ( k ) ? • write the f ( · ) function as � − exp � � − ( 1 − δ ) exp � ˜ � − exp � ˜ k � �� − γ α ˜ k k + � � k � � + ( 1 − δ ) exp � ˜ k � � − exp � ˜ k �� �� − γ f ( ˜ k �� , ˜ k � , ˜ α ˜ k ) = β exp × � � ( α − 1 ) ˜ k � � + 1 − δ � α exp

No uncertainty With uncertainty Global method Linear-Quadratic Linear, Log-linear, t(x) linear, etc • How wo get a solution that is linear in ˆ k = t ( k ) ? • Write the f ( · ) function as f ( ˆ k �� , ˆ k � , ˆ k ) = � � α � � � � t inv ( ˆ t inv ( ˆ t inv ( ˆ k � ) ) − γ + ( − k ) − ( 1 − δ ) k ) − �� � α � � � �� − γ t inv ( ˆ t inv ( ˆ t inv ( ˆ k � ) k � ) k �� ) β + ( 1 − δ ) − × � � � � α − 1 t inv ( ˆ k � ) α + 1 − δ

No uncertainty With uncertainty Global method Linear-Quadratic Numerical solution Let x solve f ( x , x , x ) = 0 That is x = h ( x ) Taylor expansion h ( x ) + ( x − x ) h � ( x ) + ( x − x ) 2 h �� ( x ) + · · · h ( x ) ≈ 2 ( x − x ) 2 = x + h 1 ( x − x ) + h 2 + · · · 2 • Goal is to find x , h 1 , h 2 , etc.

No uncertainty With uncertainty Global method Linear-Quadratic Solving for first-order term F ( x ) = 0 ∀ x Implies F � ( x ) = 0 ∀ x

No uncertainty With uncertainty Global method Linear-Quadratic Definitions Let � � ∂ f ( x �� , x � , x ) � = f 1 , � ∂ x �� x �� = x � = x = x � ∂ f ( x �� , x � , x ) � � = f 2 , � ∂ x � x �� = x � = x = x � � ∂ f ( x �� , x � , x ) � = f 3 . � ∂ x x �� = x � = x = x Note that � �� � � ∂ h ( x ) � � = h 1 + h 2 ( x − x ) + · · · x = x = h 1 � � ∂ x x = x

No uncertainty With uncertainty Global method Linear-Quadratic Key equation F � ( x ) = 0 ∀ x or ∂ h ( x � ) F � ( x ) = ∂ f ∂ h ( x ) + ∂ f ∂ h ( x ) + ∂ f ∂ x = 0 ∂ x �� ∂ x � ∂ x � ∂ x ∂ x can be written as 2 F � ( x ) = f 1 h 1 + f 2 h 1 + f 3 = 0 • One equation to solve for h 1

No uncertainty With uncertainty Global method Linear-Quadratic Key equation F � ( x ) = 0 ∀ x or ∂ h ( x � ) F � ( x ) = ∂ f ∂ h ( x ) + ∂ f ∂ h ( x ) + ∂ f ∂ x = 0 ∂ x �� ∂ x � ∂ x � ∂ x ∂ x can be written as 2 F � ( x ) = f 1 h 1 + f 2 h 1 + f 3 = 0 • One equation to solve for h 1 • Hopefully, the Blanchard-Kahn conditions are satisfied and there is only one sensible solution

No uncertainty With uncertainty Global method Linear-Quadratic Solving for second-order term F � ( x ) = 0 ∀ x Implies F �� ( x ) = 0 ∀ x

No uncertainty With uncertainty Global method Linear-Quadratic Definitions Let � ∂ 2 f ( x �� , x � , x ) � � = f 13 . (1) � ∂ x �� ∂ x x �� = x � = x = x and note that � �� � ∂ 2 h ( x ) � � � = h 2 + h 3 ( x − x ) + · · · x = x = h 2 . � (2) � ∂ x 2 x = x

No uncertainty With uncertainty Global method Linear-Quadratic Key equation F �� ( x ) = 0 ∀ x or F �� ( x ) = � ∂ 2 f � � ∂ h ( x � ) � ∂ h ( x � ) ∂ 2 f ∂ x + ∂ 2 f ∂ h ( x ) ∂ h ( x ) ∂ h ( x ) + ∂ x + ∂ x �� 2 ∂ x � ∂ x �� ∂ x � ∂ x �� ∂ x ∂ x � ∂ x � ∂ h ( x � ) � ∂ 2 h ( x ) + ∂ 2 h ( x � ) + ∂ f ∂ h ( x ) ∂ h ( x ) ∂ x �� ∂ x � ∂ x 2 ∂ x � 2 ∂ x ∂ x � ∂ 2 f � ∂ h ( x ) ∂ h ( x � ) + ∂ 2 f + ∂ 2 f ∂ h ( x ) ∂ h ( x ) + ∂ x � x �� ∂ x � ∂ x � 2 ∂ x � ∂ x ∂ x ∂ x ∂ x ∂ 2 h ( x ) + ∂ f ∂ x � ∂ x 2 � ∂ 2 f � ∂ h ( x � ) + ∂ 2 f + ∂ 2 f ∂ h ( x ) ∂ h ( x ) + ∂ xx �� ∂ x � ∂ x ∂ x � ∂ x 2 ∂ x ∂ x

No uncertainty With uncertainty Global method Linear-Quadratic Key equation Which can be written as � � 2 2 2 F �� ( x ) = f 11 h 1 + f 12 h 1 + f 13 h 1 + f 1 ( h 1 h 2 + h 2 h 1 ) + � � � � 2 2 f 21 h 1 + f 22 h 1 + f 23 h 1 + f 2 h 2 + f 31 h 1 + f 32 h 1 + f 33 = 0 • One linear equation to solve for h 2

No uncertainty With uncertainty Global method Linear-Quadratic Discussion • Global or local? • Borrowing constraints? • Quadratic/cubic adjustment costs?

No uncertainty With uncertainty Global method Linear-Quadratic Neoclassical growth model with uncertainty β t − 1 c 1 − γ ∞ − 1 t ∑ max E 1 1 − γ { c t , k t + 1 } ∞ t = 1 t = 1 s.t. c t + k t + 1 = exp ( θ t ) k α t + ( 1 − δ ) k t (3) θ t = ρθ t − 1 + σ e t , (4)

No uncertainty With uncertainty Global method Linear-Quadratic General specification E f ( x , x � , y , y � ) = 0. • x : n x × 1 vector of endogenous & exogenous state variables • y : n y × 1 vector of endogenous choice variable • Stochastic growth model: y = c and x = [ k , θ ] .

No uncertainty With uncertainty Global method Linear-Quadratic Essential insight #1 • Make uncertainty (captured by one parameter) explicit in system of equation E f ( x , x � , y , y � , σ ) = 0. Solutions are of the form: y = g ( x , σ ) and x � = h ( x , σ ) + σηε �

No uncertainty With uncertainty Global method Linear-Quadratic Neoclassical Growth Model • For standard growth model we get E f ([ k , θ ] , [ k � , ρθ + σε � ] , y , y � ) = 0 Solutions are of the form: c = c ( k , θ , σ ) (5) and � k � � � k � ( k , θ , σ ) � � 0 � e � . = + σ (6) θ � ρθ 1

No uncertainty With uncertainty Global method Linear-Quadratic Essential insight #2 Perturb around y , x , and σ . g ( x , σ ) = g ( x , 0 ) + g x ( x , 0 )( x − x ) + g σ ( x , 0 ) σ + · · · and h ( x , σ ) = h ( x , 0 ) + h x ( x , 0 )( x − x ) + h σ ( x , 0 ) σ + · · ·

No uncertainty With uncertainty Global method Linear-Quadratic Goal Let g x = g x ( x , 0 ) , g σ = g σ ( x , 0 ) and h x = h x ( x , 0 ) , h σ = h σ ( x , 0 ) . Goal: to find • ( n y × n x ) matrix g x , ( n y × 1 ) vector g σ , ( n x × n x ) matrix h x , ( n x × 1 ) vector h σ . • The total of unknowns = ( n x + n y ) × ( n x + 1 ) = n × ( n x + 1 ) .

No uncertainty With uncertainty Global method Linear-Quadratic More on uncertainty Results for first-order perturbation • g σ = h σ = 0 Results for second-order perturbation • g σ x = h σ x = 0 , but g σσ � = 0 and h σσ � = 0 How to model discrete support?