Advanced Tools in Macroeconomics Continuous time models (and - PowerPoint PPT Presentation

Advanced Tools in Macroeconomics Continuous time models (and methods) Pontus Rendahl August 24, 2016

Introduction ◮ In this lecture we will take a look at models in continuous, as opposed to discrete, time. ◮ There are some advantages and disadvantages ◮ Advantages: Can give closed form solutions even when they do not exist for the discrete time counterpart. Can be very fast to solve. Trendier than sourdough bread, fixed gear bicycles, and skinny jeans combined (so if you do continuous time you need neither). ◮ Disadvantages: Intuition is a bit tricky. Contraction mapping theorems / convergence results go out the window. The latter can create issues for numerical computing. Difficult to deal with end-conditions (like finite lives etc.)

Plan for today ◮ Continuous time methods and models are not as well documented as the discrete time cases. ◮ Proceed through a series of examples 1. The Solow growth model (!) 2. The (stochastic) Ramsey growth model 3. A monetary economy 4. Search and matching ◮ How to solve (turns out to be pretty easy, and we can apply methods we know from earlier parts of the course)

The Solow growth model ◮ The Solow growth model is characterized by the following equations t ( A t N t ) 1 − α Y t = K α K t +1 = I t + (1 − δ ) K t S t = sY t I t = S t A t +1 = (1 + g ) A t N t +1 = (1 + η ) N t ◮ To solve this model we rewrite it in intensive form X t x t = , for X = { Y , K , S , I } A t N t

The Solow growth model ◮ Using this and substituting in gives K t +1 = sk α t + (1 − δ ) k t A t N t ◮ We can rewrite as K t +1 A t +1 N t +1 = sk α t + (1 − δ ) k t A t +1 N t +1 A t N t A t +1 N t +1 k t +1 = sk α t + (1 − δ ) k t A t N t k t +1 (1 + g )(1 + η ) = sk α t + (1 − δ ) k t

The Solow growth model ◮ Ta-daa! sk α (1 − δ ) k t t k t +1 = (1 + g )(1 + η ) + (1 + g )(1 + η ) ◮ Balanced growth: k t +1 = k t = k 1 � g + η + g η + δ � α − 1 k = s ◮ This is not textbook stuff. Why? Discrete time. More elegant solution in continuous time.

The Solow growth model ◮ Continuous time is not a state in itself, but is the effect of a limit. A derivative is a limit, an integral is a limit, the sum to infinity is a limit, and so on. ◮ Continuous time is the name we use for the behavior of an economy as intervals between time periods approaches zero.

The Solow growth model ◮ The right approach is therefore to derive this behavior as a limit (much like you probably derived derivatives from its limit definition in high school). ◮ Eventually you may get so well versed in the limit behavior that you can set it up directly (like you can say that the derivative of ln x is equal to 1/x, without calculating lim ε → 0 (ln( x + ε ) − ln( x )) /ε ) ◮ I’m not there yet. I have to do this the complicated way. People like Ben Moll at Princeton is. Take a look at his lecture notes on continuous time stuff. They are great.

The Solow growth model ◮ Back to the model. ◮ Suppose that before the length of each time period was one month. Now we want to rewrite the model on a biweekly frequency. ◮ It seems reasonable to assume that in two weeks we produce half as much as we do in one month: t ( A t N t ) 1 − α . Y t = 0 . 5 K α ◮ It also seems reasonable that capital depreciates slower, i.e. 0 . 5 δ .

The Solow growth model ◮ Notice that we still have N t worker and K t units of capital: Stocks are not affected by the length of time intervals (although the accumulation of them will). ◮ The propensity to save is the same, but with half of the income saving is halved too (and therefore investment) ◮ What happens to the exogenous processes for A t and N t ?

The Solow growth model ◮ Before A t +1 = (1 + g ) A t , N t +1 = (1 + η ) N t ◮ Now A t +0 . 5 = (1 + 0 . 5 g ) A t , N t +0 . 5 = (1 + 0 . 5 η ) N t or A t +0 . 5 = e 0 . 5 g A t , N t +0 . 5 = e 0 . 5 η N t ?

The Solow growth model ◮ It turns out that this choice does not matter much for our purpose ◮ Suppose that the time period is not one month but ∆ × one month. And suppose that A t +∆ = (1 + ∆ g ) A t ◮ Rearrange A t +∆ − A t = gA t . ∆ ◮ And take limit ∆ → 0 ˙ A t = gA t

The Solow growth model ◮ Suppose that the time period is not one month but ∆ × one month. And suppose that A t +∆ = e ∆ g A t ◮ Rearrange = ( e ∆ g − 1) A t +∆ − A t A t . ∆ ∆ ◮ Notice ( e ∆ g − 1) ( ge ∆ g ) lim = lim = g ∆ 1 ∆ → 0 ∆ → 0

The Solow growth model ◮ So ( e ∆ g − 1) lim A t = gA t ∆ ∆ → 0 and thus ˙ A t = gA t ◮ Therefore, it doesn’t matter if A t +∆ = (1 + ∆ g ) A t or A t +∆ = e ∆ g A t . The limits are the same.

The Solow growth model ◮ Solow growth model in ∆ units of time t ( A t N t ) 1 − α Y t = ∆ K α K t +∆ = I t + (1 − ∆ δ ) K t S t = sY t I t = S t A t +∆ = (1 + ∆ g ) A t N t +∆ = (1 + ∆ η ) N t

The Solow growth model ◮ Substitute and rearrange as before K t +∆ A t +∆ N t +∆ = s ∆ k α t + (1 − ∆ δ ) k t A t +∆ N t +∆ A t N t A t +∆ N t +∆ = s ∆ k α t + (1 − ∆ δ ) k t k t +∆ A t N t k t +∆ (1 + ∆ g )(1 + ∆ η ) = s ∆ k α t + (1 − ∆ δ ) k t

The Solow growth model ◮ Simplify and rearrange k t +∆ (1+∆ g )(1 + ∆ η ) = s ∆ k α t + (1 − ∆ δ ) k t ⇒ k t +∆ − k t = s ∆ k α t − ∆ δ k t − ∆( g + η + ∆ g η ) k t +∆ k t +∆ − k t ⇒ = sk α t − δ k t − ( g + η + ∆ g η ) k t +∆ ∆ ◮ Take limits ∆ → 0 ˙ k t = sk α t − ( g + η + δ ) k t ◮ With steady state 1 � g + η + δ � α − 1 k = s

The Solow growth model: Solution ◮ How do you solve this model? ◮ The equation ˙ k t = sk α t − ( g + η + δ ) k t is an ODE. ◮ Declare it as a function with respect to time, t , and capital, k , in Matlab as sk α − ( g + η + δ ) k ; solow = @( t , k ) ◮ The simulate it for, say 100 units of time, with initial condition k 0 as [ time , capital ] = ode45 ( solow , [ 0 100 ] , k 0 );

The Solow growth model: Solution Solow growth model: Saving like China 2 2 1 1 Output Capital 0 0 -1 -1 -2 0 50 100 0 50 100 1 Consumption 1 Investment 0 -1 0 -2 -1 -3 0 50 100 0 50 100 Time (years) Time (years)

The Solow growth model: Solution A few pointers ◮ Once you got the solution of a deterministic continuous time model, the solution will always be of the form x t = f ( x t ), whether or not x t is a vector. ˙ ◮ The matlab function ode45 (or other versions) can then simulate a transition (such as an impulse response). ◮ You could also simulate on your own through the approximation x t ≈ x t +∆ − x t ˙ ∆ and thus find your solution as x t +∆ = x t + ∆ f ( x t ). ◮ For this to be accurate, ∆ must be small if there are a lot of nonlinearities. ◮ The ODE function in matlab uses so-called Runge Kutta methods to vary the step-size ∆ in an optimal way.

The Ramsey growth model ◮ Now consider the Ramsey growth model (without growth) v ( k t ) = max c t , k t +1 { u ( c t ) + (1 − ρ ) v ( k t +1 ) } s.t. c t + k t +1 = k α t + (1 − δ ) k t ◮ In ∆ units of time v ( k t ) = max c t , k t +∆ { ∆ u ( c t ) + (1 − ∆ ρ ) v ( k t +∆ ) } s.t. ∆ c t + k t +∆ = ∆ k α t + (1 − ∆ δ ) k t

The Ramsey growth model ◮ Notice that all flows change when the length of the time period on which they are defined changes. Stocks, k , are the same. ◮ I discount the future with 1 − ∆ ρ instead of (1 − ρ ) (or with e − ∆ ρ instead of e ρ , but these are, in the limit, equivalent). ◮ One funny thing: Consumption, c , is still “monthly” consumption, but it now only cost ∆ as much, and I only get a ∆ fraction of the utility! ◮ These assumptions are for technical reasons, and it will (hopefully) soon be clear why they are made.

The Ramsey growth model ◮ Bellman equation v ( k t ) = max c t , k t +∆ { ∆ u ( c t ) + (1 − ∆ ρ ) v ( k t +∆ ) } s.t. ∆ c t + k t +∆ = ∆ k α t + (1 − ∆ δ ) k t ◮ Subtract v ( k t ) from both sides and insert the budget constraint into v ( k t +∆ ) 0 = max c t { ∆ u ( c t ) + v ( k t + ∆( k α t − δ k t − c t )) − v ( k t ) − ∆ ρ v ( k t + ∆( k α t − δ k t − c t )) }

The Ramsey growth model ◮ From before 0 = max c t { ∆ u ( c t ) + v ( k t + ∆( k α t − δ k t − c t )) − v ( k t ) − ∆ ρ v ( k t + ∆( k α t − δ k t − c t )) } ◮ Divide by ∆ c t { u ( c t ) + v ( k t + ∆( k α t − δ k t − c t )) − v ( k t ) 0 = max ∆ − ρ v ( k t + ∆( k α t − δ k t − c t )) }

The Ramsey growth model ◮ From before c t { u ( c t ) + v ( k t + ∆( k α t − δ k t − c t )) − v ( k t ) 0 = max ∆ − ρ v ( k t + ∆( k α t − δ k t − c t )) } ◮ Take limit ∆ → 0 and rearrange c t { u ( c t ) + v ′ ( k t )( k α ρ v ( k t ) = max t − δ k t − c t ) } ◮ This is know as the Hamilton-Jacobi-Bellman (HJB) equation.

The Ramsey growth model: Solving ◮ Dropping time notation we have c { u ( c ) + v ′ ( k )( k α − δ k − c ) } ρ v ( k ) = max ◮ This is simple to solve and (can be) blazing fast!