Lecture VIII Global Approximation Methods: I · 2015-10-08 · We can exploit information about f...

Lecture VIII

Global Approximation Methods: I

Gianluca Violante

New York University

Quantitative Macroeconomics

G. Violante, ”Global Methods” p. 1 /29

Global function approximation

• Global methods: function interpolation over the entire domain

• First choice: type of approximation

1. Interpolation I: finite element methods

◮ Splines (e.g., B-splines, cubic splines, Schumaker splines)

2. Interpolation II: spectral methods

◮ Orthogonal polynomials (e.g., Chebyshev, Lagrange, etc...)

• Second choice: how to minimize residuals between true functionand interpolant

1. Collocation

2. Least squares

3. Galerkin method


Interpolation

• Today, one dimension.

• We want to represent a real-valued function f on [a, b] with acomputationally tractable function

• Interpolation is a form of function approximation in which theapproximating (interpolant) and the true function must agree , i.e.they must have same value, at a finite number of points.

• In some cases additional restrictions are imposed on theinterpolant. E.g., f ′ evaluated at a finite number of points mayhave to agree with that of the underlying function. Other examplesinclude additional constraints imposed on the interpolant such asmonotonicity, convexity, or smoothness requirements.

• In general, interpolation is any method that takes information at afinite set of points, X , and finds a function which satisfies thatinformation at X .


Interpolation

• We want to approximate a known function f (x) . The interpolant fis chosen to be a linear combination of a collection of basisfunctions {φj (x)}

n

j=0where n is the order/degree of interpolation

• Basis functions: linearly independent functions that span thefamily of functions chosen for the interpolation (the function space)

◮ any function in the space can be obtained as a linearcombination of basis functions, just as every vector in a vectorspace can be obtained as a linear combination of basisvectors.

• In general we are interested in the space of continuous orcontinuously differentiable functions.


Interpolation

• Thus, for a given family of basis function {φj}n

j=0we have:

f(x) ≃ f(x) ≡n∑

j=0

wjφj (x)

• We have reduced the problem of characterizing aninfinite-dimensional object, f , to the problem of determining the n

weights (or coefficients) {wj}

• There is arbitrariness in interpolation: there are arbitrarily manyfunctions passing through a finite number of points. Which basis?

• Choices: spectral or finite element methods depending whetherthe basis functions are nonzero over the entire domain of the truefunction (except possibly at a finite number of points) or nonzeroonly on a subinterval of the domain.


Finite element methods

• Use local basis

• These are called splines

• An order n spline consists of a series of nth order polynomialsegments spliced together so as to preserve continuity ofderivatives of order n− 1 or less.

• The points at which the pieces are spliced together are calledknots or breakpoints.

• By convention the first and last knots are the two extremes of thefunction domain.


Splines

• An order n spline with K + 1 knots has K (n+ 1) parameters.

• The simplest way to proceed (Judd calls it the “kindergartenprocedure of connecting the dots”) is to use linear B-splines,where B stands for basic.

• Suppose we have a grid with knots x0 < ...xk < ... < xK and wewant to approximate a function f (x) .

• B0 splines are defined as right-continuous step functions

B0k (x) =

{

1 if xk ≤ x < xk+1

0 otherwise

• So interpolant is:

f (x) =K−1∑

k=0

[

f (xk) + f (xk+1)

2

]

B0k (x)


Linear B-splines

• B1 splines implement piece-wise linear interpolation

B1k (x) =

x−xk−1

xk−xk−1if xk−1 ≤ x < xk

xk+1−x

xk+1−xkif xk ≤ x < xk+1

0 elsewhere

and each one looks like a “tent-function” with peak at xk equal toone, thus the weight of the interpolant must be f (xk)

• The interpolant is therefore defined as:

f (x) =K∑

k=0

f (xk)B1k (x)

• You can also obtain the same weights from 2K conditions: f atknots equal to f(x) (K + 1 conditions), and interpolant continuousat interior knots (K − 1 conditions)


Linear spline Basis Functions on [0, 1].

0

1

0

1

0 10

1

0 1 0 1


Linear B-splines

• It is easy to see that for any given point x on an interval [xk, xk+1]there are only two nonzero basis at x, Bk (x) and Bk+1 (x) .

• To locate the knots (k, k + 1) that bracket x you use a bracketingalgorithm that essentially proceeds by bisection.

• Thus

f (x) = f (xk)xk+1 − x

xk+1 − xk

+ f (xk+1)x− xk

xk+1 − xk

= f (xk)xk+1 − x− xk + xk

xk+1 − xk

+ f (xk+1)x− xk

xk+1 − xk

= f (xk) + [f (xk+1)− f (xk)]x− xk

xk+1 − xk

that is a familiar formula for linear interpolation of decision rules(see later...)


Linear Spline approximation (7 knots)

x0=a x

1 x

2 x

3 x

4 x

5 x

6=b


Pros and cons of linear splines

• Pros:

◮ Preserves monotonicity and concavity of f .

◮ We can exploit information about f clustering points moreclosely together in areas of high curvature or areas where weknow a kink exists in order to increase our accuracy.

◮ We can capture binding inequality constraints very well.

• Cons:

◮ The approximated function is not differentiable at the knots(they become kinks).

◮ f ′′ = 0 where it exists (it does not exist at the knots). Functionnot smooth.


Cubic splines

• Cubic spline (n = 3), has 4K coefficients to be determined

◮ f (xk) = f (xk) (K + 1 conditions)

◮ Continuity of f at interior points (K − 1 conditions)

◮ Continuity of f ′ at interior points (K − 1 conditions)

◮ Continuity of f ′′ at interior points (K − 1 conditions)

◮ f ′′ (x0) = f ′′ (xK) = 0 (2 conditions)

• These last two can be modified to something else depending onwhat is suitable in the particular case at hand, they are ad-hocand they can be more than 2.


Cubic spline Basis Functions on [0,1].

0

1

0

1

0 10

1

0 1 0 1


Cubic spline Basis Functions on [0,1].

0

10

20

30

40

50

60

0 5 10 15 20 25

Spline 1: Piecewise Linear Interpolation

Spline 2: Piecewise Cubic Interpolation

Data


Pros and cons of cubic splines

• Pros:

◮ Easy to compute: interpolant matrix very sparse, easy toinvert.

◮ Smooth approximation

• Cons:

◮ It may not be able to handle constraints well

◮ It does not preserve monotonicity and concavity of f


Schumaker splines

• So far: linear interpolation preserves shape (monotonicity andconcavity), but not differentiability. Cubic splines preservesdifferentiability, but not shape.

• Schumaker quadratic splines preserve both (Schumaker, 1983,SIAM Journal on Numerical Analysis).

• By shape we mean that in those intervals where the data ismonotonically increasing or decreasing, the spline has the sameproperty. Similarly for convexity or concavity.

• Idea: add knots to a subinterval by means of an algorithm


Solving the income fluctuation problem

V (a, yj) = max{c,a′}

u (c) + β∑

yi∈Y

π (yi|yj)V (a′, yi)

s.t.

c+ a′ ≤ Ra+ yj

a′ ≥ a

The Euler equation, once we substitute in the budget constraint, reads:

uc (Ra+ yj − a′)− βR∑

yi∈Y

π (yi|yj)uc (Ra′ + yi − a′′) ≥ 0.

where the strict inequality holds when the constraint binds.


Policy function iteration with linear interpolation

1. Construct a grid on the asset space {a0, a2, ..., am} witha0 = a = 0.

2. Guess an initial vector of decision rules for a′′ on the grid points,call it a0 (ai, yj), where the subscript 0 denotes the initial iteration

3. For each point (ai, yj) on the grid, check whether the borrowingconstraint binds. I.e. check whether:

uc (Rai + yj − a0)−βR∑

y′∈Y

π (y′|yj)uc (Ra0 + y′ − a0 (a0, y′)) > 0.

4. If this inequality holds, the borrowing constraint binds. Then, seta′0 (ai, yj) = a0 and repeat this check for the next grid point. If theequation instead holds with the < inequality, we have an interiorsolution (it is optimal to save for the household) and we proceedto the next step.



5. For each point (ai, yj) on the grid, use a nonlinear equation solverto find the solution a∗ of the nonlinear equation

uc (Rai + yj − a∗)− βR∑

y′∈Y

π (y′|yj)uc (Ra∗ + y′ − a0 (a∗, y′)) = 0

(a) Need to evaluate the function a0 (a, y′) outside grid points:

assume it is piecewise linear.

(b) Every time the solver calls an a∗ which lies between gridpoints, do as follows. First, find the pair of adjacent grid points{ai, ai+1} such that ai < a∗ < ai+1, and then compute

a0 (a∗, y′) = a0 (ai, y

′)+(a∗ − ai)

(

a0 (ai+1, y′)− a0 (ai, y

′)

ai+1 − ai

)

(c) If the solution of the nonlinear equation is a∗, then seta′0 (ai, yj) = a∗ and iterate on the next grid point.



6. Check convergence by comparing a′0 (ai, yj)− a0 (ai, yj) throughsome pre-specified norm. For example, declare convergence atiteration n when

maxi,j

{|a′n (ai, yj)− an (ai, yj) |} < ε

for some small number ε which determines the degree oftolerance in the solution algorithm.

7. If convergence is achieved, stop. Otherwise, go back to point 3with the new guess a1 (ai, yj) = a′0 (ai, yj).

Note that the most time-consuming step in this procedure is 5, theroot-finding problem. We now discuss how to avoid it.


Endogenous grid method (EGM)

• EGM is much faster than the traditional method because it doesnot require the use of a nonlinear equation solver

• The essential idea of the method is to construct a grid on a′, nextperiod’s asset holdings, rather than on a, as is done in thestandard algorithm.

• The method also requires the policy function to be at least weaklymonotonic in the state

• Recall the Euler equation:

uc (c (a, y)) ≥ βR∑

y′∈Y

π (y′|y)uc (c (a′, y′)) .

• As usual, we start from a guess c0 (a, y) and iterate on the EulerEquation until the decision rule for consumption that we solve foris essentially identical to the one in the previous iteration.


EGM: Algorithm

1. Construct a grid for (a′, y)

2. Guess a policy function c0 (ai, yj). If y is persistent, a good initialguess is to set c0 (ai, yj) = rai + yj which is the solution underquadratic utility if income follows a random walk.

3. Fix yj . Instead of iterating over {ai} , we iterate over {a′i}. For anypair {a′i, yj} on the mesh construct the RHS of the Euler equation[call it B (a′i, yj)]

B (a′i, yj) ≡ βR∑

y′∈Y

π (y′|yj)uc (c0 (a′i, y

′))

where the RHS of this equation uses the guess c0 on the grid


EGM: Algorithm

4. Use the Euler equation to solve for the value c (a′i, yj) that satisfies

uc (c (a′i, yj)) = B (a′i, yj)

and note that it can be done analytically, e.g. for uc (c) = c−γ we

have c (a′i, yj) = [B (a′i, yj)]− 1

γ . Here algorithm becomes muchmore efficient because it does not require a nonlinear solver.

5. From the budget constraint:

c (a′i, yj) + a′i = Ra∗i + yj

solve for a∗ (a′i, yj) the value of assets today that would lead theconsumer to have a′i assets tomorrow if her income shock was yjtoday. This yields c (a∗i , yj) = c (a′i, yj), a function not defined onthe grid points. This is the endogenous grid and it changes oneach iteration.


EGM: Algorithm

6. Let a∗0 be the value of asset holdings that induces the borrowingconstraint to bind next period, i.e., the value for a∗ that solves thatequation at the point a′0, the lower bound of the grid.

7. Now we need to update our guess defined on the original grid.

• To get new guess c1 (ai, yj) on grid points ai > a∗0 we can usesimple linear interpolation methods using values for{

c (a∗n, yj) , c(

a∗n+1, yj)}

on the two most adjacent values{

a∗n, a∗n+1

}

that bracket ai.

• If some points ai are beyond a∗m, the upper bound of theendogenous grid, just extend linearly the function


EGM: Algorithm

• To update the consumption policy function on grid valuesai < a∗0, we use the budget constraint:

c1 (ai, yj) = Rai + yj − a′0

since we cannot use the Euler equation as the borrowingconstraint is binding for sure next period: the reason is that wefound it was binding at a∗0, therefore a fortiori it will be bindingfor ai < a∗0.

8. Check convergence as before.


Envelope condition method (ECM)

It is an alternative method that, in some cases, like the EGM, avoidsthe use of nonlinear solvers

1. Construct a grid on a with K + 1 points: call it A.

2. Guess a value function V0 (a, yj) =∑K

k=0w0

kjBk (a) where Bk arecubic splines.

3. Compute the derivative of the value function on the nodes of A:

V ′0 (ai, yj) =

K∑

k=0

w0kjB

′k (ai)

Note: if you wish to use B1 splines instead, you need to use adifferent grid in the above step because V0 is not differentiable onthe nodes of the original grid A


Envelope Condition Method (ECM)

5. For any pair {ai, yj} on A× Y construct the envelope condition:

V ′0 (ai, yj) = uc (Rai + yj − a0 (ai, yj))R

from which we can obtain:

a0 (ai, yj) = Rai + yj − u−1

c

(

V ′0 (ai, yj)

R

)

On each point of the grid we must verify if the borrowing constraintbinds and if it does, we set a0 (ai, yj) = 0.

6. Update the value function from the Bellman equation. For eachpoint of the grid A compute:

V1 (ai, yj) = u (Rai + yj − a0 (ai, yj))+β∑

yj′∈Y

π (yj′ |yj) V0 (a0 (ai, yj) , yj′)


Envelope Condition Method (ECM)

• In practice, need to solve (K + 1)J equations in (K + 1)J

coefficients{

w1kj

}

of the type:

K∑

k=0

w1kjBk (ai) = u (Rai + yj − a0 (ai, yj))

+β∑

yj′∈Y

π (yj′ |yj)K∑

k=0

w0kj′Bk (a0 (ai, yj))

for all gridpoints, i = 0, 1, ...K and j = 1, 2, ..., J

• It is a linear system of equations


Lecture VIII Global Approximation Methods: I · 2015-10-08 · We can exploit information about f...

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