DSGE_log-linearisation

Log-linearisation Tutorial Weijie Chen Department of Political and Economic Studies University of Helsinki 14 November 2011 Abstract To solve DSGE models, firstly we have to collect all difference equations, such as F.O.C.s, constraints, aggregate conditions, then log-linearise them around the steady-state. It sounds easy, but it sometimes requires ingenuity and proficiency. This note is written for illustrating log-linearisation and presenting several methods, which method you are suppose to use depends on the specific functional form. 1 Introduction Log-linearisation, basically it means take the natural logarithm, then lin- earise it. Nothing mysterious, but the challenges mainly come from a tech- nical consideration: how can we log-linearise an equation easily? ‘Easily’ is a relative expression, if we are about to tackle a problem, at least we are supposed to pick the right weapon. To perform log-linearisation, we often work with log-deviation, it is de- fined as x˜t = lnXt − lnX (1) where x˜t is log-deviation, X is the stead-state of Xt. To see why we need to work with log-deviation, we can manipulate equation above, x˜t = ln ( Xt X ) = ln ( 1 + Xt X − 1 ) = ln ( 1 + Xt X − X X ) = ln ( 1 + Xt −X X ) Now we need to employ the first-order Taylor expansion, f(xt) ≈ f(x) + f ′(x)(xt − x) (2) where x is steady-state of xt. Take derivative of ln (1 + (Xt −X)/X) w.r.t. Xt, ∂ ln ( 1 + Xt−XX ) ∂Xt = X Xt 1 X = 1 Xt Then use (2), we get x˜t = ln ( 1 + Xt −X X ) ≈ 1 X (Xt −X) (3) 1 And the right-hand side of (3) is the percentage deviation from steady state X, this is the reason we prefer to work with log-deviation. Sometimes we choose replace Xt −X with dXt, the differential of Xt, which will be clear in later examples. And we can also rearrange (3), x˜t ≈ 1 X (Xt −X) = Xt X − 1 Xt ≈ X(x˜t + 1) (4) Then use this expression to substitute into all variables. 2 Universal method We entitle this method as universal is not because it can be employed deal- ing with any question, but it reveals the most fundamental idea of log- linearisation: take natural logarithm then linearise it. Let’s use a Cobb- Douglas production function as the first example Yt = K α(AtLt) 1−α Take natural logarithm on both sides, lnYt = α lnKt + (1− α) lnAt + (1− α) lnLt (5) 2 Then we need to use Taylor expansion to the first degree around steady- state, for every terms lnYt = lnY + 1 Y (Yt − Y ) lnKt = lnK + 1 K (Kt −K) lnAt = lnA+ 1 A (At −A) lnLt = lnL+ 1 L (Lt − L) Substitute back to (5), lnY + 1 Y (Yt − Y ) = α [ lnK + 1 K (Kt −K) ] + (1− α) [ lnA+ 1 A (At −A) ] + (1− α) [ lnL+ 1 L (Lt − L) ] Expand the equation above lnY + 1 Y (Yt − Y ) = α lnK + αKt K − α+ (1− α) lnA+ (1− α)At A − (1− α) + (1− α) lnL+ (1− α)Lt L − (1− α) (6) We need to eliminate the steady-state condition out of equation, according to (5) lnY = α lnK + (1− α) lnA+ (1− α) lnL Then we simplified (6) a bit, 1 Y (Yt − Y ) = αKt K − α+ (1− α)At A − (1− α) + (1− α)Lt L − (1− α) 3 Finally, since Y , K, A and L are only constants (parameters), we get a perfect linear function Yt Y = α Kt K + (1− α)At A + (1− α)Lt L − (1− α) However, as you can see from steps above, the process is quite tedious, and the result is not log-deviation form. If the function form are compli- cated, it is sensible to turn for other methods. The all rest of methods we are about to see are actually derived from this fundamental one, but they are largely simplified. 3 Total differential method In this section, you will understand why we use dX to replace Xt −X. We use Gal´ı and Monacelli (2005) as our example, to show you how to log- linearise the function you see in a real paper. From Gal´ı and Monacelli (2005), we use equation pair (6) to illustrate the total differential method, reproduce them here CH,t = (1− α) ( PH,t Pt )−η Ct, CF,t = α ( PF,t Pt )−η Ct (7) They are optimal consumption allocation to domestic goods and foreign goods. Take logarithm upon both sides for the first equation, lnCH,t = ln (1− α)− η lnPH,t + η lnPt + lnCt Then take total differential of each term on its steady-state value 1 CH dCH,t = −η 1 PH dPH,t + η 1 P dPt + 1 C dCt 4 where CH , PH and P without subscripts t are steady-state value. Then rewrite them into log-deviation form c˜H,t = −η(p˜H,t − p˜t) + c˜t (8) The same procedure for another equation, take logarithm on both sides lnCF,t = lnα− η lnPF,t + η lnPt + lnCt Take total differential on steady-state, 1 CF dCF,t = −η 1 PF dPF,t + η 1 P dPt + 1 C dCt Then log-deviation form, c˜F,t = −η(p˜F,t − p˜t) + c˜t (9) This time, the linearisation goes considerably quick, and result is in log- deviation form. The Taylor expansion to the first degree is to linearise a nonlinear function, so it functions the same as total differential approxima- tion. And obviously taking total differential is easier than Taylor expansion to the first degree, this is why we prefer not to use Taylor expansion directly. 4 Uhlig’s method In this section, we will see another interesting method proposed by Uhlig (1999), which does not even require to take derivatives. This method is just a further derivation of (1), lnXt = lnX + x˜t 5 Take exponential on both sides, Xt = e lnX+x˜t = elnX + ex˜t = Xex˜t Then the idea is clear, we replace every variable with its according trans- formed term Xex˜t , where X is the steady-state value. We have several expansion rules to follow: ex˜t ≈ 1 + x˜t ex˜t+ay˜t ≈ 1 + x˜t + ay˜t x˜ty˜t ≈ 0 Et[ae x˜t ] ≈ Et[ax˜t] + a You get the right-hand side expression by simply taking the Taylor expansion to the first degree. But you don’t need to use Taylor expansion every time, using these rules will save your lots of troubles. One important remark is that Uhlig’s method is immune to Jensen’s inequality lnEtX > Et lnX A specific example of this inequality, we have seen in our advanced mi- croeconomic textbook-risk aversion. Since here natural logarithm is strictly concave thus the inequality always holds. Then our problem is that we can’t simply take logarithm to a function with expectation operator. One clever way to circumvent the problem is to use Uhlig’s method. The example here is still from Gal´ı and Monacelli (2005), the equation 6 (10), stochastic Euler equation, 1 = βRtEt [ (Ct+1 − hCt)−σ (Ct − hCt−1)−σ Pt Pt+1 ] For sake of easy handling, we replace ξt = (Ct − hCt−1)−σ, 1 = βRtEt [ ξt+1 ξt Pt Pt+1 ] (10) Follow method in Uhlig (1999), 1 = βRer˜tEt [ ξeξ˜t+1 ξeξ˜t Pep˜t Pep˜t+1 ] Use stationary condition 1 = βR ξ ξ P P to simplify and collecting exponential terms together, 1 = Et [ er˜t+ξ˜t+1+p˜t−ξ˜t−p˜t+1 ] Taylor expansion to the first degree on the right-hand side or simply use expansion rules, 1 ≈ 1 + r˜t + Etξ˜t+1 + p˜t − ξ˜t − Etp˜t+1 (11) It is already linearised at this step, but we need to find out what is ξ˜t+1 and ξ˜t, we have previously defined ξt = (Ct − hCt−1)−σ, take logarithm on both sides, ln ξt = −σ ln(Ct − hCt−1) 7 then take total differential at steady-state value, 1 ξ dξt = −σ 1 C(1− h) [ dCt − hdCt+1] ξ˜t ≈ −σ 1 C(1− h) [ C dCt C − hC dCt+1 C ] ξ˜t ≈ −σ 1 1− h(c˜t − hc˜t+1) Substitute back to (11), 0 ≈ r˜t − Et [ σ 1− h(c˜t+1 − hc˜t) ] + σ 1− h(c˜t − hc˜t+1)− Et[p˜t+1 − p˜t] Define p˜t+1 − p˜t = p˜it+1 the inflation rate, then multiply both sides by (1− h)/σ Et(c˜t+1 − hc˜t) + 1− h σ Et[p˜it+1 − r˜t] ≈ c˜t − hc˜t−1 (12) 5 Substitution method Strictly speaking, this is not a method, it simply omits the first step of Uhlig’s replacement, then directly come to (4). There is only one step needs attention, we will see in next example: Xt + a = (1− b) Yt LtZt Taking natural logarithm on both sides, ln (Xt + a) = ln (1− b) + lnYt − lnLt − lnZt (13) Use steady-state condition, ln (X + a) = ln (1− b) + lnY − lnL− lnZ (14) 8 Subtract (14) away from (13), ln (Xt + a)− ln (X + a) = lnYt − lnY − lnLt + lnL− lnZt + lnZ They all become log-deviation form X˜t + a = y˜t − l˜t − z˜t (15) We need to find out what x˜t is, then replace X˜t + a. Here comes the step needs attention, we use (3) x˜t ≈ Xt −X X (16) Then X˜t + a becomes X˜t + a ≈ Xt + a− (X + a) X + a = Xt −X X + a (17) Since the numerator of right-hand sides of (16) and (17) are equal, we can make use of equality and set, X˜t + a(X + a) = x˜tX X˜t + a = x˜tX X + a Substitute back to (15), x˜tX X + a = y˜t − l˜t − z˜t This function actually can be easily linearised by total differential method, you can try by yourself, only needs two steps, everything will be done. 9 6 Taylor Approximation 6.1 Single variable case We can dig further to see whether we can find some tricks to largely simpli- fied the universal method. We will work on a general case first, a nonlinear difference equation Xt+1 = f(Xt) (18) where f is any nonlinear functional form you can imagine. First-order Taylor expansion of right-hand side around the steady-state is Xt+1 ≈ f(X) + f ′(X)(Xt −X) (19) If we set (18) to its steady-steady condition, X = f(X) Then (19) becomes Xt+1 ≈ X + f ′(X)(Xt −X) Dividing by X, Xt+1 X ≈ 1 + f ′(X)Xt −X X The left-hand side can be replaced by (4), 1 + x˜t+1 = 1 + f ′(X)x˜t x˜t+1 = f ′(X)x˜t (20) 10 Follow this formula, we simply take derivative at the steady-state, everything will be done. Try an example, kt+1 = (1− δ)kt + skαt Use formula (20), k˜t+1 = [sαk α−1 + (1− δ)]k˜t It is now linearised, since the term in the square brackets are simply param- eters. 6.2 Multivariable case Taylor polynomial has a vector version as well as scalar version, actually most of functions you encounter will be multivariable rather what we had seen in last section. Again, let’s start at the general case1, Xt+1 = f(Xt, Yt) (21) where f is still any nonlinear function you can imagine. The vector version (bivariate) of first-order Taylor polynomial around the steady-state is Xt+1 ≈ f(X,Y ) + fX(X,Y )(Xt −X) + fY (X,Y )(Yt − Y ) (22) As you can guess, the bivariate Taylor expansion closely relates to total derivative/differential of bivariate function. Again set steady-state condition of (21): X = f(X,Y ), (22) becomes, Xt+1 ≈ X + fX(X,Y )(Xt −X) + fY (X,Y )(Yt − Y ) 1 We use bivariate examples here for sake of simplicity. 11 Dividing by X, Xt+1 X ≈ 1 + fX(X,Y )(Xt −X) X + fY (X,Y ) (Yt − Y ) X 1 + x˜t+1 ≈ 1 + fX(X,Y )x˜t + fY (X,Y )Y X (Yt − Y ) Y x˜t+1 ≈ fX(X,Y )x˜t + fY (X,Y )Y X y˜t (23) (23) is the formula we are seeking for. Try an example again, kt+1 = (1− δ)kt + sztkαt Then calculate partial derivatives, fz(z, k) = sk α fk(z, k) = αszk α−1 + (1− δ) Use formula (23), k˜t+1 ≈ skαz˜ + [αszkα−1 + (1− δ)]k z k˜t Slightly rearrange, k˜t+1 ≈ skαz˜ + αskα + (1− δ)k z k˜t There are still many situations we haven’t covered in this note, you can find some published papers and try to log-linearise those key functions by yourself then compare the results from those provided by authors. The End. 12 References [1] McCandless, G. (2008): The ABCs of RBCs: An Introduction to Dy- namic Macroeconomic Models, Harvard University Press [2] Uhlig, H. (1999): ‘A Toolkit for Analyzing Nonlinear Dynamic Stochas- tic Models Easily’, Computational Methods for the Study of Dynamic Economies, Oxford University Press, Oxford, 30-61 13 Introduction Universal method Total differential method Uhlig's method Substitution method Taylor Approximation Single variable case Multivariable case