Notes on Foerser, Sarte, Watson (2011) "Sectoral vs. Aggregate Shocks: A Structural Factor Analysis of Industrial Production"

Transcription

1 Notes on Foerser, Sarte, Watson (2011) "Sectoral vs. Aggregate Shocks: A Structural Factor Analysis of Industrial Production"

2 Research questions 1. How correlated are shocks to industries productivities? 2. What fraction of industrial production volatility is due to common shocks? Industry-specific shocks? 3. Are the answers to (1) and (2) different for different points in the sample? Were common shocks or industry-specific shocks less volatile during the Great Moderation (after 1983)?

3 Outline Data Statistical factor analysis Model and structural factor analysis

4 Data BEA: Input/Output Table & Capital Flow Table, from Federal Reserve Board: Quarterly data on industrial production, from 1972 to Quarterly data 117 industries in manufacturing, mining, energy, and publishing.

5 Growth over previous quarter Industrial production and its components Year IP Power Gen Semiconductors MV Parts

6 Growth over previous quarter Industrial production tracks GDP Year

7 Principal component analysis Define g t as the vector of sectoral growth rates, log ( Yt+1 Y t ), and w t as the weight of each industry within the industrial sector. How can we best measure the fraction of variation in w t g t that is due to "common shocks"? Suppose that g t = Λ F t + u t where F t is some small (e.g., two) number of common factors, and u t are idiosyncratic shocks (the covariance matrix of u has zero off-diagonal terms), and F t and u t are uncorrelated. Use principal component analysis to choose Λ, F t so that ΛF explains the maximum possible variance of the g t vector. These columns of F will represent the common shocks.

8 Principal component analysis From the last slide g t = Λ F t + u t Note that Λ and F t are not separately identified. ΛF t = }{{} Λϑ ϑ 1 F t. We will normalize Λ so that the lengths of }{{} Λ F t each column equal 1. Useful formulas: Σ gg = ΛΣ FF Λ + Σ uu σ 2 g = w ΛΣ FF Λ w + w Σ uu w R 2 (F ) = w ΛΣ FF Λ w σ 2 g

9 Output Growth: Grain Milling Principal component analysis: 2-d to 1-d Output Growth: Animal Food

10 Principal component analysis The idea is to find the linear combination of the data that explains the greatest possible variance First order conditions: Note that max Λ 1 =1 Λ 1Σ Animal,Grain Λ 1 = max Λ 1 Λ 1Σ Animal,Grain Λ 1 + µ 1 ( 1 Λ 1 Λ 1 ) 2Σ Animal,Grain Λ 1 = 2µ 1 Λ 1 Σ Animal,Grain Λ 1 = µ 1 Λ 1 Λ 1Σ Animal,Grain Λ 1 = Λ 1µ 1 Λ 1 = µ 1 To maximize the left hand side, choose the unit-length eigenvector associated with the largest eigenvalue of Σ Animal,Grain.

11 Output Growth: Grain M illing Principal component analysis: 2-d to 1-d Output Growth: Animal Food Σ Animal,Grain = ( ) ( 0.78 ; Λ 1 = 0.62 )

12 Output Growth: Grain Milling Output Growth Rate Principal component analysis: 2-d to 1-d Retrieving the common factor, F Output Growth: Animal Food Σ Animal,Grain = ( Year ) ; Λ 1 = ( )

13 Output Growth: Grain Milling Output Growth Rate Principal component analysis: 2-d to 1-d Retrieving the common factor, F Output Growth: Animal Food Σ Animal,Grain = ( Year ) ; Λ 1 = ( )

14 Output Growth: Grain Milling Output Growth Rate Principal component analysis: 2-d to 1-d Retrieving the common factor, F Output Growth: Animal Food Σ Animal,Grain = ( Year ) ( ) 0.62 ; Λ 2 = 0.78

15 Output Growth: Grain Milling Output Growth Rate Principal component analysis: 2-d to 1-d Retrieving the common factor, F Output Growth: Animal Food Σ Animal,Grain = ( Year ) ( ) 0.62 ; Λ 2 = 0.78

16 Side note: Bai and Ng (2003) look at how to choose the number of factors. Principal component analysis Can extend this idea to many (say 117) data series and multiple (say 2) factors. Suppose we have 117 data series and we have computed the first factor F 1 = g Λ 1. The problem is now to find a vector Λ 2 that is orthogonal to Λ 1 and explains the greatest possible variance: max Λ 2 Λ 2Σ gg Λ 2 + µ 2 ( 1 Λ 2 Λ 2 ) + κλ 2 Λ 1 First order conditions: Σ gg Λ 2 = µ 2 Λ 2 The solution to this maximization problem, Λ 2 will be the eigenvector associated with the second largest eigenvalue of Σ gg

17 Loadings: First Factor The two columns of Λ Copper Heavy Trucks Iron Nonferrous Metal MV Bodies MV Parts Plywood Saw mills Coffee/Tea Animal Food Logging Loadings: Second Factor

18 Growth over previous quarter Industrial production and its factor component Year R 2 (F ) 0.9

19 Partial summary The story so far: there is a strong common component to industrial production. Is this because there are common shocks? Or because there are independent shocks transmitted via input-output relationships? Rest of the paper: Use a model with input-output linkages to back out productivity shocks for each industry-quarter. Perform factor analysis on the productivity shocks. N perfectly competitive sectors, which produce using capital, labor, and the output of other sectors. Consumers have preferences over leisure and consumption of the goods produced by the N industries. Productivity growth is distributed N (0, Σωω ); ω will admit an factor representation.

20 Model: Market Clearing Output can be used for consumption, as an intermediate input, or to increase one of the N capital stocks: N Y tj = C tj + M t,j i + N i=1 i=1 X t,j i j {1,...N}

21 Model: Preferences Consumers lifetime utilities are given by: U = [ N β t t=0 ψ: disutility from work i=1 (C ti ) 1 σ 1 1 σ ψl ti ] σ: preference elasticity of substitution, intertemporal elasticity of substitution.

22 Model: Production The production technology of each sector is given by: Y tj = A tj (K tj ) α j M tj (L tj ) 1 α j i γ ij The intermediate input bundle of sector j consists of the purchases from the other sectors: M tj = i M γ ij t,i j γ ij is the share of good i used in the production of the good-j intermediate input.

23 Model: Evolution of Capital, Productivity Industry-j-specific capital evolves according to: K t+1,j = (1 δ) K tj + Z tj where Z tj = i X θ ij t,i j θ ij is the share of good i used in the production of the good-j capital input. Productivity in each sector evolves according to a random walk: log A t+1,j = log A tj + ω t+1,j, ω t+1 0, Σ ωω

24 Model: Solution Outline Goal: Recover the vector of ω shocks given data on output growth. Solution: Write out Lagrangian of social planner. Take FOC Solve for steady state allocation Log linearization around the steady state System reduction Blanchard and Kahn Obtaining the model filter

25 Model: Lagrangian L= t=0 β t N j=1 C 1 σ tj 1 ψl tj + µ tj [K t+1,j (1 δ) K tj + Z tj ] 1 σ + P tj [A tj (K tj ) α j M tj (L tj ) 1 α j i γ ij C tj + ]} N M t,j i + X t,j i Write out the first order conditions for M t,j i, X t,j i, C tj, and L tj : [M t,j i ] : M t,j i = P ti P tj γ ji Y ti ; [C tj ] : C σ tj [L tj ] : ψ = = P tj ( 1 α j i γ ij ) i=1 [X t,j i ] : X t,j i = µ ti P tj θ ji Z ti P tj Ytj L tj [K tj ] : µ tj = βµ t+1,j (1 δ) + βp t+1,j α j Ytj K tj

26 Model: Steady state Plug in [K], [L], [M] FOC Y j = A j (K j ) α j M j (L j ) 1 α j i γ ij }{{} =1 in steady sstate ( ) (( α j βy j P αj j Y j = M j 1 α j ) P j Y j γ ij µ j (1 β (1 δ)) ψ i ( ) µj (1 β (1 δ)) αj ( ) γij ( Pi ψ P j = α j β γ ij 1 α j i γ ij Note µ j is also a function of the prices P j i ( ) θij Pi ) 1 αj i γ ij ) 1 αj i γ ij µ j = i θ ij Plug this in: we have a system of linear equations for log P. Now use market-clearing conditions to solve for quantities.

27 Model: Log linearization [M t,j i ] : M t,j i = P ti P tj γ ji Y ti ˆM t,j i = ˆP ti ˆP tj + Y ti [X t,j i ] : X t,j i = µ ti P tj θ ji Z ti ˆX t,j i = ˆµ ti ˆP tj + Ẑ ti [C tj ] : C σ tj = P tj σĉ tj = ˆP tj ψ [L tj ] : 1 α j i γ = P tjy tj ˆL tj = ˆP tj + Y tj ij L tj Y t+1,j [K tj ] : µ tj = βµ t+1,j (1 δ) + βp t+1,j α j K t+1,j ] ˆµ tj = β ˆµ t+1,j (1 δ) + (1 β (1 δ)) [ˆP t+1,j + Ŷ t+1,j + ˆK t+1,j Also market clearing, production function, and K-evolution equatons need to be log linearized. Then write these in vector form...

28 Model: System reduction Substitute out vectors to get things down to the 2N 2N system of equations: [ ] [ ] Ĉt+1 Ĉt E t = M 1 + M 3 E t  t+1 + M 4  t ˆK t+1 ˆK t Worries: More than N eigenvalues of M 1 are greater than 1 in absolute value. There is no stable solution Fewer than N eigenvalues of M1 are greater than 1 in absolute value. There are multiple stable solutions From here on, we will use our random walk assumption and write M 2 = M 3 + M 4 E t  t+1 =  t

29 Model: Blanchard and Kahn Start with Write [ ] [ ] Ĉt+1 Ĉt E t = M 1 + M 2 Â t ˆK t+1 ˆK t M 1 = VDV 1 D is diagonal N explosive (>1) eigenvalues are ordered first [ ] [ ] V 1 Ĉt+1 E t = DV ˆK 1 Ĉt + M 2 Â t t+1 ˆK t [ ] C E t+1 t = D K t+1 [ C t K t ] + V 1 M 2 Â t

30 Model: Blanchard and Kahn From last slide [ ] C E t+1 t = K t+1 ( D1 0 0 D 2 ) [ ] C t + V K 1 M 2  t t Writing each of the components: Using C t = D1 1 C t+1 + D1 1 V 1 M 2  t = D1 2 C t+2 + D1 2 V 1 M 2  t + D1 1 V 1 M 2  t = τ=1 D1 τ V 1 M 2  t = D1 1 K t+1 = D 2 K t + V 1 M 2  t ( ) I D 1 1 V 1 M 2  t [ ] [ ] Ĉt C = V t, we can solve for ˆK t+1 (and also Ĉ t ) in ˆK t K t terms of ˆK t and  t.

31 Model: Obtaining the model filter From before, these equations describe model dynamics ˆK t+1 = Π kk ˆK t + Π ak  t ; Ĉ t = Π kc ˆK t + Π ac  t Within our log linearized equations we can also write out output in terms of the state/co-state Ŷ t = Π ky ˆK t + Π ay  t + Π cy Ĉ t = Π ky ˆK t + Π ay  t + Π cy [ Π kc ˆK t + Π ac  t ] Look at an adjacent period Ŷ t+1 = [Π ky + Π cy Π kc ] ˆK t+1 + [Π ay + Π cy Π ac ]  t+1 ] = [Π ky + Π cy Π kc ] [Π kk ˆK t + Π ak  t + [Π ay + Π cy Π ac ]  t+1 = [Π ky + Π cy Π kc ] Π kk ˆK t + [Π ky + Π cy Π kc ] Π ak  t + [Π ay + Π cy Π ac ]  t+1 Now we can sub out ˆK t

32 Solution: Summary The model yields the following (log-linear-approximate) expression for the evolution of output: log Y t+1,1 log Y t+1,2 = Q... ( t,1 t,2... ) + R ω t1 ω t2... +S Q, R, and S are specified, given (β, δ, ψ, σ, α i, θ ij, γ ij ) Given these parameters, one can back out innovations to productivity (setting ω 0 = 0): ω t+1,1 ω t+1,2... =S 1 R ω t1 ω t2... log Y t+1,1 log Y t+1,2... Q log Y t1 log Y t2... ω t+1,1 ω t+1,2...

33 Solution: Summary From last slide ω t+1,1 ω t+1,2... Worries: =S 1 R ω t1 ω t2... log Y t+1,1 log Y t+1,2... Q log Y t1 log Y t2... The first few ω s will depend on our (ad-hoc) imposition that ω 0 = 0. Some of the eigenvalues of R may be greater than 1 in absolute value. Next step: Calibrate model s parameters, which will allow us to back out ω tj.

34 Calibration Parameter Value/Source β-discount factor 0.99 δ-capital depreciation rate ψ-disutility from work 1 σ-consumers elasticity, across goods 1 α i -capital share in production of i 1997 BEA I.O. Tables 1 α i j γ ij-share of 1997 BEA I.O. Tables capital/labor in production of i γ ij -share of good i in production of j s intermediate input 1997 BEA I.O. Tables θ ij -share of good i in production of j s capital input 1998 Capital Flow Tables

35 Downstream Industry Calibration of Θ Ups tream Industry

36 Downstream Industry Calibration of Γ Ups tream Industry McGrattan and Schmitz: Add 0.25 to the diagonal elements of Θ.

37 Calibration of Σ ωω Given the other parameters, we know ω tj for all time periods industries. Two calibrations: Σ ωω is diagonal, with the j, j th entry equal to the sample variance of ω tj Perform principal component analysis on the ωtj : Σ ωω = Λ S Σ SS Λ S + Σ uu, with a 2-dim. common factor, S t With Σ ωω in hand, we compute the following statistics: ρij : average correlation in the growth rates for two industries σ g : standard deviation of the growth rate of industrial production R 2 (S): fraction of the variation in industrial production growth explained by the common factors.

38 Results Period ρ ij σ g R 2 (S) Data Uncorrelated Shocks Common Factors

39 Comparison to other models Foerster et al.: g t+1 = Qg t + Sω t+1 Rω t Long and Plosser (1983): Materials arrive with a one-period lag, no capital, log preferences for consumption and leisure. g t+1 = Γ g t + ω t+1 ( Σ gg = Γ ) i Σωω Γ i=0 Carvalho (2007), Acemoglu et al (2012): No capital, log preferences for consumption, perfectly elastic labor supply g t+1 = ( I Γ ) ω t+1 Σ gg = ( I Γ ) Σ ωω (I Γ)

40 Model performance with independent productivity shocks ρ ij σ g σ g (diag) σ g (scaled) σ g,bench (scaled) Data Benchmark Long-Plosser Carvalho Benchmark, θ = I Benchmark, Σ ωω = σ 2 I Benchmark, Γ, α: average σ g (scaled) is defined as the σ g computed in an alternative calibration in which Σ ωω is chosen so that "model-implied variance of IP growth associated with the diagonal elements of Σ gg correspond to the value in the data."

41 Do the industry definitions matter? Period 2-digit 3-digit 4-digit 26 inds. 88 inds. 117 inds. Data ρ ij Independent Error ρ ij R 2 (S)

42 Conclusion Summary: Industry-specific shocks explain about 40% of the variation in industrial production Lower (20%) in the pre-great Moderation period; higher (50%) in the Great Moderation. Common shocks became less volatile during the great moderation Extensions: Apply this model to the whole economy, not just the goods-producing sectors (Ando 2014) Decompose output variation into firm-specific, industry-specific, and common shocks.