Practical DSGE modelling

Practical DSGE modelling

Simulation techniques Martin Ellison University of Warwick and CEPR Bank of England, December 2005 Baseline DSGE model yt P22* 1 P21* wt wt 1 ( P11* P12* P22* 1 P21* ) 1 1 ( P11* P12* P22* 1 P21* ) wt ( P11* P12* P22* 1 P21* ) 1 R1vt 1 Recursive structure makes model easy to simulate Numerical simulations Stylised facts Impulse response functions Forecast error variance decomposition Stylised facts Variances Covariances/correlations Autocovariances/autocorrelations Cross-correlations at leads and lags Recursive simulation 1. Start from steady-state value w0 = 0 2. Draw shocks {vt} from normal distribution 3. Simulate {wt} from {vt} recursively using

wt 1 ( P11* P12* P22* 1 P21* ) 1 1 ( P11* P12* P22* 1 P21* ) wt ( P11* P12* P22* 1 P21* ) 1 R1vt 1 Recursive simulation * 1 * 22 21 4. Calculate {yt} from {wt} using yt P P wt 5. Calculate desired stylised facts, ignoring first few observations Variances Standard deviation Interest rate 0.46 Output gap 1.39 Inflation 0.46

Correlations Interest rate Output Inflation gap Interest rate 1 -1 -1 Output gap -1 1 1 Inflation -1

1 1 Autocorrelations t,t-1 t,t-2 t,t-3 t,t-4 Interest rate 0.50 0.25 0.12 0.06 Output gap 0.50

0.25 0.12 0.06 Inflation 0.50 0.25 0.12 0.06 Cross-correlations Correlation with output gap at time t t-2 t-1 t t+1 t+2

Output gap 0.25 0.50 1 0.50 0.25 Inflation 0.25 0.50 1 0.50 0.25 Interest rate

-0.25 -0.50 -1 -0.50 -0.25 Impulse response functions What is effect of 1 standard deviation shock in any element of vt on variables wt and yt? 1. Start from steady-state value w0 = 0 2. Define shock of interest 0 {vt } 1 0 0 0 0

0 0 0 0 0 0 Impulse response functions 3. Simulate {wt} from {vt} recursively using wt 1 ( P11* P12* P22* 1 P21* ) 1 1 ( P11* P12* P22* 1 P21* ) wt ( P11* P12* P22* 1 P21* ) 1 R1vt 1 4. Calculate impulse response {yt} from {wt} yt P22* 1 P21* wt using Response to vt shock 0.6 0.4 0.2 0 -0.2 0 1 2 3 4 5 6 7 8 9 1011 12 -0.4 t

-0.6 -0.8 -1 -1.2 -1.4 interest rate output gap inflation Forecast error variance decomposition (FEVD) Imagine you make a forecast for the output gap for next h periods Because of shocks, you will make forecast errors What proportion of errors are due to each shock at different horizons? FEVD is a simple transform of impulse response functions FEVD calculation Define impulse response function of output gap to each shocks v1 and v2 1.5 1 0.5

0 -0.5 0 t 1 2 3 4 5 6 7 8 11 12 21 22

31 32 41 42 51 52 61 62 71 72 81 82 -1 -1.5 response to v1 response to v2 response at horizons 1 to 8 FEVD at horizon h = 1

At horizon h = 1, two sources of forecast errors Shock Impulse response at horizon 1 1 t v 1 1 Contribution to 1 2 2 variance at 1 1 horizon 1 2 t v

2 1 2 2 1 2 v2 FEVD at horizon h = 1 Contribution of v1 1 2 1 ( ) 1 2 1 2 v1 2 v1

2 2 1 ( ) ( ) 2 v2 FEVD at horizon h = 2 At horizon h = 2, four sources of forecast errors Shock Impulse response at horizon 2 1 t v 1 2 2 t

v 2 2 Contribution to 1 2 2 2 2 2 variance at 2 1 2 v2 horizon 2 1 t 1 2 t 1 v v

1 1 2 1 1 2 1 2 2 1 2 1 2 v2 FEVD at horizon h = 2 Contribution of v1 1 2 2 1

v1 1 2 2 2 v1 1 2 2 2 2 1 ( ) ( ) 1 2 1 2 v1 2 v1 2 v2 2 2 2 ( ) ( ) ( ) ( )

2 v2 FEVD at horizon h At horizon h, 2h sources of forecast errors Contribution of v1 h 1 2 i ( ) i 1 h h i 1 i 1 2 v1 1 2 2 2 2 2

( ) ( i v1 i ) v2 FEVD for output gap 1 0.9 0.8 0.7 0.6 interest rate shock 0.5 co st- push shock 0.4 0.3 0.2

0.1 0 h 0 1 2 3 4 5 6 7 8 9 10 FEVD for inflation 1 0.9

0.8 0.7 0.6 interest rate shock 0.5 co st- push shock 0.4 0.3 0.2 0.1 0 h 0 1 2 3

4 5 6 7 8 9 10 FEVD for interest rates 1 0.9 0.8 0.7 0.6 interest rate shock 0.5 co st- push shock

0.4 0.3 0.2 0.1 0 h 0 1 2 3 4 5 6 7 8 9 10

Next steps Models with multiple shocks Taylor rules Optimal Taylor rules

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