Impulse response functions help examine the effects of a unit innovation shock to future values of the response of a time series model, without accounting for the effects of exogenous predictors. For example, if an innovation shock to an aggregate output series, e.g., GDP, is persistent, then GDP is sensitive to such shocks. The examples below show how to plot impulse response functions for regression models with various ARIMA error model structures using `impulse`

.

This example shows how to plot the impulse response function for a regression model with AR errors.

Specify the regression model with AR(4) errors:

$$\begin{array}{l}{y}_{t}=2+{X}_{t}\left[\begin{array}{l}5\\ -1\end{array}\right]+{u}_{t}\\ {u}_{t}=0.9{u}_{t-1}-0.8{u}_{t-2}+0.75{u}_{t-3}-0.6{u}_{t-4}+{\epsilon}_{t}.\end{array}$$

Mdl = regARIMA('Intercept',2,'Beta',[5; -1],'AR',... {0.9, -0.8, 0.75, -0.6})

Mdl = regARIMA with properties: Description: "Regression with ARMA(4,0) Error Model (Gaussian Distribution)" Distribution: Name = "Gaussian" Intercept: 2 Beta: [5 -1] P: 4 Q: 0 AR: {0.9 -0.8 0.75 -0.6} at lags [1 2 3 4] SAR: {} MA: {} SMA: {} Variance: NaN

The dynamic multipliers are absolutely summable because the autoregressive component is stable. Therefore, `Mdl`

is stationary.

You do not need to specify the innovation variance.

Plot the impulse response function.

impulse(Mdl)

The impulse response decays to 0 since `Mdl`

defines a stationary error process. The regression component does not impact the impulse responses.

This example shows how to plot a regression model with MA errors.

Specify the regression model with MA(10) errors:

$$\begin{array}{l}{y}_{t}=2+{X}_{t}\left[\begin{array}{l}5\\ -1\end{array}\right]+{u}_{t}\\ {u}_{t}={\epsilon}_{t}+0.5{\epsilon}_{t-2}-0.4{\epsilon}_{t-4}-0.3{\epsilon}_{t-6}+0.2{\epsilon}_{t-8}-0.1{\epsilon}_{t-10}.\end{array}$$

Mdl = regARIMA('Intercept',2,'Beta',[5; -1],... 'MA',{0.5,-0.4,-0.3,0.2,-0.1},'MALags',[2 4 6 8 10])

Mdl = regARIMA with properties: Description: "Regression with ARMA(0,10) Error Model (Gaussian Distribution)" Distribution: Name = "Gaussian" Intercept: 2 Beta: [5 -1] P: 0 Q: 10 AR: {} SAR: {} MA: {0.5 -0.4 -0.3 0.2 -0.1} at lags [2 4 6 8 10] SMA: {} Variance: NaN

The dynamic multipliers are absolutely summable because the moving average component is invertible. Therefore, `Mdl`

is stationary.

You do not need to specify the innovation variance.

Plot the impulse response function for 10 responses.

impulse(Mdl,10)

The impulse response of an MA error model is simply the MA coefficients at their corresponding lags.

This example shows how to plot the impulse response function of a regression model with ARMA errors.

Specify the regression model with ARMA(4,10) errors:

$$\begin{array}{c}{y}_{t}=2+{X}_{t}\left[\begin{array}{l}5\\ -1\end{array}\right]+{u}_{t}\\ \left(1-0.9L+0.8{L}^{2}-0.75{L}^{3}+0.6{L}^{4}\right){u}_{t}=\left(1+0.5{L}^{2}-0.4{L}^{4}-0.3{L}^{6}+0.2{L}^{8}-0.1{L}^{10}\right)\end{array}$$

Mdl = regARIMA('Intercept',2,'Beta',[5; -1],... 'AR',{0.9, -0.8, 0.75, -0.6},... 'MA',{0.5, -0.4, -0.3, 0.2, -0.1},'MALags',[2 4 6 8 10])

Mdl = regARIMA with properties: Description: "Regression with ARMA(4,10) Error Model (Gaussian Distribution)" Distribution: Name = "Gaussian" Intercept: 2 Beta: [5 -1] P: 4 Q: 10 AR: {0.9 -0.8 0.75 -0.6} at lags [1 2 3 4] SAR: {} MA: {0.5 -0.4 -0.3 0.2 -0.1} at lags [2 4 6 8 10] SMA: {} Variance: NaN

The dynamic multipliers are absolutely summable because the autoregressive component is stable, and the moving average component is invertible. Therefore, `Mdl`

defines a stationary error process.

You do not need to specify the innovation variance.

Plot the first 30 impulse responses.

impulse(Mdl,30)

The impulse response decays to 0 since `Mdl`

defines a stationary error process.

This example shows how to plot the impulse response function of a regression model with ARIMA errors.

Specify the regression model with ARIMA(4,1,10) errors:

$$\begin{array}{c}{y}_{t}=2+{X}_{t}\left[\begin{array}{l}5\\ -1\end{array}\right]+{u}_{t}\\ \left(1-0.9L+0.8{L}^{2}-0.75{L}^{3}+0.6{L}^{4}\right)(1-L){u}_{t}=\left(1+0.5{L}^{2}-0.4{L}^{4}-0.3{L}^{6}+0.2{L}^{8}-0.1{L}^{10}\right){\epsilon}_{t}.\end{array}$$

Mdl = regARIMA('Intercept',2,'Beta',[5; -1],... 'AR',{0.9, -0.8, 0.75, -0.6},... 'MA',{0.5, -0.4, -0.3, 0.2, -0.1},... 'MALags',[2 4 6 8 10],'D',1)

Mdl = regARIMA with properties: Description: "Regression with ARIMA(4,1,10) Error Model (Gaussian Distribution)" Distribution: Name = "Gaussian" Intercept: 2 Beta: [5 -1] P: 5 D: 1 Q: 10 AR: {0.9 -0.8 0.75 -0.6} at lags [1 2 3 4] SAR: {} MA: {0.5 -0.4 -0.3 0.2 -0.1} at lags [2 4 6 8 10] SMA: {} Variance: NaN

One of the roots of the compound autoregressive polynomial is 1, therefore `Mdl`

defines a nonstationary error process.

You do not need to specify the innovation variance.

Plot the first impulse responses.

quot = sum([1,cell2mat(Mdl.MA)])/sum([1,-cell2mat(Mdl.AR)])

quot = 1.2000

impulse(Mdl,50) hold on plot([1 50],[quot quot],'r--','Linewidth',2.5) hold off

The impulse responses do not decay to 0. They settle at the quotient of the sums of the moving average and autoregressive polynomial coefficients (`quot`

).

$$\text{quot}=\frac{1+0.5-0.4-0.3+0.2-0.1}{1-0.9+0.8-0.75+0.6}=1.2.$$