Simulate responses using filter and simulate. Then compare the simulated responses.

Both filter and simulate filter a series of errors to produce output responses y, innovations e, and unconditional disturbances u. The difference is that simulate generates errors from Mdl.Distribution, whereas filter accepts a random array of errors that you generate from any distribution.

Specify the following regression model with ARMA(2,1) errors:

where $${\epsilon }_t$$ is Gaussian with variance 0.1.

Mdl = regARIMA('Intercept',0,'AR',{0.5 -0.8}, ...
    'MA',-0.5,'Beta',[0.1 -0.2],'Variance',0.1);

Simulate data for the predictors and from Mdl using Monte Carlo simulation.

rng(1);           % For reproducibility
X = randn(100,2); % Simulate predictor data
[ySim,eSim,uSim] = simulate(Mdl,100,'X',X);

Standardize the simulated innovations and filter them.

z1 = eSim./sqrt(Mdl.Variance);
[yFlt1,eFlt1,uFlt1] = filter(Mdl,z1,'X',X);

Confirm that the simulated responses from simulate and filter are identical using a plot.

figure
h1 = plot(ySim);
hold on
h2 = plot(yFlt1,'.');
title '{\bf Filtered and Simulated Responses}';
legend([h1, h2],'Simulate','Filter','Location','Best')
hold off

Alternatively, simulate responses by randomly generating your own errors and passing them into filter.

rng(1);
X = randn(100,2);
z2 = randn(100,1);
yFlt2 = filter(Mdl,z2,'X',X);
figure
h1 = plot(ySim);
hold on
h2 = plot(yFlt2,'.');
title '{\bf Filtered and Simulated Responses}';
legend([h1, h2],'Simulate','Filter','Location','Best')
hold off

This plot is the same as the previous plot, confirming that both simulation methods are equivalent.

filter multiplies the error, Z, by sqrt(Mdl.Variance) before filtering Z through the model. Therefore, if you want to specify your own distribution, set Mdl.Variance to 1, and then generate your own errors using, for example, random('unif',a,b) for the Uniform(a, b) distribution.

Simulate the impulse response of an innovation shock to the regression model with ARMA(2,1) errors.

The impulse response assesses the dynamic behavior of a system to a one-time shock. Typically, the magnitude of the shock is 1. Alternatively, it might be more meaningful to examine an impulse response of an innovation shock with a magnitude of one standard deviation.

In regression models with ARIMA errors,

Specify the following regression model with ARMA(2,1) errors:

where $${\epsilon }_t$$ is Gaussian with variance 0.1.

Mdl = regARIMA('Intercept', 0, 'AR', {0.5 -0.8}, ...
    'MA', -0.5,'Variance',0.1);

When you construct an impulse response function for a regression model with ARIMA errors, you must set Intercept to 0.

Simulate the first 30 responses of the impulse response function by generating a error series with a one-time impulse with magnitude equal to one standard deviation, and then filter it. Also, use impulse to compute the impulse response function.

z = [sqrt(Mdl.Variance);zeros(29,1)]; % Shock of 1 std
yFltr  = filter(Mdl,z);
yImpls = impulse(Mdl,30);

When you construct an impulse response function for a regression model with ARIMA errors containing a regression component, do not specify the predictor matrix, X, in filter.

Plot the impulse response functions.

figure
subplot(2,1,1)
stem((0:numel(yFltr)-1)',yFltr,'filled')
title...
    ('Impulse Response to Shock of One Standard Deviation');
subplot(2,1,2)
stem((0:numel(yImpls)-1)',yImpls,'filled')
title 'Impulse Response to Unit Shock';

The impulse response function given a shock of one standard deviation is a scaled version of the impulse response returned by impulse.

Simulate the step response function of a regression model with ARMA(2,1) errors.

The step response assesses the dynamic behavior of a system to a persistent shock. Typically, the magnitude of the shock is 1. Alternatively, it might be more meaningful to examine a step response of a persistent innovation shock with a magnitude of one standard deviation. This example plots the step response of a persistent innovations shock in a model without an intercept and predictor matrix for regression. However, note that filter is flexible in that it accepts a persistent innovations or predictor shock that you construct using any magnitude, then filters it through the model.

Specify the following regression model with ARMA(2,1) errors:

where $${\epsilon }_t$$ is Gaussian with variance 0.1.

Mdl = regARIMA('Intercept', 0, 'AR', {0.5 -0.8}, ...
    'MA', -0.5,'Variance',0.1);

Simulate the first 30 responses to a sequence of unit errors by generating an error series of one standard deviation, and then filtering it.

z = sqrt(Mdl.Variance)*ones(30,1);...
    % Persistent shock of one std
y = filter(Mdl,z);            
y = y/y(1);  % Normalize relative to y(1)

Plot the step response function.

figure
stem((0:numel(y)-1)',y,'filled')
title('Step Response for Persistent Shock of One STD')

The step response settles around 0.4.