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## Fit VAR Model to Simulated Data

This example simulates data from an arbitrary 3-D VAR(2) model, and fits a VAR(2) model to the simulated data.

Create the following VAR model.

`${y}_{t}=\left[\begin{array}{c}1\\ 0.5\\ -0.5\end{array}\right]+\left[\begin{array}{ccc}0.3& -0.1& 0.05\\ 0.1& 0.2& 0.1\\ -0.1& 0.2& 0.4\end{array}\right]{y}_{t-1}+\left[\begin{array}{ccc}0.1& 0.05& 0.001\\ 0.001& 0.1& 0.01\\ -0.01& -0.01& 0.2\end{array}\right]{y}_{t-2}+\left[\begin{array}{c}0.5\\ -0.2\\ 0\end{array}\right]t+\text{L}{\epsilon }_{t}.$`

${\epsilon }_{t}$ is a 3-D standard Gaussian random variable and $L$ is the 3-by-3 identity matrix.

```Constant = [1; 0.5; -0.5]; AR1 = [0.3 -0.1 0.05; 0.1 0.2 0.1; -0.1 0.2 0.4]; AR2 = [0.1 0.05 0.001; 0.001 0.1 0.01; -0.01 -0.01 0.2]; Trend = [0.5; -0.2; 0]; L = eye(3); TrueMdl = varm('Constant',Constant,'AR',{AR1 AR2},'Trend',Trend,... 'Covariance',L);```

`TrueMdl` is a fully specified `varm` model object, which means all parameters of its corresponding VAR(2) model are known.

Generate a time series path of length 100 using `simulate`.

```rng(1); % For reproducibility numobs = 100; Y = simulate(TrueMdl,numobs);```

Create a 3-D VAR(2) model template for estimation. Specify the presence of a linear trend component.

```Mdl = varm(3,2); Mdl.Trend = nan(Mdl.NumSeries,1);```

`Mdl` is a `varm` model object serving as a template for estimation.

Fit the VAR model to the simulated data by calling `estimate`.

`EstMdl = estimate(Mdl,Y);`

`EstMdl` is a fully specified, estimated `varm` model object. The estimation sample size is 98 because `estimate` requires 2 presample observations for initialization.

Compare the estimated model with the true model.

`results = summarize(EstMdl)`
```results = struct with fields: Description: "AR-Stationary 3-Dimensional VAR(2) Model with Linear Time Trend" SampleSize: 98 NumEstimatedParameters: 24 LogLikelihood: -396.0032 AIC: 840.0063 BIC: 902.0456 Table: [24x4 table] Covariance: [3x3 double] Correlation: [3x3 double] ```
`Lhat = chol(EstMdl.Covariance,'lower')`
```Lhat = 3×3 0.9632 0 0 -0.0926 0.9555 0 0.0337 -0.0190 0.8755 ```
```trueValues = [Constant; AR1(:); AR2(:); Trend]; compCoeff = abs(results.Table.Value - trueValues); twoSE = compCoeff > 2*results.Table.StandardError```
```twoSE = 24x1 logical array 1 0 0 0 0 0 0 0 0 0 ⋮ ```

The estimate of `L`, the innovations scale matrix, is fairly close in magnitude to the identity matrix. Only the estimated constant in the first response series is two standard errors further from its true value.