VECM

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Derivation of VECM MODEL

\[\begin{align*} Y_{1,t} = \beta_{1,1} Y_{1,t-1} + \beta_{1,2} Y_{2,t-1} + \epsilon_{1,t}\\\\ Y_{2,t} = \beta_{2,1} Y_{1,t-1} + \beta_{2,2} Y_{2,t-1} + \epsilon_{2,t} \end{align*}\]

So \[\begin{align} Y_{1,t} - Y_{1,t-1} = (\beta_{1,1}-1) Y_{1,t-1} + \beta_{1,2} Y_{2,t-1} + \epsilon_{1,t}\\\\ \Delta Y_{1,t} = (\beta_{1,1}-1) Y_{1,t-1} + \beta_{1,2} Y_{2,t-1} + \epsilon_{1,t} \end{align}\]

Let \[(\beta_{1,1}-1) =\phi_1\] and \[\beta_{1,2}=-\phi_1 \lambda\] we have

\[\begin{align} \Delta Y_{1,t} &= (\beta_{1,1}-1) Y_{1,t-1} + \beta_{1,2} Y_{2,t-1} + \epsilon_{1,t}\\\\ &= \phi_1 Y_{1,t-1} -\phi_1 \lambda Y_{2,t-1} + \epsilon_{1,t}\\\\ &= \phi_1 (Y_{1,t-1} - \lambda Y_{2,t-1}) + \epsilon_{1,t} \end{align}\]

With the same technique, we can get \[\begin{align} \Delta Y_{2,t} &= \phi_2 (Y_{1,t-1} - \lambda Y_{2,t-1}) + \epsilon_{2,t} \end{align}\]

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Johansen Test Derivation - Unit Root Test

Characteristic Equation - AR polynomial equal to 0: \[F(x) = 0\]

• Univariate AR(1)

\[ Y_{t} = \beta_{1} Y_{t-1} + \epsilon_{t}\\\\ F(x) = 1-\beta_1 x =0 \]

\[ \Delta Y_{t} = Y_{t} - Y_{t-1} = (\beta_{1}-1) Y_{1,t-1} + \epsilon_{t}\\\\ \] Let \[ \gamma = \beta_1 - 1 \rightarrow \gamma=-F(1) \] So \[ Unit\ root \rightarrow F(1) =0 \rightarrow \gamma=0 \]

• Univariate AR(2) \[ Y_{t} = \beta_{1} Y_{t-1} + \beta_{2} Y_{t-2} + \epsilon_{t}\\\\ F(x) = 1-\beta_1 x - \beta_2 x^2=0 \]

\[\begin{align} \Delta Y_{t} &= Y_{t} - Y_{t-1} \\\\ &= (\beta_{1}-1) Y_{t-1} +\beta_{2} Y_{t-2} + \epsilon_{t}\\\\ &= (\beta_{1}-1) Y_{t-1} -\beta_{2} Y_{t-1} +\beta_{2} Y_{t-1} + \beta_{2} Y_{t-2} + \epsilon_{t}\\\\ &= (\beta_{1} +\beta_2 -1) Y_{t-1} -\beta_{2} (Y_{t-1} - Y_{t-2}) + \epsilon_{t}\\\\ &= (\beta_{1} +\beta_2 -1) Y_{t-1} -\beta_{2} \Delta Y_{t-1} + \epsilon_{t} \end{align}\]

Let \[ \gamma = \beta_1 + \beta_2 - 1 \rightarrow \gamma=-(1-\beta_1 - \beta_2)=-F(1) \] So \[ Unit\ root \rightarrow F(1) =0 \rightarrow \gamma=0 \]

This means to check unit root, we just need to check if \[ \gamma = 0 \]

• VARMA(p,q)

For matrix form, we use I to replace 1. \[ Y_t = \sum_{j=1}^{p}a_jY_{t-j} + \epsilon_t +\sum_{j=1}^{q}b_jY_{t-j} \\\\ F(x)=I-\sum_{j=1}^{p}a_j x^j \] Using the similar technique like AR(2) model, we can rewrite the model as

\[ \Delta Y_t = \gamma \centerdot Y_{t-1}+ \sum_{j=1}^{p-1}\phi_j \Delta Y_{t-j} + \epsilon_t +\sum_{j=1}^{q}b_jY_{t-j} \] Using the same result in univariate scenario: \[ Unit\ root \rightarrow |F(I) |=0 \rightarrow |\gamma|=0 \]

For matrix form, we use \[rank(\gamma) = 0\] to replace \[\gamma = 0\]

If x=I, we also have \[ |F(I)|=|I - \sum_{j=1}^{p}a_j |= |\gamma| \]

• rank=0 Y is I(1).
• rank=d Y is I(0).
• 0< rank < d Y is cointegrated!