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I am studying Structural VECM models using Lutkepohl's book. VECM process has the following Beveridge-Nelson MA representation:

$$y_t = \Xi\sum_{i=1}^{t}u_i + \sum_{j=0}^{\infty}\Xi_j^*u_{t-j} + y_0^*,$$

where the $\Xi_j^*$ are absolutely summable and $y_0^*$ are some initial values. Hence, long-run effects of shocks ($u_t$) are captured by the term $\Xi\sum_{i=1}^{t}u_i $ . Importantly, $\Xi$ is $K\times K$ matrix with $rank(\Xi) = K-r$. We are interested in finding such $K\times K$ matrix $B$ that $$u_t = B\varepsilon_t,$$ $$\Sigma_u = BB'$$ and now (plugging for $u_t$ into the model) long term effects are represented by matrix $\Xi B$ which also has rank $K-r$.

Long story short, one has to impose $K(K-1)/2$ restrictions on $B$ in order to be able to estimate it. Suppose we know $r$. Then, at most $r$ columns in $\Xi B$ can be zero ($r$ transitory shocks) which we consider as additional source for restrictions. But the author says that since $\Xi B$ has rank $K-r$, then each column of zeros stands for $K-r$ independent restrictions only. Thus, $r$ transitory shock represent $r(K-r)$ restrictions.

I do not understand his point. Why zero column gives only $K-r$ independent restrictions and not $K$ (since we impose $K$ zeros)?

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  • $\begingroup$ I added the (obvious) vecm tag as there was one vacant slot. $\endgroup$ – Richard Hardy Jun 2 '17 at 13:41
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I think I found an answer, so I post it here, may be it could help others. So, we have that $rank(\Xi B) = K-r$. For simplicity I will stick to the case with $K=3$ and $r=1$. Hence, $rank(\Xi B) = 1$. We need $K(K-1)/2 = 3$ independent restrictions in order to be able to identify matrix $B$. We know that at most one column of $\Xi B$ is a zero column, which corresponds to the transitory shock that has no long-run effect on the variables of the system. Hence, we could restrict one column if $\Xi B$ to zeros. The issue is that one should be cautions when counting how many restrictions it gives (not $3$). Suppose we've put two zeros in the last column. We have $$ \Xi B = \begin{bmatrix} * &* &0 \\ * &* &0 \\ * &* &\cdot \end{bmatrix} $$ where $*$ denotes unrestricted element and $\cdot$ denotes element of our interest. Since we know that the rank of the above matrix is $2$, then one row can always be represented as a linear combination of other rows. Thus, row 3 is a linear combination of row 1 and row 2. Any linear combination will result in the $(3,3)$ element ot be zero. Hence two zeros imply the third zero and we have $2$ (not $3$) independent restrictions.

Another way to think of it is by using the definition of cointegration. Suppose we are still in the case where $r=1$ which means that there is one independent cointegrating relationship. Without loss of generality assume that $y_{1t}$ and $y_{3t} $ are cointegrated. That is $$y_{1t} - \gamma y_{3t} \sim I(0),$$ where $\gamma$ is some constant and we are again in the case of having $$ \Xi B = \begin{bmatrix} * &* &0 \\ * &* &0 \\ * &* &\cdot \end{bmatrix} $$

Zero in position $(1,3)$ implies that $y_{1t}$ is not affected in the long-run by shock in $\varepsilon_{3t}$. But since $y_{1t}$ and $y_{3t}$ are cointegrated, $y_{3t}$ also can not be affected by shock in $\varepsilon_{3t}$. Otherwise cointegrating relationship would break down (i.e. $y_{3t}$ would be driven by a stochastic trend in a long run while $y_{1t}$ not).

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