# Identification restrictions in Structural VECMs

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)?

• I added the (obvious) vecm tag as there was one vacant slot. – Richard Hardy Jun 2 '17 at 13:41

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).