Solution. With \(G = 3\), each equation needs at least \(G - 1 = 2\) exclusions. Equation 1 excludes \(y_3, x_2, x_3\): three exclusions, so it is overidentified by the order condition (it has one endogenous regressor and two excluded exogenous variables). Equation 2 excludes \(y_1\) and \(x_3\): two, exactly identified (two endogenous variables included, one on the right, and one excluded exogenous variable). Equation 3 excludes \(y_2, x_1, x_2\): three, overidentified. The rank condition must still be checked. For equation 2 the excluded variables are \(y_1\) and \(x_3\), and their coefficients in equations 1 and 3 form the matrix \(\begin {bmatrix} b_{11} & 0\\ b_{31} & \gamma _{33}\end {bmatrix}\), whose determinant \(b_{11}\gamma _{33}\) is non-zero because both variables are included where marked. The rank is \(2 = G - 1\), so equation 2 is identified.

Solution. The reduced form is \(Y_t = \alpha /(1 - \beta ) + I_t/(1 - \beta ) + v_t\) and \(C_t = \alpha /(1 - \beta ) + \beta I_t/(1 - \beta ) + v_t\). The ratio of the \(I_t\) coefficients is \(\beta = 1.5/2.5 = 0.6\), and the multiplier \(1/(1 - \beta ) = 2.5\) is consistent with it. Then \(\alpha /(1 - \beta ) = 120\) gives \(\alpha = 48\). The equation is exactly identified, with one excluded exogenous variable and one endogenous regressor, and in that case ILS and IV are the same estimator: \(\hat \beta _{\text {IV}} = \sum i_tc_t/\sum i_ty_t\) is exactly the ratio of the reduced-form slopes \(\sum i_tc_t/\sum i_t^2\) and \(\sum i_ty_t/\sum i_t^2\).

Solution. Take any \(\lambda \neq 0\) and form a weighted average of the two equations, each normalised on \(q\): the combination \((1 - \lambda )(\text {demand}) + \lambda (\text {supply})\) is \begin {align*} q_t = {}&[(1 - \lambda )\alpha _0 + \lambda \beta _0] + [(1 - \lambda )\alpha _1 + \lambda \beta _1]p_t\\ &+ (1 - \lambda )\alpha _2\text {inc}_t + \text {disturbance}. \end {align*}

It has exactly the form of the demand equation (it contains \(p\) and income), and together with the supply equation it implies the same reduced form, so the data cannot distinguish it from the true demand curve. The supply equation excludes income, which appears in demand, so it is exactly identified: income shifts demand along it.

Solution. (a) The characteristic equation is \(\lambda ^2 - 0.9\lambda + (0.2 - 0.02) = \lambda ^2 - 0.9\lambda + 0.18 = 0\), with roots \(0.6\) and \(0.3\), both inside the unit circle: stable.

(b) \(\Psi _1 = A_1\), and \(\Psi _2 = A_1^2 = \begin {bmatrix} 0.27 & 0.18\\ 0.09 & 0.18\end {bmatrix}\). A unit shock to \(y_2\) raises \(y_1\) by \(0.2\) after one period and \(0.18\) after two.

(c) Yes: the coefficient of \(y_{2,t-1}\) in the equation for \(y_1\) is \(0.2 \neq 0\).

Solution. (a) The determinant is \(0.16 - 0.16 = 0\) and \(\Pi \neq 0\), so the rank is \(1\). The second column is \(-2\) times the first, so \(\Pi = \alpha \beta '\) with \(\beta = (1, -2)'\) and \(\alpha = (-0.4, 0.2)'\).

(b) The long-run relation is \(y_1 = 2y_2\). When \(y_1\) is above \(2y_2\), the disequilibrium \(y_{1,t-1} - 2y_{2,t-1}\) is positive, and \(y_1\) falls (\(\alpha _1 = -0.4\)) while \(y_2\) rises (\(\alpha _2 = 0.2\)): both variables move to close the gap, \(y_1\) faster.

Solution. \(A \otimes B\) is the block matrix with \((i, j)\) block \(a_{ij}B\). The \((i, k)\) block of the product is \(\sum _ja_{ij}B\,c_{jk}D = \bigl (\sum _ja_{ij}c_{jk}\bigr )BD = (AC)_{ik}BD\), which is the \((i, k)\) block of \(AC \otimes BD\). Then \((A \otimes B)(A^{-1} \otimes B^{-1}) = AA^{-1} \otimes BB^{-1} = I \otimes I = I\), which proves the inverse formula, and with \(A = \Sigma \), \(B = I_n\) it gives \((\Sigma \otimes I_n)^{-1} = \Sigma ^{-1} \otimes I_n\).