Solution. (i) Not true. Adding \(X_4\) cannot reduce \(R^2\) (Theorem 6.3.2), so \(R^2_{1\cdot 234} \geq R^2_{1\cdot 23}\).

(ii) Not true. By the Cauchy–Schwarz inequality the expression equals \(\sum x^2\sum y^2(1 - r^2) \geq 0\) (Corollary 1.3.3).

(iii) True; it is Theorem 6.5.2. Adding \(X_3\) explains the fraction \(r_{13\cdot 2}^2\) of what \(X_2\) left unexplained.

(iv) It can be true. By (iii) with the roles of \(X_2\) and \(X_3\) exchanged, \(R^2_{1\cdot 23} = r_{13}^2 + r_{12\cdot 3}^2(1 - r_{13}^2) \geq r_{13}^2 = 0.626\), and also \(\geq 0.227\); \(0.71\) satisfies both bounds. It requires \(r_{12\cdot 3}^2 = (0.71 - 0.626)/0.374 = 0.225\), which is possible. For example, if \(r_{12}\) and \(r_{13}\) are both positive, the identity \(R^2_{1\cdot 23} = (r_{12}^2 + r_{13}^2 - 2r_{12}r_{13}r_{23})/(1 - r_{23}^2)\) gives \(R^2 = 0.71\) at \(r_{23} = 0.247\). What would be impossible is \(R^2_{1\cdot 23}\) below \(0.626\).

Solution. \(\bar R^2 = 1 - \frac {\text {RSS}/(n - k)}{\text {TSS}/(n - 1)}\) rises exactly when \(\text {RSS}/(\text {residual degrees of freedom})\) falls, that is when \(\text {RSS}_2/(n - k - 1) < \text {RSS}_1/(n - k)\). Write \(d = n - k - 1\). The condition is \((d + 1)\text {RSS}_2 < d\,\text {RSS}_1\), or \(\text {RSS}_2 < d(\text {RSS}_1 - \text {RSS}_2)\), or \((\text {RSS}_1 - \text {RSS}_2)/(\text {RSS}_2/d) > 1\), which is \(t^2 > 1\).

Solution. By Theorem 6.4.1, \(b_j\) and the residuals come from the simple regression of \(\tilde y = M_1Y\) on \(\tilde x = M_1X_j\), where \(M_1\) removes the other regressors. In that regression \(\text {RSS} = \sum \tilde y^2(1 - r^2)\) with \(r\) the partial correlation, and \(b_j^2\sum \tilde x^2 = r^2\sum \tilde y^2\). The variance of \(b_j\) is estimated by \(s^2/\sum \tilde x^2\) with \(s^2 = \text {RSS}/(n - k)\); this is the same as in the full regression, which has the same residuals and the same degrees of freedom. Hence \[ t^2 = \frac {b_j^2\sum \tilde x^2}{s^2} = \frac {r^2\sum \tilde y^2(n - k)}{\sum \tilde y^2(1 - r^2)} = \frac {(n - k)r^2}{1 - r^2}, \] and solving for \(r^2\) gives \(r^2 = t^2/(t^2 + n - k)\).

Solution. The means are \(\bar X_2 = 4\), \(\bar X_3 = 2\) and \(\bar Y = 9.3333\). In deviations, \begin {align*} x_2 &= (-2, 0, -1, 1, 2, 0), \qquad x_3 = (-1, 0, 0, 1, 1, -1),\\ y &= (-4.333, -0.333, -1.333, 2.667, 4.667, -1.333) . \end {align*}

The sums are \[ \sum x_2^2 = 10, \quad \sum x_3^2 = 4, \quad \sum x_2x_3 = 5, \] \[ \sum x_2y = 22, \quad \sum x_3y = 13, \quad \sum y^2 = 51.333 . \] The denominator is \(10(4) - 25 = 15\), so \[ b_2 = \frac {22(4) - 13(5)}{15} = \frac {23}{15} = 1.5333, \qquad b_3 = \frac {13(10) - 22(5)}{15} = \frac {20}{15} = 1.3333, \] and \(b_1 = 9.3333 - 1.5333(4) - 1.3333(2) = 0.5333\). \(\text {ESS} = 1.5333(22) + 1.3333(13) = 51.067\), so \(R^2 = 51.067/51.333 = 0.9948\).

Solution. The first part is the last step of the proof of Theorem 6.4.1. If \(X_1'X_2 = 0\) then \(M_1X_2 = X_2 - X_1(X_1'X_1)^{-1}X_1'X_2 = X_2\), so \(b_2 = (X_2'X_2)^{-1}X_2'M_1Y = (X_2'X_2)^{-1}(X_2'Y - X_2'X_1(X_1'X_1)^{-1}X_1'Y) = (X_2'X_2)^{-1}X_2'Y\), the short regression coefficient. Leaving \(X_1\) out of the regression does not change the estimate of \(\beta _2\) when \(X_1\) is uncorrelated with \(X_2\) in the sample. Omitting a variable biases the coefficients of the included ones only through its correlation with them. That is why randomised experiments, in which the treatment is uncorrelated with everything else by design, can estimate a causal effect without controlling for other variables.

Solution. In deviations, the regression of \(y\) on \(x_2\) has slope \(25/10 = 2.5\), with residuals \(\tilde y = y - 2.5x_2 = (0, 0.5, -1, 0.5, 0)\). The regression of \(x_3\) on \(x_2\) has slope \(8/10 = 0.8\), with residuals \(\tilde x = x_3 - 0.8x_2 = (-0.4, 0.8, -1, 1.2, -0.6)\). Then \(\sum \tilde x\tilde y = 0 + 0.4 + 1 + 0.6 + 0 = 2\) and \(\sum \tilde x^2 = 0.16 + 0.64 + 1 + 1.44 + 0.36 = 3.6\), so the slope is \(2/3.6 = 0.5556 = b_3\). The residuals \(\tilde y - 0.5556\tilde x\) are \((0.222, 0.056, -0.444, -0.167, 0.333)\), with sum of squares \(0.3889\), the residual sum of squares of the full regression, \(64 - 63.611\).