Solution. (i) Observations on many units at one point or period of time; for example, the incomes and food expenditures of 5 000 households in a national survey of a given year. (ii) Observations on the same units at several points in time; for example, the annual output, employment and capital stock of the same 200 manufacturing firms for ten years. (iii) Independent cross sections from different periods combined into one data set; for example, two household surveys five years apart, each with a fresh random sample.

The difference is whether a unit is followed. In a panel, firm \(i\) in year 1 and firm \(i\) in year 2 are the same firm, so the data can compare a unit with itself and remove its permanent unobserved characteristics. In pooled cross sections the households of the two rounds are different households. The data can compare the two populations, not the units. A pooled cross section is also independent across periods by construction, while a panel’s repeated observations on one unit are typically correlated.

Solution. No. By the Cauchy–Schwarz inequality (Theorem 1.3.2), \(\bigl (\sum x_i y_i\bigr )^2 \leq \sum x_i^2 \sum y_i^2\) for any numbers, so the expression is always \(\geq 0\). Equivalently, it equals \(\sum x_i^2 \sum y_i^2 (1 - r^2)\), which is non-negative because \(r^2 \leq 1\). A negative value means an arithmetic error. It is zero only if the points lie exactly on a line.

Solution. Each change of units is an increasing linear transformation. The dollar price per bag is \((50P - 15)/e\) for an exchange rate of \(e\) kwacha per dollar, of the form \(a + bP\) with \(b = 50/e > 0\), and kilograms are \(1000\) times tonnes. By Theorem 1.3.4 the correlation is multiplied by \(\operatorname {sgn}(bd) = +1\). The sign changes only if exactly one variable is transformed with a negative slope, for example if quantity were replaced by unsold stock, \(S - Q\) for a fixed supply \(S\).

Solution. (a) The needed moments are \(E(X) = 1\), \(E(X^2) = \frac 53\) and \(E(X^3) = \frac {0 + 1 + 8}{3} = 3\). By iterated expectations \(E(Y) = E(X^2) = \frac 53\) and \(\Cov (X, Y) = \Cov \bigl (X, E(Y \mid X)\bigr ) = \Cov (X, X^2) = 3 - 1\cdot \frac 53 = \frac 43\). Also \(\Var (X) = \frac 53 - 1 = \frac 23\). Hence \(b^* = \frac {4/3}{2/3} = 2\) and \(a^* = \frac 53 - 2 = -\frac 13\).

(The step \(\Cov (X, Y) = \Cov \bigl (X, E(Y \mid X)\bigr )\) holds because \(E(XY) = E\bigl [X\,E(Y \mid X)\bigr ]\) by iterated expectations. The best linear predictor depends on \(Y\) only through its CEF.)

(b) At \(X = 0, 1, 2\) the CEF is \(0, 1, 4\) and the predictor is \(-\frac 13, \frac 53, \frac {11}{3}\). The differences are \(\frac 13\), \(-\frac 23\) and \(\frac 13\), so the mean squared difference is \(\frac 13\bigl (\frac 19 + \frac 49 + \frac 19\bigr ) = \frac 29\). By the proof of Theorem 1.4.3, this is exactly how much larger the mean squared prediction error of the best linear predictor is than that of the CEF.

Solution. (a) \(E(\varepsilon ) = E(X^2) - E(X^2) = 0\). By symmetry \(E(X) = 0\) and \(E(X^3) = 0\), so \(\Cov (X, \varepsilon ) = E(X\varepsilon ) = E(X^3) - E(X)E(X^2) = 0\).

(b) \(E(\varepsilon \mid X) = X^2 - E(X^2)\), which is a non-constant function of \(X\) because \(X\) is not degenerate. It is not identically zero.

(c) Zero conditional mean implies zero correlation with every function of \(X\) (Theorem 1.4.2), and in particular with \(X\) itself. The converse fails: \(\varepsilon \) here is uncorrelated with \(X\) but perfectly predictable from \(X^2\). The error of a best linear predictor is always uncorrelated with \(X\), by Theorem 1.4.4, but it has conditional mean zero only when the CEF is linear. The distinction matters later: unbiasedness of least squares needs the stronger condition, while consistency needs only the weaker one.

Solution. (a) Holding capital fixed, a 1% increase in labour is associated with a \(0.7\%\) increase in output. Holding labour fixed, a 1% increase in capital is associated with a \(0.2\%\) increase in output.

(b) Scaling both inputs by 1% raises output by about \(\hat \beta _L + \hat \beta _K = 0.9\%\), which points to decreasing returns to scale.

(c) \(0.9\) is an estimate, so the question is whether it differs from \(1\) by more than sampling error. That requires the standard error of \(\hat \beta _L + \hat \beta _K\), which is \(\sqrt {\Var (\hat \beta _L) + \Var (\hat \beta _K) + 2\Cov (\hat \beta _L, \hat \beta _K)}\). The covariance of the two estimates is needed as well as their standard errors. It is usually negative, because labour and capital tend to move together, and it can change the answer. The test of \(\beta _L + \beta _K = 1\) is developed with inference in the \(k\)-variable model.

Solution. (a) Both series grow steadily over the sixteen years. Any two trending series are highly correlated with each other, and each is highly correlated with time, whether or not they are related. A large part of \(0.984\) reflects the common upward trend rather than any connection between employment and output.

(b) Differencing removes a linear trend: if \(Y_t = \alpha + \delta t + e_t\) then \(\Delta Y_t = \delta + \Delta e_t\), and the trend leaves only a constant. The correlation of the changes asks whether employment rises more than usual in years when output rises more than usual. Here it is still \(0.892\), so the co-movement is real and not only a shared trend. That conclusion was not available from the levels. The section on unit roots makes this precise.

(c) Stata: tsset Year, then correlate D.Employed D.GNP. EViews: group g d(employed) d(gnp), then g.cor.