Solution. The averages at \(t = 2, \dots , 6\) are \((5 + 8 + 6)/3 = 6.33\), \((8 + 6 + 9)/3 = 7.67\), \((6 + 9 + 7)/3 = 7.33\), \((9 + 7 + 10)/3 = 8.67\) and \((7 + 10 + 8)/3 = 8.33\). The series zig-zags, and the moving average smooths it into a gently rising trend. At \(t = 1\) there is no \(x_0\), and at \(t = 7\) no \(x_8\), so the symmetric window does not fit.

Solution. The raw indices add to \(4.04\), an average of \(1.01\), so divide each by \(1.01\): \(0.891\), \(1.040\), \(1.188\), \(0.881\). The adjusted value of the third-quarter observation is \(48/1.188 = 40.4\): third quarters run about \(19\%\) above trend.

Solution. The two twelve-term averages straddling \(t\) cover \(x_{t-6}, \dots , x_{t+5}\) and \(x_{t-5}, \dots , x_{t+6}\). Averaging them, \(x_{t-5}, \dots , x_{t+5}\) appear in both, with weight \(\frac 12(\frac 1{12} + \frac 1{12}) = \frac 1{12}\), and \(x_{t-6}\), \(x_{t+6}\) in one only, with weight \(\frac 1{24}\). For the seasonal part, \(s_{t+6} = s_{t-6}\), so the weighted sum is \(\frac 1{12}(s_{t-6} + s_{t-5} + \dots + s_{t+5})\), a twelfth of the sum over a whole year, which is \(0\).

Solution. \(\bar x = 5.5\), and \(u_t = t - 5.5\) runs from \(-4.5\) to \(4.5\) in steps of \(1\), with \(\sum u_t^2 = 82.5\). For lag \(1\), \(\sum _{t=1}^9u_tu_{t+1} = 57.75\), so \(r_1 = 0.70\). For lag \(5\), \(\sum _{t=1}^5u_tu_{t+5} = (-4.5)(0.5) + (-3.5)(1.5) + (-2.5)(2.5) + (-1.5)(3.5) + (-0.5)(4.5) = -21.25\), so \(r_5 = -0.258\). The coefficients of a short trending series fall steadily (here they turn negative because with \(n = 10\) the two ends of the series are paired); for a long trending series they stay positive and large for many lags. Either way the correlogram mainly reflects the trend, which should be removed before the correlogram is interpreted.