Computing a lag-1 sample autocorrelation

Question

You are handed a short series x = [3, 1, 4, 1, 5]. Compute the lag-1 sample autocorrelation \hat{\rho}1 by hand, and say what its sign tells you about the series.

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The sample autocorrelation at lag 1 is \hat{\rho}1 = \frac{\sum{t=1}^{n-1}(xt - \bar{x})(x{t+1} - \bar{x})}{\sum{t=1}^{n}(xt - \bar{x})^2}. Mean: \bar{x} = \dfrac{3 + 1 + 4 + 1 + 5}{5} = \dfrac{14}{5} = 2.8. Centered values: 0.2,\ -1.8,\ 1.2,\ -1.8,\ 2.2. Denominator (sum of squares): 0.2^2 + 1.8^2 + 1.2^2 + 1.8^2 + 2.2^2 = 0.04 + 3.24 + 1.44 + 3.24 + 4.84 = 12.8. Numerator (consecutive products): (0.2)(-1.8) + (-1.8)(1.2) + (1.2)(-1.8) + (-1.8)(2.2) = -0.36 - 2.16 - 2.16 - 3.96 = -8.64. Therefore \hat{\rho}1 = \frac{-8.64}{12.8} = -0.675. The strong negative lag-1 autocorrelation reflects th…

Computing a lag-1 sample autocorrelation

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