8.1 The Mann–Kendall Test

Let \(x_1, x_2, \ldots , x_n\) be observations in time order. Compare every pair and count how often the later value exceeds the earlier one.

\[S = \sum ^{n-1}_{i=1}\sum ^{n}_{j=i+1}\operatorname {sgn}\left (x_j - x_i\right ), \hspace {0.8cm} \operatorname {sgn}(u) = \begin {cases} +1 & u > 0,\\ \ 0 & u = 0,\\ -1 & u < 0. \end {cases}\]

A large positive \(S\) says later observations tend to exceed earlier ones: an upward trend. A large negative \(S\) says the reverse. Under \[H_0: \text {the observations are independent and identically distributed}\] every ordering of the data is equally likely, so \(S\) is symmetric about zero and \(E(S) = 0\).

Notice what \(S\) does not use: the sizes of the differences, only their signs. This is exactly why an outlier cannot drag the result — one absurd value contributes at most \(n-1\) to a sum that ranges over \(\binom {n}{2}\) pairs, whereas in least squares it would move the fitted slope by an unbounded amount.

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