6  Trend analysis

Climate change is manifested in the modification of the statistical variables used to characterize the climate, the best known being the mean (for example, the mean temperature or mean total precipitation). We say that a series is non-stationary if these variables change over time (see Fact Sheet 3). Climate change most often manifests itself as a gradual increase in the mean, but other variations are possible, such as an increase in variability.

Series of climate variables can be considered in the same way as random series (see Fact Sheet 5). An infinite number of series are possible, all with the same statistical characteristics (such as the same mean). That means it’s realistic to randomly generate a series with a trend. In order to assess whether a trend in a series is simply the result of chance or, on the contrary, results from changes in the statistical characteristics, a trend test is applied. Several trend tests have been developed in the past, with the best known being the Mann-Kendall test.

The Mann-Kendall test is used to determine whether the detected trend is statistically significant. For a series of n values \(\left\{ x_{1},x_{2},\ldots,x_{n} \right\}\), the Mann-Kendall test allows us to verify the null hypothesis H0 that no monotonic trend is present in the series; the alternative hypothesis H1 is that a monotonic trend is present. The S statistic of the test is calculated as follows: \[S = \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \text{sgn} \left( x_{j} - x_{i} \right),\] where \[\text{sgn}\left( \phi \right) = \left\{ \begin{matrix} + 1 \text{ if } \phi > 0 \\ 0 \text{ if }\phi = 0 \\ - 1 \text{ if } \phi < 0 \\ \end{matrix} \right.\ \]

The S statistic is a count of the number of times pairs of values in the series are increasing, equal, or decreasing. So the more increasing pairs there are in the series, the more positive and larger S will be, while the more decreasing pairs the series has, where \(x_{i} > x_{j}\), the more negative S will be. S = 0 corresponds to the case where there is no net increase or decrease. Assuming that the values of the series are independent and identically distributed, that S has a normal distribution, that the mean of this distribution is zero and that there is no equality (\(x_{i} \neq x_{j}\forall i,j\)), we can show that the variance V is given by1: \[V(S)= \frac{n (n-1)(2 n+5)}{18}\]

The \(Z_{\text{MK}}\) statistic of the test is then calculated:

\[Z_{\text{MK}} = \left\{ \begin{matrix} \frac{S-1}{\sqrt{V(S)}} & \text{if } S > 0 \\ 0 & \text{if } S = 0 \\ \frac{S+1}{\sqrt{V(S)}} & \text{if } S < 0 \end{matrix} \right.\ \]

A positive \(Z_{\text{MK}}\) value suggests an upward trend, while a negative value of \(Z_{\text{MK}}\) is associated with a downward trend.

If \(Z_{\text{MK}} > 0\), the null hypothesis H0 of no monotonic trend is compared to the alternative hypothesis H1 of a monotonic increasing trend. Thus, for a threshold α, we reject the H0 hypothesis in favour of the hypothesis of an upward trend if \(Z_{\text{MK}} > Z_{\left( 1 - \alpha \right)}\), where \(Z_{\left( 1 - \alpha \right)}\) is the (1 - α)th percentile of the standard normal distribution2.

If \(Z_{\text{MK}} < 0\), the null hypothesis H0 of no monotonic trend is compared to the alternative hypothesis H1 of a monotonic decreasing trend. Thus, for a threshold α, we reject the hypothesis H0 in favour of the hypothesis of a downward trend if \(Z_{\text{MK}} < - Z_{\left( 1 - \alpha \right)}\). For \(\alpha = 0.05\), where \(Z_{\left( 1 - \alpha \right)} = 1.645\), we reject H0, knowing that there is a 5% probability of this series being obtained if H0 is true.3


  1. The expression for V(S) is different when there is equality between values of the series (Hamed 2008).↩︎

  2. But note that we can never accept H0 because we can never reject H1.↩︎

  3. There has been an ongoing debate for several years about the threshold α to be considered to reject the null hypothesis and about the interpretation of the significance threshold. Without going into details, if you are interested in these questions, consult Colquhoun (2014).↩︎