Solved – Mann-Kendall’s trend test result

hypothesis testingkendall-taup-valuertest-for-trend

I'm running the Mann-Kendall trend test on my data (35 different vectors, 5 elements each). What I'm doing for each one is:

install.packages("Kendall")
library("Kendall")
x=c(170.192, 179.397, 171.199, 108.399, 144.964)
MannKendall(x)

where x is a sample vector. I'm getting a result:

tau = -0.4, 2-sided pvalue =0.46243

Now, my question is – is it believable if I obtain exactly the same result for other vectors? For example:

y=c(2.63075, 2.01657, 1.39161, 2.51637, 1.79445)
MannKendall(y)
tau = -0.4, 2-sided pvalue =0.46243

z=c(7.12617, 6.01319, 4.64843, 6.42136, 5.36868)
MannKendall(z)
tau = -0.4, 2-sided pvalue =0.46243

Is it about the data type or I'm doing/interpreting something wrong? Or maybe it's ok and I shouldn't worry?
Thanks.

Best Answer

The test statistic will be obtained by calculating the Kendall correlation of the time series with the sequence $1,2,...,n$. Getting identical Kendall correlations with very small samples is not a surprise.

When there's no ties, this correlation corresponds to counting the number of increases (times $y_j>y_i$ when $j>i$) minus number of decreases (times $y_j<y_i$ when $j>i$) divided by number of such pairs.

Your 3 sets of ranks are

rank(x);rank(y);rank(z)
[1] 3 5 4 1 2
[1] 5 3 1 4 2
[1] 5 3 1 4 2

the last two sets are identical, so the Kendall correlation for z must be the same as for y. Let's just look at the first two, then. "3 5 4 1 2" has 3 increases (3 vs 5, 3 vs 4 and 1 vs 2) and the rest are decreases. "5 3 1 4 2" has 3 increases (3 vs 4, 1 vs 4 and 1 vs 2) with the rest decreases. This means the Kendall correlation with time in each case is $(3 - 7)/10 = -0.4$

... which is what your test says they are; the p-values are then the same because the sample sizes are the same.

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Feature Importance – Trend Analysis of Feature Importance Over Time in R

So here's my approach so far... which could definitely be improved upon.

Here is some fake data that represents something like one of the time series signals in my data

 data = jitter(c(rep(0, 100), c(1:100, rep(100, 100))), factor=3)
 plot(data, type="l")

enter image description here

The correct answer to this simulated dataset should be that the signal changed at point 101 and then settled back out again at 200. What we are going to do is fit two lines to the data. One line is a linear line that starts at 101 and goes to x. The next line is a flat line (y = ?) that goes from x to 300. Here is the function we will use to represent a line

fn.line = function(x1,y1,x2,y2,x) {
   # uses point-slope formula for a line
   y1 + (y2-y1)*(x-x1)/(x2-x1)
}

What we want to do is pick x (our response time to a steady state) by minimizing the sum of squared errors

sse = function(actual, fitted) {
    sum((actual - fitted)^2)
}

We are going to use some optimization magic in R that tries to minimize fitting those two lines

data = data[101:300]
obj.fn = function(response.time) {
    line.selection = 1:response.time
    # fit a linear line
    line.fnc = function(x) {
         fn.line(1,data[1],response.time,data[response.time], x)
    }
    linear.actual = data[line.selection]
    linear.fitted = line.fnc(line.selection)
    # Assume a flat line
    flat.actual = data[(response.time + 1):length(data)]
    flat.fitted = line.fnc(line.selection[length(line.selection)] + 1)
    mean(c(sse(linear.actual, linear.fitted), sse(flat.actual, flat.fitted)))
}

which almost gives me the right answer:

round(optimize(obj.fn, c(0, 300-101))$minimum)

Solved – Interpretation of Mann-Kendall statistics

P value < 0.05 tells that there is (monotonic)trend and if $\tau$ is +ve, increasing trend, and if $\tau$ is -ve, decreasing trend.

P value >0.05, tells no monotonic trend, away from monotonic trend.

Best Answer

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Feature Importance – Trend Analysis of Feature Importance Over Time in R

Solved – Interpretation of Mann-Kendall statistics

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