How bad are the last 11 dry years in a row?

Question

How bad are the last 11 dry years in a row?

In the following graph there are 61 years of records of the annual energy that could be transformed in electricity using water-driven power plants in Chile:

In order to explain there is a trend of dry years, I am comparing that the last 11 years are below the average of the previous 50 years with records, but to give a sense of how bad it is, I would like to take the probability of having these 11 years below the average in a row at the end just for chance (expecting it to be very low, so it tell we have indeed a trend), but I don't know how to do it, since the time variable should have a meaning: the problem is somehow like having 11 tails in a row after 50 random previous results on 61 coin tosses, different of just counting 11 tails out of 61 tosses, but since the average change with each year, it is different from the coin example (which are there plenty of examples in Google).

Here you could find the table with the data: Google Sheets, but I preferred as an answer, the explanation of how to get the mentioned probability than just the final number.

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What I have tried so far (added later)

I have test the data visually with a Gaussian distribution by taking the data Empirical Distribution and its look its fit quite good:

So since it don't look skew, I think is a good assumption consider that there is a $50/50$ chance of being above or below the average. With this, the problem "looks" similar to having coin tosses, so the total amount of alternative results are $2^{61}$, and having $11$ results below the average at the end becomes a string of $\underbrace{2\cdot 2\cdots 2\cdot 2}_{50\,\text{times}}\cdot \underbrace{1 \cdot 1\cdots 1\cdot 1}_{11\,\text{times}}=2^{50}$ alternatives fitting the required pattern, so I could estimate the probability of having the shown results just by chance as: $$ P\biggr(\text{the last 11 years below the average in a row}\biggr) = \frac{2^{50}}{2^{61}} = 0.0488 \%$$

Where I used that having exactly the average is a zero-measure point so it can be ignored.

But the issue in my line of thought, I think, is that the time variable is not considered, and the obtained probability thinking in coin tosses is valid for 11 heads (or tails) in a row independently of where the row is located, so having it exactly at the end could lead to a probability even lower.

So far, I think the result could be used as an upper bound for the real probability, but since I don't know nothing about stochastic time series, I would like to see if what I have done make any sense or not.

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Added later

After the comment by @TickaJules I realize the complications: a change in the average could drop my assumptions since I have implicitly assumed the process is stationary.

But in the same line, if I compare the same statistics for the previous 50 years, which looks quite Normal/Gaussian, with the final 11 years were both the average and the standard deviation have decreased, they don't look Gaussian at all, and even when they are too few points to say something meaningful, the first graph kind of show a decreasing tendency, which is somehow supported but their cumulative distribution skewed to lower values:

So I think that the previous calculated probability could still be a valid calculation considering it as a measuring a probability of $\mathbf{99.95\%}$ chance that something structural have changed in the last 11 years compared with the previous 50 years... or it is just too "cherry picking" as an assertion?

My arguments are:

• If I believe that nothing have change in the last 61 years, since the aggregated data looks Normal, then the procedure for taking the probability $p = 0.0488 \%$ is valid, so is far more probable that indeed something "bad" have happened - and ironically this dismissed the assumption that nothing have happened.
• And if I believe that something have change in the last 11 years, since these dry years in a row skewed the data to lower values, thinking in a moving average, the average now is lower making even harder to these years to be below it, but indeed it still happening. Since all the data is been considered in the analysis, I think is not skewed enough to say it have so much weight to lower values now, than it is more probable to have lower values than above the average, even when it is indeed what is happening so far, which somehow supports the hypothesis that something structural is different now. Following this argument, then the probability $p = 0.0488 \%$ is an upper bound for the "true probability", so considering all the years registered I have that at least with a probability of $99.95\%$ something structural have change on the final 11 years.

Does it make sense? or there are still something missing in what could be explaining these 11 dry years?

Answer 1

The last 11 years have very low ranks. If the data is such that every year is independently distributed, then there is a strong statistical significant effect that the recent 11 years are lower than the 50 years before that.

Asside from comparing the numbers above/below the mean (whose worst case give the p-value 0.000488) you could also use a rank test, which gives an even lower p-value.

    Wilcoxon rank sum test

data:  y[1:50] and y[51:61]
W = 492, p-value = 4.686e-05
alternative hypothesis: true location shift is not equal to 0

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An important question is, "does it make sense to assume that the years are independent?".

Clearly the hypothesis that you have a steady state distribution where every year is independent is wrong. However, this does not need to mean that 'something structural have changed'. It can be that you have random fluctuations over larger time scales that influence multiple years. It can be normal to have longer periods of years that are high or low.

or it is just too "cherry picking" as an assertion

This is always a risk with observational studies. Black swans happen and will be cherry picked.

More data, experiments, and theory can improve your believes.

then the probability p=0.0488% is an upper bound for the "true probability", so considering all the years registered I have that at least with a probability of 99.95% something structural have change on the final 11 years.

The p-value indicates the probability of a type-I error (the probability of falsely rejecting the null hypothesis when it is actually true). It is not the probability that a certain effect is present.

Answer 2

This will be a two-part answer, the first part a direct answer to the question, and the second part a commentary.

Part 1: A simple, and exact, way to do it is to use the Hypergeometric distribution, as follows.

I am going to translate your problem into an "urn" model. We have 61 balls, corresponding to the 61 years of observations. 32 of these balls are "above" the average, and 29 are "below" the average. If I choose 11 balls without replacement - corresponding to the last 11 observations - what is the probability that they are all "below" balls?

The probability is easily calculated using any number of stat packages, or a calculator, as approximately $0.01\%$.

Part 2: However, this isn't really telling you what you want to know, in a formal statistical sense. To see this, consider whether you even would have done this test had, say, 5 of the last 11 observations been below the average, or what test you would have done if it had been the last 9 observations below average instead of the last 11 observations. The fact that you observed what appeared to be a highly unusual result, then tested the significance of exactly that result, pretty much cancels out the value of the significance test - as it's based on "samples from a finite population that I think are highly unusual" rather than "random samples from a finite population" as the test calculation assumes.

In an informal sense, it's OK to say "I thought this was a highly unusual result, and it is!" But it shouldn't be cited as a formal statistical test result.

Edit in response to comments:

To lend support to the validity of the Hypergeometric, I've constructed a simple example in code. We have 61 observations, 29 of which are "below" and 32 "above". We randomly rearrange them a million times, count the number of times that the last 11 observations have $0, 1, 2, \dots, 11$ "below" values, and compare to what the Hypergeometric distribution tells us to expect:

obs <- c(rep("above",32), rep("below", 29))

p0_to_11 <- rep(0,12)

for (i in 1:1e6) {
  x <- sample(obs)   # randomly rearranges the elements of "obs"
  nbelow <- sum(x[51:61] == "below")
  p0_to_11[nbelow+1] <- p0_to_11[nbelow+1] + 1
}

p0_to_11 <- p0_to_11 / 1e6

plot(p0_to_11 ~ c(0:11), type="b", pch=16, lwd=2, col=2,
     ylab = "Probabilities & frequencies",
     xlab = "# of 'below' observations")
lines(dhyper(0:11,29,32,11) ~ c(0:11), type="l", lwd=2, col=1)

The red dots indicate the observed frequencies, and the black lines are the Hypergeometric probabilities. There would be red lines too, except that the black lines overlay them. This is at least supporting evidence for the statement that the Hypergeometric is indeed the distribution to use in this circumstance.

Answer 3

The easiest excel-friendly way to show the trend is to do a polynomial interpolation,

a more sophisticated version of this would be to use a digital filter.

However, apart from the approach @TickaJules suggested, I don't think there is a sensible way to define or interpret the probability of $11$ specific below-average data points.