Before us are the observations X1 to X156. Recall we are assuming that each of these X has been measured without any error. Given that we observe X1 = 0.43, the probability X1 = 0,43 is 1 or 100%. Now let’s have some fun and ask some more complicated questions of our data. We can ask anything we like.
How about this? What is
(9) Pr(X156 > X1| Observations)?
Well, all we have to do is look. It appears from the plot, and a glance at the data confirms, that this probability is 0, or 0%. Why? Well, because X156 was not larger than X1. How could that cause any controversy?
But notice that (9) is an entirely different question than this one:
(10) Pr(X156 > X1| M1)
(11) Pr(X156 > X1| M2),
It could very well be that (10) is greater than (say) 90% but (11) is less than 10%. It could even be, depending how rigid the models are, that (10) equals 1 and (11) equals 0; that is, the models could say the exact opposite of one another. And this is not a rare situation: indeed, we saw it last week when M1 was a climate model and M2 was a probability model. Don’t forget that we already agreed with eq. (3): different models produce different probabilities for Xi in general.
How about this question?
(12) What is the probability that temperatures (as measured by X) increased?
I don’t know and neither do you. Why? Because this question is ambiguous. Just what exactly does “increased” mean? Is it asking if X156 > X1? That’s a kind of increase. If so, then we can compute an answer (given our observations, it is 0, or 0%). Does “increased” mean that X increased at least once? If so, then we can compute an answer, which in this case is 1, or 100%, the opposite of the first definition. If you say that “increased” means “increased generally” then you have to supply the definition of “generally.”
Suppose “increased generally” means that X increased more often than it decreased or stayed the same. We can easily compute this, given our observations. And this is a different probability than if “increased generally” means that X increased or stayed the same more often than it decreased. If you do not see this, stop here and ponder until you do.
We could go on and on. “Increased” could be taken to me increased by a certain amount, or never decreased more than another fixed number. There are more possibilities, too, but we’ll skip them. All we need remember is that the probability of these questions can all be different. There is just no evading these kinds of definitions (even if you want to). So before you find your blood boiling and you hear yourself shouting “Denier!” make sure you understand what question you and your enemy are answering.
Let’s pick one of these definitions so that we can move on: let’s say that “increased generally” means “X increased or stayed the same more often than it decreased.” Now, what is (12) with this definition in mind? Well, all we have to do is glance at our observations: the probability, given these observations, is 0, or 0%. Let’s write this out in notation, to make it crystal clear:
(13) Pr( X increased | Observations) = 0.
Of course, (13) assumes our definition of “increased”, which to be precise we should write inside the parentheses; however, we’ll let it slide for ease of reading.
We seem to be done. We wanted to know if X “increased”, we agreed on a definition of what “increased” meant, we looked at the observations, and we knew the answer. We even agreed that each X is measured without error. Except for disagreements about the word “increased”, there would seem to be no room whatsoever for controversy.
There is, of course, plenty of space, but only because people confuse just what probability is being computed. Because some people, when hearing “What is the probability X increased”, instead of computing (13), compute this
(14) Pr(X increased | M1)
(15) Pr(X increased | M2),
etc. There is no contradiction for using the past tense here, for we can just assume, as people do, that “increased” means “would have increased”. Now it is perfectly possible, as before, that (14) could be greater than, say, 90%, and that (15) could be less than, say, 10%—even though, in fact, X did not increase (by our definition). And we don’t have to settle for just these two models: we have an infinite supply of models, so that we can compute
(16) Pr(X increased | Mj)
for j = 1, 2, … Each (or most) of these probabilities will be different. Which is the correct one?
Well, that is a separate question. For real temperatures, I have no idea which model is best. Many people, however, claim they know exactly which model is the best and truest and most pure. Well, maybe they are right. They say that they have a model which takes into account, in just the proper way, forced versus unforced feedback, the effects of variable sunlight, the thermodynamics of this and that gas, and on and on. Suppose the folks making this claim are right and that their model is the “best”, where we can leave the notion “best” float for now. Further suppose that
(17) Pr(X increased | Mbest) > 90%;
that is, given the best model, the chance that X would have increased over this series is 90%. Given Mbest, this probability is true. But, alas, so is (13). It is still the case, no matter what (17) tells us, that (13) holds, that temperatures did not, in actual fact, increase. This is just too bad for Mbest. Since we are supposing that Mbest is best, all we have learned is that our model has rather severe limits and that X is not predictable to the extent we would like it to be. Tough luck for the model! Sometimes the universe is predictable and sometimes she isn’t.
But then, it is unlikely that Mbest really is best (just as it is even more likely that the model you, the reader, have in mind is best). I said I don’t know what model really is best, but I do claim to know, and I will shortly prove, that the way to think about temperatures is flawed, because we don’t consider that X is measured with error. We’ll tackle that next time. We’ll also talk about “significant” and “linear” increases, etc.