не в моей компетенции, конечно
Jan. 4th, 2020 04:51 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png)
Но я наслаждаюсь аргументами противников теории ГП. Красиво гонят, такое ощущение, что это какая-то умственная паника.
- на Марсе тоже ледники тают;
- всех интересует только температура на поверхности, а что творится на высоте 10км, никого не интересует;
- при динозаврах вообще стояла жарища;
- инерция поведения океана: вода, что сейчас выходит к поверхности на Бермудах (и дальше идет в качестве Гольфстрима), шла от Антарктиды, вдоль Южной Америки, примерно тысячу лет;
- так нам в Калифорнии чего конкретно ожидать-то, засухи или наводнений? А то каждый год новости;
- кто-нибудь вообще изучал вопрос изменения поведения Солнца за последние 50-100 лет?
Типа почему никакого ГП нету вообще:
- да последние пять лет самые холодные за наблюдаемую историю вообще;- у нас на Магадане морозы надоели уже, пусть уже потеплее будет;
- в 1500-м 97% ученых считали, что Земля неподвижна, а кто был не согласен, того на костре сжигали;- на Марсе тоже ледники тают;
- всех интересует только температура на поверхности, а что творится на высоте 10км, никого не интересует;
- при динозаврах вообще стояла жарища;
- инерция поведения океана: вода, что сейчас выходит к поверхности на Бермудах (и дальше идет в качестве Гольфстрима), шла от Антарктиды, вдоль Южной Америки, примерно тысячу лет;
- Грета Тунберг в школу давно не ходила;
- так нам в Калифорнии чего конкретно ожидать-то, засухи или наводнений? А то каждый год новости;
- кто-нибудь вообще изучал вопрос изменения поведения Солнца за последние 50-100 лет?
no subject
Date: 2020-01-08 04:57 pm (UTC)The only way to deal with the question of trend is, in my view, is to assume that T(t) = a + b* t + U(t) and to estimate "b". We can also assume other models, e.g. T(t) = a + b*t + c*t*t + U(t), etc.
no subject
Date: 2020-01-08 05:05 pm (UTC)I probably haven't formulated what the sticking point is for me with estimating "b".
We start with stating that our model here is a good estimate of T(t), which we show is non-linear, and want to compute "a trend". Then we build another function that is linear, "a trend", and essentially we want to prove that that is also a good estimate of T(t)! But ok, maybe the difference is that T(t) is for the whole century, and "a trend" is for a smaller period of time.
no subject
Date: 2020-01-08 06:01 pm (UTC)That's my question, again: what exactly do we compute in order to reject this as a null hypothesis? I don't understand how we could do that. We can certainly ask whether the mean of T(t) is zero or not, but that would be the same as asking whether a = 0 or not. It's not asking whether T(t) is "just a" or not "just a". Can you describe what calculation needs to be performed with the T(t) data in your excel table in order to decide this first null-hypothesis?
We start with stating that our model here is a good estimate of T(t), which we show is non-linear, and want to compute "a trend".
I don't understand what you are saying. What exactly does it mean that T(t) is "non-linear"?
no subject
Date: 2020-01-08 08:17 pm (UTC)"What exactly does it mean that T(t) is "non-linear"?"
Maybe I am confused. Let me try and untangle what I was thinking.
So, we have a weather model, M(t). This model is a function that is not linear - that is, M(x+y)=M(x)+M(y) ∧ M(k*x)=k*M(x) does not hold. In other words, there is no f(x)=a+b*x such that M(x)=f(x)+random noise. Put yet another way, we've spent so much effort modelling weather systems because they are far more complex than just a straight line.
Now we want to show that temperature has "a trend". The proposal is to estimate a and b such that the trend T(t)=a+b*t. Then we have some test procedure (which I understand).
My question is, aren't the two propositions at odds with each other? If we prove M(x) is the best fit, why do we bother with "a trend", T(t)? If we prove T(t) is the best fit, why do we bother with the model, M(x)?
Now: of course we can find a and b using least mean squares; this will estimate the slope that describes the temperature growth with the least error. But isn't this supposed to not be a good fit, ever, so it is meant to have a non-random noise added to it ("the least error" is not guaranteed to be distributed normally with mean 0)? We've got to reject "a trend" every time we accept the model to be non-linear!
no subject
Date: 2020-01-08 08:23 pm (UTC)Which two distributions are you going to compare? I only see one distribution, namely the observed T(t).
Now, as for non-linear models, of course M(t) should be a weather model that is based on some equations of atmospheric physics. But we are not considering the question of how to compute M(t). We are simply asking the question: does the observed temperature T(t) exhibit a growth trend or not, i.e. can we say that, on average, the temperature grows by X degrees per century. This implies a simple linear model, T(t) = a + b*t + U(t) where U(t) is some zero-based random noise.
no subject
Date: 2020-01-08 08:32 pm (UTC)The essence of the question about linear trend remains the same. If we manage to accept T(t)=a + b*t + random noise, we should be using T(t) as the weather predictor, not a more complex function. So I am a bit at a loss why there is some hope of having a suitable "trend" defined after some 100 years, or ever.
A completely different way of looking at it is: the question of temperature having a trend is like a question of knowing the slope of a derivative. It has no predictive power. (Like, "what's the trend of a sine at 2019?")
no subject
Date: 2020-01-08 09:14 pm (UTC)Now, of course, sin(t) is a highly repetitive function, so if M(t) = sin(t) is the correct model of temperature, we would have a lot of predictive power. But this is not the case with temperature: we can't even predict ordinary weather for more than a week in advance, let alone for 100 years. We don't have M(t).
I am asking a very limited question: given the observed T(t), can we see a growth trend, and can we say with certainty that the temperature is growing, and that it is growing faster after 1950 than it was growing before 1950?
If somebody says that today we have "global warming" and that it has accelerated in recent years, can we verify this with observational data? What exactly do these words mean, "the temperature is growing" and "it is growing faster than before", in terms of observational data? That's all I'm asking. And we can answer that question simply by doing statistical analysis on the observed data for T(t), with no need for complicated physics.
Now, of course, that analysis will tell us nothing about predicting T(t) for the next 1000 years. It will have no predictive power for the next 1000 years. But predicting for the next 1000 years is a very hard question - and not the question I'm asking.
no subject
Date: 2020-01-09 08:58 am (UTC)no subject
Date: 2020-01-09 01:40 pm (UTC)You performed a calculation where you estimated "b" linearly from different time intervals. Let us first assume that the true value "b" is the same for all time from 1900 to 2020. Then you can perform linear fit for "b" with different time intervals. For example, take the 20-year intervals 1900-1920, 1901-1921, 1902-1922 and so on until 2000-2020. The result will be 100 different estimates of "b". They are not independent, of course, but highly correlated. Nevertheless, you can look at the resulting distribution of estimates and see if there is evidence that the mean of "b" is not zero.
You can compute the mean and the standard deviation of the set of 100 estimates of "b". Roughly, if the mean is > 2 stdev then the mean is nonzero with high confidence. You can also use other statistical tests for nonzero mean, of course.
no subject
Date: 2020-01-09 03:37 pm (UTC)I think the criticism remains in force. If the "b"s are not iid, then "the mean is > 2 stdev" may not apply. The problem is not only the correlation between "b"s (one flavour of "dependent"), but also how they are going to be distributed (another flavour of "dependent"). Put differently, if you were to draw samples of 20 normally distributed values, and computed "b"s, would such "b"s be distributed normally? If not, then why would the two-sigma rule be meaningful?
no subject
Date: 2020-01-09 03:43 pm (UTC)