Juan-Carlos Gandhi (![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png)
juan_gandhi) wrote2020-01-04 04:51 pm
juan_gandhi) wrote2020-01-04 04:51 pm
Entry tags:
не в моей компетенции, конечно
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?
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I'd like to understand this bit more. On what grounds do we use "two sigma" here? Is the slope ("the trend") meant to be normally distributed?
I mean, I can't wrap my head around double assumption: If T were a normally distributed value, then "the trend" would be some f(T) that is not necessarily normally distributed, but would be for some f, and won't be for some other f. But for "the trend" to be non-zero, T must be not a normally distributed value; why do we assume f(T) is normally distributed?.. How do we know that's the right hypothesis to test?
I mean, even the absence of normal distribution for f(T) doesn't mean there is "a trend" - T may still be a normally distributed random value, but f is such that f(T) is not normally distributed - because f(T) are not i.i.d. (not "independent").
no subject
The least-squares estimator for the trend coefficient "b" is a linear function of T(t). So, the trend estimator is then also a Gaussian distributed value with some (possibly nonzero) mean and some standard deviation. Our goal is to compute the mean and the standard deviation of the estimator for "b". This was the main goal in my Fourier-based analysis that I pointed out before.
The assumption of Gaussian or normal distributions is not at all important. We will conclude that the trend is nonzero not because the distribution is not normal or not Gaussian, but because the mean value is far enough from zero so that it can't be just a natural fluctuation. The mean value and the standard deviation are defined just as well for non-Gaussian distributions.
Even if the distribution of noise is not Gaussian, we still assume that it has zero mean (by definition, it is "noise" and has no trend). So, we can still assume that U(t) has some fixed auto-correlation function. The estimator for "b" is still a linear function of T(t), so we can compute the standard deviation of b.
The usual procedure is to require that some value is beyond 2 sigma away from zero. The null hypothesis is that there is zero trend. We can refute the null hypothesis at high confidence if we show that the mean of "b" is larger than 2 sigma.
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Of course drilling is less smooth than the atmospheric process, but it has something in common. Unpredictability of what happens next.
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"the mean value is far enough from zero so that it can't be just a natural fluctuation"
A natural fluctuation of what? Of the trend? Do we conclude "b" is nonzero? I think we will only find out whether the hypothesis that the function is linear of some b must be rejected.
no subject
If we imagine performing this observation many times (for different time epochs, or for different imaginary Earths), we will get different values of "B" - even though "b" remains the same. This is what I called the "natural variability". We can then plot the histogram of different values of "B" and see how likely it is that actually b = 0. A simple way of doing this is assuming Gaussianity and checking 2*Sigma. If the distribution is very far from Gaussian, we would have to, say, find "B" for many 10-year intervals throughout the data set and plot a histogram of its values. But I don't think it's far from Gaussian.
We can also check whether the distribution of "noise" U(t) looks like a Gaussian. Take the entire data set T(t) from 1901 to 2010, estimate a + b*t as A + B*t using least squares, and subtract from the temperature. The process T(t) - A - B*t has zero mean by construction, and is close to U(t). We can then plot the histogram of its values and see if the distribution is sufficiently close to Gaussian. I would expect that it will be.
Your calculation so far, with 10-year intervals, shows that "B" is distributed roughly within the interval [-0.2, 0.2]. If you plot a histogram of "B" for all 10-year intervals, you will probably see something like a Gaussian curve with mean close to zero. So, one will have to conclude that 10-year intervals have too much natural variability to show that the true trend "b" is nonzero; the null hypothesis cannot be refuted.
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Before estimating "b" in a + b*t, we test whether it is just a. This is not the same as estimating "b", because the result is not a choice between a non-zero "b" and a zero "b". (A "zero" b is already an estimate of b, right?) This is a test whether temperature is just a random quantity. (Perhaps, with some autocorrelation)
Then we estimate "b" to see whether a + b*t is a good description of a "trend". But it can also be seen as a test to see whether a + b*t is a good estimator of temperature. This is where I am stuck. Consequently, when we reject "b" as not a good estimate, we are not rejecting a concrete value of it, we are rejecting the hypothesis that the temperature estimator is a linear function.
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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
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
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"?
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"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!
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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.
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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
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.
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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.
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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)
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> T(t) as a function of time (t) is a sum of a zero-mean unknown "noise" U(t) and a linear trend a+b*t.
нет, если брать за основу осредненную за год температуру, то, что обычно печатается наблюдателями и моделями, там нужно учесть многолетние известные изменения.
Речь, прежде всего, идет об осцилляциях (мальчик-девочка). Грубо говоря, годовое осреднение убирает сезонные колебания, но оставляет многолетние регулярные эффекты.
Т.е. функция дб a + O(t) + GW(t) + noise
Реально в первом приближении, емнип, достаточно учитывать осцилляции и 11-тий солнечный цикл.
Нельзя известные нам явления запихивать в шум
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Then once we extract that, how is the confidence interval estimated? Surely this should work when we don't have any model?
(Of course, knowing some cycles and oscillations is going to help, but shouldn't be a prerequisite, right? Otherwise we can't tell the growth rate until we have the exact model!)
Then the growth rates that are alarming, did they publish confidence intervals? I can't find that easily.
no subject
Солнечный циклъ понятно, а что конкретно вы здѣсь назвали словомъ "осцилляцiи"? Каковъ перiодъ этой функцiи и каковъ физическiй механизмъ, порождающiй эти колебанiя температуры?
Сезонныя колебанiя температуры уже были убраны передъ моими вычисленiями, т.к. я смотрѣлъ на графикъ среднегодовой температуры. На этомъ графикѣ видны большiя и нерегулярныя осцилляцiи.
Если была бы хорошо предсказываемая, регулярная осцилляцiя перiода 11 лѣтъ, то я полностью согласенъ, ее надо было бы вычесть, а не учитывать какъ шумъ, и я согласенъ, что sigma получилась бы завышена, если этого не сдѣлать. Однако, я не встрѣчалъ упоминанiя солнечнаго цикла или какихъ-либо другихъ извѣстныхъ циклическихъ несезонныхъ осцилляцiй въ статьяхъ объ оцѣнкѣ sigma, прочитанныхъ мной. Какiя статьи вы читали? Дайте ссылки.
Главный вкладъ въ мою оцѣнку sigma, насколько я помню, вносило большое и нерегулярное колебанiе температуры на масштабѣ около 30 лѣтъ. Глобальная температура падала въ началѣ 20 вѣка, потомъ выросла въ 1930-1940 (тогда уже предсказывали полное отсутствiе арктическаго льда къ 1950-му году), потомъ опять падала до 1970-хъ (тогда климатологи писали о томъ, что насъ всѣхъ убъетъ глобальное похолоданiе), потомъ опять росла до 2000-го (климатологи, соотвѣтственно, предсказывали исчезновенiе арктическаго льда къ 2020-му). Ничего изъ предсказаннаго климатологами въ 20-мъ вѣкѣ не случилось, т.е. они сильно занижали sigma въ своихъ расчетахъ.
no subject
ENSO, El Nino - La Nina, мальчик-девочка
https://www.weather.gov/mhx/ensowhat
no subject
On periods ranging from about three to seven years, the surface waters across a large swath of the tropical Pacific Ocean warm or cool by anywhere from 1°C to 3°C, compared to normal.
Это какъ разъ и есть одна изъ компонентъ хаотическаго шума, т.е. нѣчто такое, которое мы не въ состоянiи предсказать, но при этомъ оно "can have a strong influence on weather across the United States and other parts of the world". Поэтому оно какъ разъ даетъ какой-то вкладъ въ "sigma".
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<удивившись>
не так чтобы совсем и не можем
есть люди, занимающиеся осцилляциями, есть модели, есть реконструкции прошлых осцилляций
> Это какъ разъ и есть одна изъ компонентъ хаотическаго шума, т.е. нѣчто такое, которое мы не въ состоянiи предсказать
<удивившись еще больше>
почему явление, которое мы, скажем так, не очень хорошо понимаем, сразу становится хаосом?
вполне регулярное явление, реконструируется в прошле века, на графиках температуры зачастую помечают "год мальчика", "год девочки"
> Поэтому оно какъ разъ даетъ какой-то вкладъ въ "sigma".
нет, конечно
осцилляции - это то, что мы знаем, регулярное явление, а не иррегулярный шум. Потому при определении форсинга мы должны осцилляции учитывать отдельным фактором, отдельным членом в уравнении, а не записывать их скопом в шум
касательно солнечного цикла, вот одна из основополагающих, кяп, статей
https://pubs.giss.nasa.gov/docs/2008/2008_Rind_ri07700f.pdf
no subject
It doesn't. But when you don't have a function for it, you need to test whether there is anything apart from noise. That's the purpose of testing whether we should reject null hypothesis (that all there is, is just white noise).
You start with:
1. ok, this is normal distribution, no function, nothing, just random temperature. Oops, this is so for p-value of 0.5.
2. ok, here is a function that removes solar cycles, now the rest is white noise. Oops, the remainder is white noise for p-value of 0.2
3. ok, here is a function that removes (some other cycle), now the rest is white noise. Oops, the remainder is white noise for p-value of 0.15.
4. ...
5. ok, here is a linear trend of 0.8 C per century, and 0.2 C in the last decade, which removes ever speeding up global warming, now the rest is white noise. Oops, ...what p-value do we get?
(ermmm.... well, I don't do this stuff for living, so I am allowed to make imprecise statements like above. In reality you'd need to turn them "inside out" - because p-value means "if we reject the hypothesis that so and so is white noise, what's the probability of making a mistake", so as we get more and more accurate theories, p-value should be increasing to indicate that what we get is closer and closer to white noise)