What makes an unsteady graph unsteady or why oil is oil? - page 32
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Мощьное заявление, и главное что все подсознательно хотят чтоб оно было правдой.
I don't know how you look at it - but it is quite obvious that the processes are stationary, and the RMS is the same to the sixth digit. In general, it is a stationary process, and statistical methods confirm this with very good accuracy (and it works on smaller scales). Another thing is that this in itself does not make the process predictable.
Thanks, colleague. I will try to repeat your calculations. But I would like to change the course of these discussions and move from theory to practice. I believe that profitable forex trading is possible. Lilliput with his ruler has proved it. The question everyone has is the same: how do we find and use the latent regularity of the market. There are three basic methods of constructing trading systems
The big problem in points 2 and 3 is the selection of input data in such a way as to describe the market condition uniquely and succinctly. This is where dimensionality reduction methods are needed.
Has anyone wondered how Lilliputa's system works? In his interview he said he uses RIPPER algorithm to find entry and exit rules. Anyone familiar with this algorithm?
(1) Я только хотел сказать, что приведенная мной методика проверки ряда на независимость приращений дает практически однозначный и теоретически на 99,99% обоснованный результат - ценовой ряд не является рядом с независимыми приращениями (даже если они мало или вообще не коррелируют). (2) А это, в свою очередь, говорит о том, что все модели работы с ценой, подразумевающие независимость соседних отсчетов - неадекватны.
(1) Yes, the increments are not independent. This has already been calculated before us. GARCH model is a famous and rather simple method to approach this problem. More advanced methods are possible. Or even just raise the order of the GARCH model, and it's already very sophisticated.
(2) Any model is inadequate, this follows from the model definition. Only the market itself will be perfectly adequate to the market. I.e. there is a constant choice between increasing the model adequacy and increasing its complexity, and even a small increase of adequacy requires a serious increase of complexity. The question is, therefore, how inadequate the model is. A simple model is often better than a sophisticated one. The increments can be considered independent and even normally distributed and the price itself a random walk. In reality this is not the case, but it can nevertheless be a good model to go with.
I didn't doubt that, the question is different, what exactly have you calculated, I'm trying to figure it out (only for myself).
The point is that clear (to me) and proven methods of verification require more segments, it simply requires a number. The obtained series of parameters by segments is analysed for consistency with a certain (depending on the method or its variant) distribution and only after that one can apply the trend criteria. For two points, it is difficult to draw such conclusions.
Of course, if one wants to, one can. Here is a simple example: EURUSD series, M15, with 200 000 samples in history. I divide the series into two parts of 100 000 and plot frequencies of the first differences (the second image is a logarithm):
I think you will smile, but the visual analysis for stationarity estimation also applies as the first information. Let's see how the RMS of the two pieces relate:
As Shiryaev said, volatility itself is volatile. Dispersion is actually one way of measuring it. Yes, it has an average value and on long chunks of history the hospital average will be the same, but that does not mean the same on shorter chunks. It is statistically proved that volatility is clustered and autoregressive that is why ARCH/GARCH models are quite adequate (it is proved in "Fundamentals of Financial Mathematics" by Shiryaev).
Of course, the stationarity and variance invariance model does not consider such properties of real series.
And from a purely visual point of view, the analysis of the wave shows that there are periods of increasing volatility (like now) that have a tendency to continue. It is the same with MO: if we count on large chunks of data, the average temperature in the hospital will be 0. However, this does not exclude periods of trends within. Therefore, coincidence of Mo and variance on long sections does not indicate stationarity of the series. If we are to estimate variance change, it should be statistically, not by two points. For example if you want to have 200 samples, break it into series of 1000 and check dispersion distribution.
Для тех кто по прежнему в танке - М.О. случайного блуждания(цены) равно нулю.
To estimate the m.o. of this process (random walk) at least in the time domain, one would have to calculate whether there is a limit to the arithmetic mean of all the members of the series since the beginning of history. But this quantity has no limit, neither classical nor by probability (l.i.m.). What limit can we talk about, if the price in its movement can deviate as far and as long as it wants from the value at the beginning of the trajectory?
We can only talk about m.o. when averaging over realisations at a given point. But in this case, as timbo pointed out, it is equal to the previous price.
Чтобы оценить м.о. этого процесса (случайного блуждания) хотя бы во временно й области, пришлось бы вычислить, существует ли предел среднего арифметического всех членов ряда с начала истории. Но у этой величины предела не существует - ни классического, ни по вероятности (l.i.m.). О каком пределе можно говорить, если цена в своем блуждании может сколь угодно далеко и на достаточно долгое время отклоняться от значения в начале траектории?
Об м.о. можно говорить только при усреднении по реализациям в заданной точке. Но в этом случае, как и указывал timbo, оно равно предыдущей цене.
Yes, the variance goes to infinity. If we consider this formula: x(t) = x(t-1) + e(t), where e(t) ~ N(0,1), then yes, М.О. equals the previous price. I.e. the price yesterday was 1.18, noise is zero, then the price today is x(t) = 1.18+0=1.18 - our profit is zero, minus commission. But I meant М.О. not at the next readout, but expected SB movement in future.Look picture timbo c "bell" - well, how many realizations, what equals the MO? And the absolute scale is not involved. Ie if we trade EUR / USD - I trade without any chart and the current price, I press on the buy and sell, and you trade by TA, clustering chips - in the long term result will be the same.
Да,дисперсия уходит в бесконечность.
If you are talking about gold, maybe.
But not on currency pairs, again inserting my opinion.
What would it look like on an inverse currency pair? To zero? ;)
Look at the process binocularly.
Through forward and reverse quotes (e.g. EURUSD and USDEUR)...
Many illusions will disappear.
Посмотри картинку timbo c "колоколом" - ну и,сколько там реализаций,чему равно М.О.?И абсолютная шкала здесь не причём.Тоесть если мы будем торговать евро/долл - я торгую вообще без графика и текущей цены,давлю на бай и селл,а ты торгуешь по ТА,кластеризуешь фибы - в долгосрочке результат будет одинаковым.
Если вы о золоте - может быть. Но не на валютных парах, опять вставляю свое мнение. Как это на обратной паре будет выглядеть? К нулю? ;)
Currency pairs are not normal assets, there are different rules for them. If you look at monthly bars, you can say that the process is mean-reverting - returning to the mean. However, on the other hand, considering that no one trades on monthly bars and no deposit is long enough to sit out years of drawdowns, i.e. no one can even come close to the notion of not only infinity, but even just "long". And on a shallow scale, even currency pairs behave like random wandering.