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Fine, we have found such a solution for some time interval. >> What's next?
Think! And then, we are operating with values averaged over a large interval, so the obtained solution is not local and is true for any part of BP.
IMHO, the type of function Y is missing. Or am I missing something?
We don't have a priori knowledge of the right kind of curve Y. Everything we need has to be buried in the function:
S=w1*(X[i]-Y[i])^2+w2*(Y[i]-Y[i-1])^2-w3*{(Y[i]-Y[i-1])*(Х[i]-Х[i-1])}^2-->min
From it, with any luck, we will get the form of MA, or rather its recursive form.
Like what? Earning -- profitability is embedded in the target function.
On a past profitable function, are you going to profit? How is this different from trying to profit after optimisation?
On a profitable former feature are you going to make a profit? How is this different from trying to profit after optimising?
We are going to do it better than others! Because we are exploiting the best (maximising profitability). Of course, if there is nothing in the original BP that can be exploited (in a certain sense), then we will get zero. But if there is... then we'll get maximal profitability!
We are going to do it better than others! Because we are exploiting the best (maximising profitability). Of course, if there is nothing in the original BP that can be exploited (in a sense), then we get zero. But if there is... we'll get the most out of it!
I'm picking on the words you said.
Take a time period and calculate the maximum possible profit on it. Will that be the same as the result of your function on that time frame?
Or do you have a maximum-operating-profit? I think the evaluation of the degree of possible exploitation is subjective.
It's easier to criticise (that's me) than to create. But in this case I think you are trying to develop a dead end.
Any averaging leads to a lag. That's a fact. At one time, I too was euphoric about the "possibility" of applying "Mashas".
Then I realized that very often I was just wishful thinking. It is impossible to improve the averaging! A simple calculation of the profit
on whole bars. At the best case profit/loss = 5/6. And this after dancing around Take Profit, Stop Loss .... Too late signals.
Otherwise, the flag in your hands.
Take a time frame and calculate the maximum possible profit on it. Will this be the same as the result of applying your function to this timeframe?
No, it won't.
Because we are working on the right border of BP (without looking into the future), while what you have formulated (maximum-opportunity profit) is a history work or adjustment. In contrast, we maximize profit "without knowing the history", analyzing only the latest reading of the quote and its one previous value Х[i]-Х[i-1] and that's it. It seems to be like this.
Any averaging leads to a lag. This is a fact. At one time I too had euphoria at the "possibility" of applying "Machs".
Then I realized that I often wishful thinking. And this after dancing around Take Profit, Stop Loss .... Too late signals.
The difference is that you are using the scientific method, while we are solving the problem theoretically. Your result is not a proof due to its lack of generality. What we get is the best (in a sense) and general.
P.S. I have repeatedly stated my, what I think is a reasonable point of view on the applicability of the moving average method in trading. And the attempt to find optimal moove is not dictated by changing my opinion, just, the problem seemed to me interesting in itself, as well as its definition and possible solutions.
No, it won't.
In contrast, we maximize the profit "without knowing" the history, analyzing only the latest reading of the quote and its one previous value X[i]-X[i-1] and that's it. That's kind of how it is.
The necessary historical component will be taken from X[i - 1] and Y[i - 1] . So knowing the history, i.e. the necessary part of it.
It's easier to criticize (that's me) than to create. But in this case I think you are trying to develop a dead-end path.
I haven't seen the task in this form yet, and it looks quite nice. Why not try it?
The necessary historical component will be taken from X[i - 1] and Y[i - 1]. So knowing the history, i.e. the necessary part of it.
In recursive filters, each count of muv, contains information about ALL previous values of quotient, taken with decaying coefficients... It turns out that we do take into account the history and optimize the profit with an eye on it... It may be a fit, but the fit may be "right", not as dumb optimization in the tester.
Somebody take a derivative of S function on parameter Y[i] and equate it to zero! Because I'm already so-so...
In recursive filters, each muv count contains information about ALL previous quotient values, taken with decaying coefficients... It turns out that we do take history into account and optimize the profile with an eye on it... It's possible that it's a fitting, but the fitting is probably "correct" not as in dumb optimization in temter.
Yup, that's what I'm talking about.
The perfect MA kind of does. 'Author's dialogue. Alexander Smirnov.'
see post ANG3110 06.02.2008 20:48