A probability theory problem - page 11

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Stanislav Korotky:

Let's go through it again in order.

The formula proposed above (I will deliberately write it differently - through X, A, B, C):

P(X) = 1 - (1 - P(A)) *(1 - P(B)) *(1 - P(C))

will give the probability of a signal from at least one indicator. This is why the result is so high - three indicators signal more often. But this is essentially not what the problem statement is looking for.

By Bayes:

P(D|ABC) = P(ABC|D) * P(D) / P(ABC)

Here P(ABC) = P(A) * P(B) * P(C)

where a priori indicator probabilities are calculated as the number of signals of each indicator among the total sum of all indicators.

P(D) = 0.5 by default, when there is no super-trend, i.e. the probability of buying and selling signals are equal.

But I have doubts how to calculate P(ABC|D). The simplest way (due to independence):

P(ABC|D) = P(A|D) * P(B|D) * P(C|D)

and each such conditional probability must be calculated as the number of signals of each indicator on the set of all bars where the buy was correct.

But all this is not a truth in the last resort. ;-/

firstly, 3 signals is too much :-)

it is enough to solve the problem for 2 signals.

Secondly, even without knowing the initial a priori signal probabilities, you can make assumptions about them that are very close to the truth.

For example, we can introduce the relation P(X)=f(P(D|X)), i.e. consider that the a priori probability = a function of the known probability of "takeProfit after the signal". quite a lot is known about this very f:

  • it is symmetric with respect to 0.5 (like P(D|X) by the way)
  • it is inversely proportional to P(D|X) - most likely more accurate signals are rare
  • it contains exp, i.e. is non-linear - for any signal is a composition in itself, you can't get away from Gauss :-)

That is, you can choose a convenient function for calculations and calculate what there is approximately and on what "what you get" depends more strongly.

 
Maxim Kuznetsov:

firstly 3 signals is too much :-)

It is enough to solve the problem for 2 signals.

The overshoot is probably a joke, judging by the smiley face. It is desirable to have a system in analytical form for N signals, where 2 of course also includes, but according to my observations 3 is quite a popular number (at least, "dog breeders recommend" - main, confirming and filter).

And what is the analytical solution for 2 signals, if I am wrong?

So far it is only clear to me that we are looking at D here as the only outcome, but there are actually several: buy (Db), sell (Ds) and wait (Dw), and they form a complete group and may affect the calculation of P(A), P(B), P(C).
 
The radio operators are pushing back. The signal is being searched for.
 
Stanislav Korotky:

Overkill is probably a joke, judging by the smiley face. It is desirable to have a system in analytical form for N signals, which of course includes 2, but according to my observations 3 is quite a popular number (at least, "dog breeders recommend" - main, confirming and filter).

And what is the analytical solution for 2 signals, if I am wrong?

So far it is only clear to me, that we are looking at D as the only outcome, but in fact there are several: buy (Db), sell (Ds) and wait (Dw), and they form a complete group and may influence the calculation of P(A), P(B), P(C).

having solution for the simplest case 2=(1+1) signal, the system for N is quite easy to build: 3 signals are (1 + 1) + 1 and so on.

I don't have a ready-made solution in my pocket, so as soon as an idea appears, I propose it here...

We look at the outcomes quite correctly - within the framework of the original problem and without trying to overcomplicate everything.

In real life of course the signal X signals: "in time not longer than T the price will reach +Profit points rather than -Loss points with the probability P", and adding real signals, in which at least one characteristic differs from T, Profit, Loss, is a great pleasure :-)

 
Maxim Kuznetsov:

In real life of course signal X signals: "in time not longer than T the price will reach +Profit points rather than -Loss points with the probability P", and adding real signals with at least one different T,Profit,Loss characteristic is a real pleasure :-)

Often TakeProfit, StopLoss and time horizon T are determined by the strategy, i.e. are equal for all signals collected in statistics. Let us not complicate things prematurely. ;-)
 
Stanislav Korotky:
Often TakeProfit, StopLoss and time horizon T are determined by strategy, i.e. are equal for all signals collected in the statistics. Let's not complicate things prematurely. ;-)

I am calling you not to complicate the task, but to simplify it as much as possible - to consider only an abstract task with only 2 signals.

A final digression into reality: TakeProfit, StopLoss set in strategies and the mentioned characteristics of Loss/Profit signals are sort of "two big differences" :-) Generally speaking, real signals have some non-linear characteristic (you can consider it as a diagram) F(t) "probability of reaching Profit before Loss in time t from the signal" that t increases and t t tends to be similar to the one for an arbitrary entry on a "random walk" chart

And one last digression: it's a pity we can't experimentally verify the analytical solution. Or does anyone know independent signals with 55,60,65% reliability ?

 
Maxim Kuznetsov:

One last digression: it is a pity that we cannot experimentally verify the analytical solution. Or is anyone aware of independent signals with 55,60,65% reliability?

Of course we will be able to check the analytical solution. We can take a pair of poorly correlated indices and calculate for them all a priori probabilities and probabilities of signals that coincide with the wins. It does not matter what the values are for checking. Even if 30%, 40% - this will also do for testing the formulas ;-) . To evaluate the behavior of this analytical solution we can simply calculate N-dimensional function for different probabilities, and the subsequent search for indices with higher reliability is a separate issue.
 
Stanislav Korotky:
We will of course be able to check the analytical solution. We can take a pair of poorly correlated indices and calculate for them all a priori probabilities, and probabilities of signals coinciding with wins. It does not matter what the values are for checking. Even if it were 30%, 40% - that would also be fine for testing the formulas ;-) ....
If it were 30, 40 you would be a billionaire by now. The real number is 50. Don't take them all together or separately, it's the same 50.
 
Alexander:

Suppose we have three indicators that periodically give buy/sell signals and their readings are independent of each other. Let's denote the event when the first indicator gives a signal to buy an asset by A, the second one by B and the third one by C.

Let's denote the increase of price as event D.

Let P(D/A)=0.55 - the probability of price increase if the A indicator gives a signal to buy.

P(D/B)=0.6 and P(D/C)=0.65.

Find P(D/ABC) - the probability that the price will rise if all three indicators gave a buy signal.


The answer is:

P(D|ABC) = [P(D|A) * P(D|B) * P(D|C)] / [P(D|A) * P(D|B) * P(D|C) + (1 - P(D|A)) * (1 - P(D|B)) * (1 - P(D|C))

An article has been published on this subject.

 
Stanislav Korotky:

The answer to the question is:

P(D|ABC) = [P(D|A) * P(D|B) * P(D|C)] / [P(D|A) * P(D|B) * P(D|C) + (1 - P(D|A)) * (1 - P(D|B)) * (1 - P(D|C))

Gut!
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