A probability theory problem - page 8

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Let me bring up the subject.

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 the asset as A, the second one as B and the third one as 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 the signal to buy.

Solve through the probability of inverse events:

1-0.55=0.45 - probability that price will not rise if event A occurs,

1-0.6=0.4 - probability that price will not increase upon occurrence of event B,

1-0.65=0.35 - probability that the price will not increase in case event C occurs,

Then the probability that the price will not rise when A&B&C occur simultaneously will be equal: 0.45x0.4x0.35 = 0.063

Then the required probability P(D/ABC) = 1-0.063 = 0.937

Questions:

1. Did I calculate correctly?

2. Is the probability of P(D/ABC) too high, considering rather low probabilities P(D/A), P(D/B) and P(D/B)? It turns out that if P(D/A)=P(D/B)=P(D/B)=0.5 (actually a finger in the sky) then P(D/ABC)=0.875 which imho is not logical.

 
Alexander:

Questions:

1. Has the calculation been correct?

2. Is the probability of P(D/ABC) too high, given the rather low probabilities of P(D/A), P(D/B) and P(D/B)? It turns out that if P(D/A)=P(D/B)=P(D/B)=0.5 (actually a finger in the sky) then P(D/ABC)=0.875 which imho is not logical.

IMHO, everything makes sense. If 3 independent events gave signals, then it is no longer a "finger in the sky".
 
Stanislav Korotky:
IMHO, it all makes sense. If 3 independent events give signals, then it is not a finger in the sky.

But the probability of these events is 0.5
 
Alexander:

But the probability of those events is 0.5.


We roll the die. If it is odd, we have a signal to buy, if it is even, we have a signal to sell.

Roll three times. If three times odd, we buy. Three times even, we sell.

 
Alexander:

Let me bring up the subject.

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 the asset as A, the second one as B and the third one as 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 the signal to buy.

Solve through the probability of inverse events:

1-0.55=0.45 - probability that price will not rise if event A occurs,

1-0.6=0.4 - probability that price will not increase upon occurrence of event B,

1-0.65=0.35 - probability that the price will not increase in case event C occurs,

Then the probability that the price will not rise when A&B&C occur simultaneously will be equal: 0.45x0.4x0.35 = 0.063

Then the required probability P(D/ABC) = 1-0.063 = 0.937

Questions:

1. Did I calculate correctly?

2. Is the probability of P(D/ABC) too high, considering rather low probabilities P(D/A), P(D/B) and P(D/B)? It turns out that if P(D/A)=P(D/B)=P(D/B)=0.5 (actually a finger in the sky) then P(D/ABC)=0.875 which imho is not logical.

It's kinda weird. IMHO it should be about 0.6, but you need to calculate the full probability field and the tree of outcomes, and this is so at a glance - the average. The final value can't be more than the maximal and less than the minimal - they are independent. Otherwise you will get that by making a random independent sample from a random value of any of them you will improve the result.
 
Maxim Kuznetsov:
it's kind of weird. IMHO it should be around 0.6


Intuitively I would say around 0.7

Maxim Kuznetsov:
But we need to calculate the full probability field and the tree of outcomes,


This is hardly possible. (

Maxim Kuznetsov:
And this is just a rough guess - the average. The final value can't be greater than the highest and less than the lowest - they're independent. Otherwise you get that by making a random independent sample from a random value of any of them you'll improve the result.


Why an average? Why do traders look for confirmation of signals from other sources? Why in the court they question more than one witness (if any), why do they accept as evidence the material evidence, examination results and etc.? All these factors tilt the odds in favour of a correct decision, increasing its probability. It should be the same with indicators (signals). One is good, two is better, three is even better. The question is how much better it is and how to calculate it analytically?



 
Alexander:

But these events have 0.5 probability.

So what? Three times 0.5 is a very "strong" coincidence - clearly the total value must be much higher.
You gave the correct formula.

It is also desirable to consider the probabilities of P(A), P(B), P(C) themselves. After all, indicators must generate signals with different frequency.

 
Stanislav Korotky:

So what? Three times 0.5 is very "strong" coincidence - clearly the total value must be much higher.
You gave the correct formula.


Thank you. I will have to believe it. )

Stanislav Korotky:

It is also desirable to consider the probabilities of P(A), P(B), P(C) themselves. After all, indicators must generate signals with different frequency.


Yes, of course. In general, with different frequency and at different times. But that is another task.

The moment when the signals coincide was of interest. I am interested in the moment when the signals coincide and what is more profitable:

  • To wait for the moment when three signals coincide. This happens notably less frequently, but it significantly increases the chances of success.
  • To be satisfied with coincidence of two signals. It occurs more frequently, but with a higher probability of failure.
It seems that MM rules here: with two signals we open with a standard lot, with three - with a larger one.

 
Stanislav Korotky:

So? Three times 0.5 is a very "strong" coincidence - clearly the total value must be significantly higher.
You have given the correct formula.

It is also desirable to consider the probabilities of P(A), P(B), P(C) themselves. After all, indicators must generate signals with different frequencies.

Three times 0.5 is no coincidence at all. This probability of an increase (0.5) occurs after any event, it coincides with the probability of a decrease. That is, expectations do not shift in any direction. Such events, which do not affect (are not correlated) the course, can be counted a hundred per second (tram passing, three passengers entering it, etc.).
 
Vladimir:
Three times 0.5 is no coincidence at all. This probability of increase (0.5) occurs after any event, it coincides with the probability of decrease. That is, expectations do not shift in any direction. It is possible to count a hundred such events that do not influence in any way (are not correlated) on the course (a tram passing by, three passengers getting on it, etc.).


I agree. That is why I wrote that 0.5*0.5*0.5 is a finger in the sky.

Do you have an alternative solution to the problem or at least a hint?

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