What is the cumulative probability? - page 6
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Not like that. 1 minus the probability of getting sick. The answer is 0.94 probability of getting sick.
I see. I couldn't see which words belong to the formula.
Regarding the independence of bulls and bears opinions. Are you, by putting in shorts, in any way dependent on your opponent standing in the long? Not to mention the crowd, which is swarming with conflicting opinions about the future course of the price, which is why the price is here at the moment and not wherever you would like it to be.
I know how to count too. Where did the last two addends come from?
I quote again:
we obtain a system of
up P1*(1-P2)
down P2*(1-P1)
up + down -- a complete group of events whose sum of probabilities is 1.
we get --
P1*(1-P2) + P2*(1-P1) == 1
Waiting for an explanation.
P1*P2 of course at the end, good for noticing, but you could see it by the numbers anyway. Substitute the values and calculate what you get. You should get 1.
Let me explain. We have two oracles. The space of events is as follows:
P1*(1-P2)+(1-P1)*P2+(1-P1)*(1-P2)+P1*P2
The first says "up" the second says "up" P1 and (1-P2), (first event)
+
First says "down" second says "down" (1-P1) and P2, (second event)
etc.
that is, all the outcomes are taken into account, of which there are four.
P1*P2 is of course at the end, it's good to notice, but you could see it from the numbers. Substitute the values and calculate what you get. You should get 1.
Let me try to explain. We have two oracles. The space of events is as follows:
P1*(1-P2)+(1-P1)*P2+(1-P1)*(1-P2)+P1*P2
The first says "up" the second says "up" P1 and (1-P2), (first event)
+
First says "down" second says "up" (1-P1) and P2, (second event)
etc.
>> that means all the outcomes, of which there are four, are taken into account.
So why are you twisting the subject? And discretising into up and down.
An analogue series problem.
We have two oracles! The first one says: price will cross or touch the level of 1.5000 with the probability of 0.6 in the current day.
The second oracle disagrees and says: price will cross or touch 1.5000 with probability 0.2 within the current day.
What is the final probability that the price will cross or touch 1.5000 within the current day ???????????.
Note that if the first oracle's prediction were the same as the second: p1=p2=0.2, the final probability would be 0.2. How simple it is.
But if the first oracle still predicts p1=0.6 ? How do you calculate the final probability ???????
I see. I didn't see which words belong to the formula.
Regarding the independence of bulls and bears. When you put in shorts, do you depend on your opponent, who is in a long position? Not to mention the crowd, which is full of contradictory opinions about the future course of the price, that is why the price is here at the moment and not somewhere where you want it to be.
>> Sure! And me and my opponent -- we have the same raw data, there is no point in talking about independence, there is none!
I have a question for the mathematicians. Although it looks like an off-topic, it is applicable to MTS.
Problem:
Let there be an event X whose probability of occurrence is equally dependent separately on two events A and B independent of each other.
If the probability of A-dependent event X is P(A)=0.4,
and the probability of occurrence of event X, dependent on B, is defined as P(B)=0.2,
then the question is:
What is the final probability of occurrence of event X: P(A && B) ???
So, the final conclusion, I think, is still obtained. There is no solution due to the incorrectness of the condition.
But we can look at the problem from the other side.
We have 2 forecast series for the same data -- bullish series and bearish series.
What prevents you from building statistics? There are only THREE dimensions -- a bullish row, a bearish row, a resultant row.
Thus we get some discrete (if you want continuous) function P(A && B) = F(P(A), P(B)).
Which, by the way, will confirm or refute the conclusions above.
Good luck.
Why are you twisting the subject? And discretise it up and down.
An analogue series problem.
We have two oracles! The first one says: Price will cross or touch the 1.5000 level with a probability of 0.6 for the current day.
The second oracle disagrees and says: price will cross or touch 1.5000 with probability 0.2 within the current day.
What is the final probability that the price will cross or touch 1.5000 within the current day ???????????.
Note that if the first oracle's prediction were the same as the second: p1=p2=0.2, the final probability would be 0.2. How simple it is.
But if the first oracle shows p1=0.6? How to calculate the final probability ??????? ?
The problem statement could be like this?
We have two oracles! The first one says: "The price will cross or touch the level of 1.5000", with the probability that he is right, 0.6 for the current day.
The second oracle disagrees and says: "Price will cross or touch 1.5000", with 0.2 probability he is right within a day.
What is the final probability that the price will cross or touch 1.5000 within the current day if both oracles are touching?
If both oracles are independent, then, as mentioned above, in order to calculate the probability of a joint event, the probabilities must be multiplied. p1*p2=0.12. So the second oracle will not improve your result, because it is wrong in most cases. About "...if p1=p2=0.2, then the final probability would be 0.2" pull up the terver textbook and see for yourself that this is not true.
Thanks for the formulas. Only I don't get the right answer in the output from any of the formulas.
Below p1 and p2 are probability values in the range (0;1) not included:
1.1 If P(A)=1 and P(B)=p1, then P(A && B)=1.
Look again carefully at the formulas I have given:
P(A & B) = P(A) * P(B) = 1 * p1 = p1, but not 1
As I understand from the graph: we plot the value (0.5+1)/2=0.75 on the x-axis and get the probability value on the y-axis. Question: what is this function? I want to write down the final formula.
Option - Y=3*X^2-2*X^3
How about this?
We have two oracles! The first one says: "Price will cross or touch the 1.5000 level", with a probability that he is right, 0.6 for the current day.
The second oracle disagrees and says: "Price will cross or touch 1.5000", with 0.2 probability he is right within a day.
What is the final probability that the price will cross or touch 1.5000 within the current day, if both oracles show a touch?
If both oracles are independent, then, as stated above, in order to calculate the probability of a joint event, the probabilities must be multiplied. p1*p2=0.12. So the second oracle will not improve your result, because it is wrong in most cases. About "...if p1=p2=0.2, then the final probability would be 0.2" pick up the terver textbook and see for yourself that this is not true.
You are still missing the point. If the forecasts are both 50/50. Then, according to you, the total forecast would be 0.5*0.5=0.25 ?????. I.e., the more analysts, the worse the outlook for the event?! :)
You're just throwing around formulas from a book, which are absolutely irrelevant to this case. This is not an event where you calculate the probability of two sixes falling together. If you don't think about it, you'd better read it, there's no need to write for nothing. Thousands of analysts will make probability predictions about the odds of the pair hitting 1.5000 and all the mathematicians will say: "Such event will happen with probability P(1)*P(2)*...*P(1000)*......., in short - the event will not happen, as we are many and we are the power".
Take another close look at the formulas I have given:
P(A & B) = P(A) * P(B) = 1 * p1 = p1, but not 1
Your formulas do not solve the problem at hand. Again, think carefully about why.
Option - Y=3*X^2-2*X^3
Thanks for the function. I'll let you know the results later.