Calculate the probability of reversal - page 3
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Generally, little is known about the process, here I purposely generated a sequence, in which the next step depends on the previous one and the probability of continuation is about 65%, I do not remember exactly. That is, I set probability of continuation-> generated sequence-> got distribution, now I want to get back parameter of continuation probability from distribution.
It is unlikely to be possible to calculate it analytically. You could try a Monte Carlo simulation to see approximately how the distribution (e.g. its variance) depends on the probability of continuation.
Generally, I don't know much about the process, I've intentionally generated a sequence, where the next step depends on the previous one and continuation probability is about 65%, I don't remember exactly. In other words, I set probability of continuation-> generated sequence-> got distribution, now I want to get back parameter of continuation probability from distribution.
In the original post it was: "hence the question how, having only a probability density plot, to calculate the probability of reversal at each step."
So you want to find one number (65% in the example) common to all steps? You don't want the probabilities of reversal (not necessarily the same) at each step?
In the original post it was: "hence the question of how, with only a probability density plot, to calculate the probability of reversal in each step."
So you want to find one number (65% in the example) common to all steps? You don't want the probabilities of reversal (not necessarily the same) at each step?
The meaning of the histogram is as follows: we take a sample of 10 steps (1 step may be up or down) and measure the distance by which the process moved from the starting point for these 10 steps. Then we take 10 000 samples of such samples and calculate how many percent have gone for -10 steps from the starting point (downwards), then -8, -6 and so on. These percentages are written on the histogram, and values from -10 to 10 are written at the bottom of the histogram.
Why did you limit it to the ten innermost points. For the edges of the non-zero probability range P <> 0 (reachable points) at each step number i, the equality P(max) = k^i is true, where k is the required constant fraction of step up directions. Accordingly P(min) = (1-k)^i. From these perturbation propagation fronts we can also estimate k. Only you should not take the middle (10 out of 10,000) but the edges.
You can use a range of 10 steps, then your histogram shows Pmax=0.0217, k = 0.0217^0.1=0.68178, Pmin=0.0225, k = 0.0225^0.1=0.684255. It is not much different from 0.65. But here you can see that you have k exactly the probability of trend continuation, while I was talking about the probability of a step up in the post above.
The estimation error will decrease if you take more steps. But you need the probabilities Pmax and Pmin to still have a reasonable order of magnitude, they decrease rapidly as i increases. At 30 steps their values will be for k=0.7 about 0.00002, for k=0.3 about 2.00E-16 (k is step up probability).
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And hence the question of how, with only a probability density plot, to calculate the probability of reversal at each step.
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The sum of one side of the central bar + half of the central bar divided by the total sum of all bars. Probability.
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Suppose we have the following probability density graph
Here, on the x-axis, you can see how many steps a person took from the starting point, from -10 (to the left) to +10 (to the right) and it is signed with what probability he did it in %. How do you find what was the probability of turning at each step?
And what do you mean by a U-turn? - One step in the opposite direction or all subsequent steps in the opposite direction?
On the face of it, the usual problem from the realm of Markov chains is the evolution of the initial distribution over time. Some complication is due to the fact that the chain is of the second order (the probability of the price at the moment n depends not only on the price at the moment n-1, but also at the moment n-2).
The calculation has to be done numerically. Elegantly (analytically) it would be possible except to calculate stationary distribution, but here it is obviously not defined.
Alexey, and given graph of probabilities of finite steps and fact that next step p=50%, can not be solved as stationary table distribution?
ap: got it that it's not 50%. But all the same, if we consider that the distribution remains normal, and consider that this very probability is constant on this sample then I think it is possible to calculate it analytically.
And if it is not constant, then the problem has many solutions.
You can use a range of 10 steps, then your histogram shows Pmax=0.0217, k = 0.0217^0.1=0.68178, Pmin=0.0225, k = 0.0225^0.1=0.684255. It is not much different from 0.65. But here you can see that you have k exactly the probability of trend continuation, while I was talking about the probability of a step up in the post above.
The estimation error will decrease if you take more steps. But you need the probabilities Pmax and Pmin to still have a reasonable order of magnitude, they decrease rapidly with increasing i. At 30 steps their values will be for k=0.7 about 0.00002, for k=0.3 about 2.00E-16 (k is step up probability).
What do you mean by a U-turn? - One step in the opposite direction or all subsequent steps in the opposite direction?
Alexey, and the given graph of finite step probabilities and the fact that the next step p=50%, can't you solve as a stationary table distribution?
ap: understood that it is not 50%. But all the same, if we consider that the distribution remains normal, and consider that this very probability is constant on this sample then I think it is possible to calculate it analytically.
And if it is not constant, then the problem has many solutions.
What do you mean by a U-turn? - One step in the opposite direction or all subsequent steps in the opposite direction?
Alexey, and the given graph of finite step probabilities and the fact that the next step p=50%, can't you solve as a stationary table distribution?
ap: understood that it is not 50%. But all the same, if we consider that the distribution remains normal, and consider that this very probability is constant on this sample then I think it is possible to calculate it analytically.
And if it is not constant, then the problem has many solutions.
By definition, the stationary distribution should not change at each step. In this case, any distribution will "spread out" at each step, increasing the variance.