Put in a good word about the occasional wanderer... - page 22
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Curious, has anyone calculated the average duration of a Brownian bridge on kotirs?
Counting the intersection with the average exit at 0...
;)
I checked by direct calculation. Well, of course, if you introduce a very long waiting period (over-sitting), it will be somewhere around 50%, but if you introduce, say, 10 bars of waiting (that will be the time limit), it will not be 50% - much less.
Anyone here or anyone at all?))
here or elsewhere. because we don't generally know the number of "princes"/bars.
So the distribution of "total choice time" is what's interesting...
There's a lot of scriptwriters here - maybe someone will write one.
;)
Well, that depends on what to count. We need a priori probability estimations before the price starts moving from the set point, otherwise there is no sense in counting it all post factum... Therefore, the parameters will have to be set beforehand - how many n points it is, for how much time, how many bars to count afterwards.
It's understandable. But here the question itself, as you probably realise, does not apply to the market. If the price was 1.0000 when the long position was opened and then it became 0.0050 after n bars, then the probability that "the princess will see the best of the candidates", i.e. the maximum of already seen prices, while limiting the time of position holding, surely will not exceed 50%.
Let's measure it, if you're interested. But I'm no longer interested. We are not dealing with the equal probability of occurrence event described in the article. The conditions are different.
If the price was 1.0000 when you opened a long position and then after n bars it was 0.0050
Are you a forex trader or an equity trader?
;)
Are you a forex trader or a stock trader?
;)
It's understandable. But here the question itself, as you probably realise, does not apply to the market. If the price was 1.0000 when the long position was opened and then it became 0.0050 after n bars, then the probability that "the princess will see the best of the candidates", i.e. the maximum of already seen prices, while limiting the time of position holding, surely will not exceed 50%.
Let's measure it, if you're interested. But I'm no longer interested. We are not dealing with the equal probability of occurrence event described in the article. The conditions are different.
Here the problem condition will be as follows: each of 1000 princes has a number theoretically from minus infinity to infinity, exponentially distributed. The princess counts the sum of all the numbers and tries to guess the moments of reaching the maximum and minimum values.
Not like that, I think. The position has to be open and then the counting starts. How else to implement the selection of the best number?
And where does the exponential distribution come from? Where does counting the sum of the numbers come from?
We are talking about comparing the numbers with those already seen.
Here the problem condition will be as follows: each of 1000 princes has a number theoretically from minus infinity to infinity, exponentially distributed. The princess counts the sum of all the numbers coming in and tries to guess the moments when the maximum and minimum values are reached.
is a dead end, I think. Seems to me that it starts looking for the first extremum of the optimum strategy when the "trand" has supposedly started. That said, if it has not spawned a trade - not bad either...
;)
Not like that, I think. The position has to be open and then the counting starts. How else to implement the selection of the best number?
And where does the exponential distribution come from? Where does counting the sum of the numbers come from?
We are talking about comparing the numbers with those already seen.