and wandering around randomly again... - page 54
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Which graph with which cloud? What do mine have to do with yours? Have you performed an infinite number of throws?
Yes... You've done yours, I've done mine.
And the graphs of our outcomes will go EVERYWHERE along their own trajectories.
Therefore, your assertion that "There is only one trajectory" is untrue.
Your words unambiguously mean that (!) the probability of getting heads on the 101st roll is MORE than 50%, if there were 60 heads and 40 tails in the previous 100 rolls (for example).
But this is at variance with the generally accepted view that ON ANY shot in the series (!) heads and tails are EQUALLY likely to fall out, since the results of the previous shots do not affect any of the subsequent ones.
Where exactly did you see this ambiguity?
Yes... You did yours, I did mine.
And the graphs of our outcomes will go EVERYWHERE along their own trajectories.
Therefore, your assertion that "There is only one trajectory" is untrue.
Eventually with an infinite number of throws, yours will go to zero and mine will go to zero.
One coin has one trajectory. Two coins - two trajectories - each one has its own trajectory. It would be foolish to think anyone would claim otherwise. In any case, with an infinite number of tosses, the trajectory from each coin will go to zero.
Here is a list of papers (5087 pieces) which try to get the residual from the model (model fitting error) to represent a random walk.
5087 papers matched the search for GARCH, GJR-GARCH, EGARCH, in titles and keywords.
Where exactly did you see this ambiguity?
I said"unambiguous."
Eventually, with an infinite number of flips, yours will go to zero and mine will go to zero.
One coin has one trajectory. Two coins - two trajectories - each has its own trajectory. It would be foolish to think anyone would claim otherwise. In any case, with an infinite number of flips, the trajectory from each coin will go to zero.
Not necessarily, for I have exactly the same right to hit the "2 down, 1 up..." trajectory or the like as you do on a dangling around zero. They are equal probability.
Not necessarily, because I have exactly the same right to hit the "2 down, 1 up..." trajectory or the like as you do on a dangling around zero. They' re equal probability.
They are equally likely for a single coin on an infinity of a smaller order than the infinity of the total number of flips.
There is no way a coin can produce an infinite trajectory of +1 and -2.
You are an ignoramus who does not want to understand anything.
Here is a list of papers (5087 pieces) which try to get the residual from the model (model fitting error) to represent a random walk.
5087 papers matched the search for GARCH, GJR-GARCH, EGARCH, in titles and keywords.
They are equally likely for a single coin on an infinity of a smaller order than the infinity of the total number of flips.
There is no way a coin can produce an infinite trajectory of +1 and -2.
You are an ignoramus who does not want to understand anything.
"... There is no way a coin can produce an infinite trajectory of +1 and -2..."
This is your categorical assertion that the probability of the sequence "+1, -2" and similar sequences is EQUAL to ZERO and there are no other sequences.
As a "dormant ignoramus", I conclude that your coin knows how to control the process, stores the history of throws... and falls out in a STRICTLY DETERMINED pattern (not 50/50) whenever a series starts to add up in a way that doesn't satisfy you.
(you can get punched in the eye for such a coin from casino customers
)