It's impossible to make money on Forox!!! - page 38
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Can you be a little more specific here, Oleg?
In a nutshell it's hard... but I'll try... :) I'll figure out how to present it better.
Starting with Einstein and Wiener, the highbrow know very well what Brownian motion is. It does not help them to predict it.
It depends on what section? If one predicts the deviation of the distance of the current point from the starting point as a function of time, then the function is quite accurate and has a high approximation with a large number of trials. That is, if Brownian motion had anything to do with trading, I would always bet on the distance of the point from the initial movement, because this very distance is strictly proven and has a clear formula.
But when it comes to SB, Brownian motion has about as much to do with trading as I have to do with the Bolshoi Theatre - I've never been there.
The theoretical basis of SB used in trading for applied purposes is referred to as: "Random walk on a straight line corresponding to Bernoulli's scheme". The mathematical apparatus is quite elaborate, both for symmetric wandering - the sideways trend, and for asymmetric - the trend. For example, for symmetric random walk on a straight line, there is a strict proof that the point will return to the origin with probability 1 - 100% guarantee (where it has visited at least once, it will visit again and again, and the time between returns is non-uniform).
And the applied problem answering the question about probability of triggering tees and moose (if they have been set) is called the "Brokeback Problem".
Does it depend on what section? If one predicts the deviation of the distance of the current point from the starting point as a function of time, then the function is accurate enough and has a high approximation with a large number of trials. That is, if Brownian motion had anything to do with trading, I would always bet on the removal of a point from the initial movement, because this very removal is strictly proven and has a clear formula.
You probably meant the maximum deviation from the starting point? The distance from the starting point to the current point is not predictable for martingales, which SB is. More precisely for them the best prediction for any time ahead is the last available value of the series. It is clear that the slope of this prediction increases in direct proportion to the square root of the prediction time. That's why in martingale any forecast is that nothing will change since the last observation, but the range of possible values increases as the time for which the forecast is made increases
you must mean the maximum deviation from the starting point? The distance from the starting point to the current point is not predictable for martingales, which SB is. For them, the best prediction for any time ahead is the last available value of the series. Clearly the skop from this prediction increases in direct proportion to the square root of the prediction time.
cf. Brownian motion
see. Brownian motion
where the function is described
"predict the deviation of the distance of the current point from the starting point as a function of time, then the function is sufficiently accurate and has a high approximation with a large number of trials. "
where the function is described
"predict the deviation of the distance of the current point from the starting point as a function of time, then the function is sufficiently accurate and has a high approximation with a large number of trials. "
See function (1) in the above link, i.e. which calculates the square of the displacement of a particle along any direction (the square of the change (increment) of the distance along any axis) as a function of time.
See function (1) in the above link, i.e. which calculates the square of the displacement of a particle along any direction (the square of the change (increment) of the distance along any axis) as a function of time.
this formula is the essence of the variation of dispersion (or sco) with time as I wrote about. Yes, it increases, but it is not the distance from the current point to the starting point as a function of time.
If I say that tomorrow afternoon in Moscow will be the same temperature as today, for example +5, with a possible range of +-3, then those 6 degrees is the accuracy of the forecast. And the forecast is +5. The formula you are referring to just tells how forecast accuracy decreases (or possible range expands) with time.
this formula is the essence of the variance (or sko) changing with time as I wrote about. Yes, it increases, but it is not the distance from the current point to the starting point as a function of time.
P...dx all you like, but dx is in no way dispersion or RMS, it is the distance (displacement) from one point to another as a function of time along any of the chosen axes.
see experimental data:
Brownian motion "through the eyes" of a digital microscope
To quote for the particularly gifted:
"So, if in 1 min a Brownian particle moves on average by 10 µm, then in 9 min it should on average move by -10 = 30 µm, in 25 min by -10 = 50 µm, etc."