Econometrics: one step ahead forecast - page 41

 
If m.o.s. is 10 pips and s.c.s. is 100, it is theoretically possible that the overall profitability will have to wait too long.
 
Mathemat:
If m.o.s. is 10 pips and s.c.s. is 100, it is theoretically possible that the overall profitability will have to wait too long.
At a minimum, no more than the candle length, and in general, one should compare it with the variance of the original series. Although in one of the posts above it was proved that in volatility trading this is not quite the case.
 
faa1947:

Isn't the prediction itself a mathematical expectation of the model

Everything is fine if the error is stationary. Many times I have written and given graphs of the error, which have a very convoluted appearance.


I don't quite see how the non-stationarity of the error can affect the estimate of the forecast. Doesn't the error ultimately affect only the risk of the model in the context of Value at Risk!
 
Mathemat:

In order to grasp such a statistical advantage to be able to talk about its significance, we do not need 100 transactions at all, but many times more, tens of thousands.

Let us assume that there are N deals. Stat advantage (2% * N) must be at least two times more than sqrt(N). And at the same time we will be about 95% sure of stat advantage value.

What is your 97% quality of this mach (if you are talking about HP)? Is there a formula?


The significance of statistical estimates of mo and variance depends on the variance itself and the root of the number of transactions, not on the advantage (mo). I.e. if one system's mo is 2 times larger than another system's mo, then for the same accuracy of mo and mo estimates you need 4 times as many trades in the first system. Of course, it is better to describe everything in confidence intervals (its width). Width of MDI of Mo and dispersion estimations depends on dispersion itself and root of number of deals

P.S. This of course is all for stationary distributions. With non-stationarity the MD is not defined at all - at least temporal stationarity or approximation to it is needed

 
C-4:

I don't quite see how the non-stationarity of the error can affect the estimate of the forecast. Doesn't the error ultimately affect only the risk of the model in the context of Value at Risk!

I use the following definition of stationarity: approximately the mo constant and the standard deviation. Here is one of the graphs:

What's the guarantee that on an out-of-sample forecast (and we're only considering that option) you won't get another outlier of error? Further, being confident that the error is almost a constant (is the 25 pips spread on the chart a constant?), either you get into a pose by reasoning about the risks in the form of forecast execution confidence intervals, or you assume the forecast is a constant and piously believe in that figure.

 
C-4:

I don't quite see how the non-stationarity of the error can affect the estimate of the forecast. Doesn't the error ultimately affect only the risk of the model in the context of Value at Risk!

the predicted value is an estimate mo the future series, and the error is its variance (sko). What is actually being predicted is some future distribution of price increments. If this distribution is non-stationary, then neither the estimate of mo nor the estimate of its variance can be trusted. That is, the forecast cannot be trusted
 
Avals:

the significance of statistical estimates of mo and variance depends on the variance itself and the root of the number of transactions, not on the advantage (mo). I.e. if sko of one system is 2 times more than sko of another, then for the same accuracy of estimates mo and sko you need 4 times more trades in the first system. Of course, it is better to describe everything in confidence intervals (its width). Width of MDI of mo and variance estimates depends on the variance itself and the root of the number of trades

Now I am beginning to understand. It turns out that the higher the s.c.o., the greater the requirements for the m.o. value to confirm its statistical significance. For the c.s.o. of the current model its m.o. is too insignificant to be statistically significant and such a model cannot be used.
 
Avals:


The significance of statistical estimates of mo and variance for


I completely agree with this, but for me the interesting question is what happens outside the sample?

What needs to be analysed inside the sample to increase the probability of an out-of-sample prediction being fulfilled?

Is the calculation of the error and the stationarity requirement for it sufficient?

A final question. What is the forecast horizon? One step or several steps? If several steps, how is this possibility defined?

 
C-4:

Now I am beginning to understand. It turns out that the higher the s.c.o., the greater the requirements for the m.o. value to confirm its statistical significance. For the c.s.o. of the current model, its m.o. is too insignificant to be statistically significant, and such a model cannot be used.

approximately. As a result of prediction tests (or TC) we get estimates of mo and sko - that's 2 numbers. This is actually wrong - we have two intervals, and the resulting values are their midpoints. I.e. if we got mo=10pnuts, then actually mo=10+-delta. This delta depends on sko - the bigger it is, the bigger is delta and on the number of deals (root). I.e. delta is directly proportional to sko/Root(N)
 
faa1947:

Is the calculation of the error and the stationarity requirement for it sufficient?

It is not clear to me how one can expect stationary error on non-stationary data? In the graph you presented above, the error value clearly does not have the properties of a finite variance, so it is at least questionable to apply estimates based on the variance of the results (like s.r.f. or the square root of N) to it.
Reason: