Econometrics: one step ahead forecast - page 65
You are missing trading opportunities:
- Free trading apps
- Over 8,000 signals for copying
- Economic news for exploring financial markets
Registration
Log in
You agree to website policy and terms of use
If you do not have an account, please register
for the sole purpose of comparing with the neutral option (sb) to identify trendiness/returns
Trendiness, of course, is a curious property.
But we should all be interested in the predictability of the quote at the point of the last bar. What can we take from history to make a conclusion about predictability? So far I can't add anything to what I've laid out here. These are the following requirements for the model:
1. R-squared must tend to 1
2. the error of model fitting to the quote should be less than the candle length for the forecasted time frame
3. the probability of the model coefficient being equal to zero should approach zero
4. probability of absence of autocorrelation in the residue should be more than 10% and approach 100%
5. probability of absence of heteroscedasticity of the residue has to be more than 10% and should approach 100%
6. the model residue must be close to the stationary one
7. prediction error should aspire to at least a stop level.
If you manage to gather a certain amount of models that meet all these requirements, you can look for the optimal model with the maximal profit factor among such models.
If not - you cannot enter the market, because there is no model for the current market. And if you have been in the market, you need to get out.
Maybe I'm wrong? But all that is listed has been proven by calculations in this thread and my articles and has not been refuted.
3. the probability of the model coefficient being equal to zero should tend towards zero
4. probability of absence of autocorrelation in the residual should be more than 10% and tend to 100%
5. probability of absence of residual heteroscedasticity should be more than 10 % and approach 100 %
These criteria are strange and the threshold values are more than suspicious. What if the probability of 4) is equal to, say, 75%, then we can, with a clear conscience, consider that there is no autocorrelation?
There is something wrong with the evidentiary validity of the stochastic conclusion.
С-4: Совершенно верно, отсутствие позиции - это всегда ноль, и неважно неттинг это или локи.
I wrote about a different implication: "a position with zero volume is no position and nothing else". In MT4 you can make an integral position with zero volume, but with both halves open.
Two systems are equivalent in terms of different accounting methodologies, as long as they have fully matched equities and balances. The accounting methodologies affect the logic of the system. Need proof?
But that's not the case for this system, no matter how you look at it. If I don't understand something, please explain. Maybe the balance is superfluous here, and only equity matching is sufficient?
I can tell you more about it. Can anybody recommend me a proper drawing program, so that I could put arrows, circles and other simple markings on the drawing? No Photoshop.
P.S. Found it, FastStone. I didn't find how to save edits in it.
The criteria are odd, and the boundary values are more than suspicious. What if the probability of 4) is, say, 75%, then we can safely assume that there is no autocorrelation?
There's something wrong with the evidentiality of the statistical inference here.
10% is from the null hypotheses. They sound different and 10% is the number to start the conversation with, 5% is more common.
For 4) strictly sounds like this:
But: no autocorrelation in the residual. We have a figure of 6%. conclusion: at 5% significance level we reject hypothesis But, but at 10% significance level we cannot reject hypothesis But.
My reasoning is easier to accept. Although due to the imprecision of my formulation, it can be understood that 75% is the probability of no autocorrelation. This is completely wrong. It is about the hypothesis, not the autocorrelation.
But that's not the case for this system, no matter how you look at it. If I don't understand something, explain. Maybe balance is superfluous here, and only equity matching is enough...?
I repeat. It can be done, but it is a bore. This billet affects the logic of the system.
I see. At least one netizen has admitted that it's a bummer. The logic of the system, of course, does not change. It's the implementation that changes.
I'll post what I wanted to do in the "stat.arb on muve and price" system.
Click on the illustration for a better view. Now for an explanation:
After the moment of price purchase (along with the 13th sale of a muv) it is necessary to accompany the position, waiting for the moment when the situation becomes favorable. We should always make sure that at the current moment the muv is always sold. When the next bar is closed, the earliest part of the mouve should be closed and a new one should be opened (sold). You don't need to do the same with the price.
Selling muv - in parts of 0.01, and buying price - in parts of 0.13, because here the muv period is 13.
But at netting after the 13th step we are out of the market all the time :(
And the impression is that it is possible to make a profit even on a Wiener process!
Slava, make a timeline of the swing back for the SB. I'd like to make sure it's not the tail that's leading the dog.
I checked it for Sat. For Sat copying the volatility of a real instrument and classic ones. The conclusion is as follows - it is impossible to estimate returns below a day in this way. I.e. for Sat that uses real vol vol the results are 100% coincident with real series. It shows intraday cyclical volatility and the difference from the Sat is explained by cyclical changes in volatility. The daily data on the Sat Sat using real ox or classic Sat clearly coincides with the theoretical one, unlike the real series. In general, the ox has many effects that are difficult to interpret correctly.
Click on the illustration for a better view. Now for an explanation:
After the moment of price purchase (along with the 13th sale of a muv) it is necessary to accompany the position, waiting for the moment when the situation becomes favorable. We should always make sure that at the current moment the muv is always sold. When the next bar is closed, the earliest part of the mouve should be closed and a new one should be opened (sold). You don't need to do the same with the price.
But at netting after the 13th shang we are always out of the market :(
how do you sell a muv? :)
The muv is equal to the arithmetic average of the prices on the different bars. In the initial stage of muva accumulation we sell the price :) After 13 steps it turns out that we have sold the muv. More precisely, the sum of prices is sold. But at the 13th step we buy the price in an amount exceeding the steps of accumulation exactly 13 times.
I just took a simple machine. With the others will be much more complicated.