Machine learning in trading: theory, models, practice and algo-trading - page 239
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Try to describe the candle by two numbers, each in the range [-1.0; 1.0]. These are the position of O and C in relation to H and L.
How do you do that?
In height H is 1, L is -1, respectively express O and C relative to H and L.
Here the volatility of the candle is not taken into account, all calculations are as if inside the candle, and what kind of candle it is, a gap or a small dojiq the MO does not see
I think the most normal is the % increments, but I do not count them correctly
a selection about the clustering of candles, but as normalized do not disclose, and those that do disclose, the result did not satisfy me
https://www.elitetrader.com/et/threads/statistical-analysis-of-candlesticks-patterns.285918/
http://robotwealth.com/unsupervised-candlestick-classification-for-fun-and-profit-part-1/
http://robotwealth.com/unsupervised-candlestick-classification-for-fun-and-profit-part-2/
http://intelligenttradingtech.blogspot.com/2010/06/quantitative-candlestick-pattern.html
It does not take into account the volatility of the candle, all calculations are as if inside the candle, and what kind of candle it is, a gap or a small dojiq the MO does not see
I think the most normal is the % increments, but I do not count them correctly
volatility is exactly what you don't need to take into account. And you need to get rid of gaps (move the candles to the distance of the gap).
The lesson is over)))
Thanks, I think I got it. It seems very simple, I do not believe it, but I will check.
Also strange that the sign is a separate predictor, I would just make the size of the candle negative if it is down. This should also try.
Thanks, I think I got it. It seems very simple, I do not believe it, but I will check.
Also strange that the sign is a separate predictor, I would just make the size of the candle negative if it is down. This, too, should try.
I don't get it, though.
How was the target made?
Where did the formula come from?
I continue to believe that without selecting predictors by their effect on the target variable, everything else is irrelevant. This is the very first step. Either we remove the noise predictors and then our chances of building a NOT retrained model increase, or the noise predictors remain, which will necessarily lead to retraining. And since the behavior of the retrained model in the future is in no way related to its behavior in the past, such a retrained model is not needed.
Another interesting approach to determining the importance of predictors. Multiple algorithms for determiningsignificance testing are not used here.
Here is the executed code from this post
> n <- 10000
>
> x1 <- runif(n)
> x2 <- runif(n)
> y <- -500 * x1 + 50 * x2 + rnorm(n)
>
> model <- lm(y ~ 0 + x1 + x2)
>
> # 1a. Standardized betas
> summary(model)$coe[,2]
x1 x2
0.02599082 0.02602010
> betas <- model$coefficients
> betas
x1 x2
-500.00627 50.00839
> imp <- abs(betas)/sd.betas
Ошибка: объект 'sd.betas' не найден
> sd.betas <- summary(model)$coe[,2]
> betas <- model$coefficients
> imp <- abs(betas)/sd.betas
> imp <- imp/sum(imp)
> imp
x1 x2
0.9091711 0.0908289
> imp1 <- abs(model$coefficients[1] * sd(x1)/sd(y))
> imp2 <- abs(model$coefficients[2] * sd(x2)/sd(y))
>
> imp1 / (imp1 + imp2)
x1
0.9095839
> imp2 / (imp1 + imp2)
x2
0.0904161
> # 2. Standardized variables
> model2 <- lm(I(scale(y)) ~ 0 + I(scale(x1)) + I(scale(x2)))
> summary(model2)
Call:
lm(formula = I(scale(y)) ~ 0 + I(scale(x1)) + I(scale(x2)))
Residuals:
Min 1Q Median 3Q Max
-0.0236475 -0.0046199 0.0000215 0.0046571 0.0243383
Coefficients:
Estimate Std. Error t value Pr(>|t|)
I(scale(x1)) -9.932e-01 6.876e-05 -14446 <2e-16 ***
I(scale(x2)) 9.873e-02 6.876e-05 1436 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.006874 on 9998 degrees of freedom
Multiple R-squared: 1, Adjusted R-squared: 1
F-statistic: 1.058e+08 on 2 and 9998 DF, p-value: < 2.2e-16
> abs(model2$coefficients)/sum(abs(model2$coefficients))
I(scale(x1)) I(scale(x2))
0.90958355 0.09041645