Can I use the residuals of a multiple linear regression instead of the Predicted Value (y hat)?

Question

I know this probably is a very newbie question, but I haven't been able to find anything about it elsewhere.

I'm running some OLS regressions on high frequency data of stocks in order to model some imbalances, all done in R and checked with Excel.

But, in my tests, I'm constantly seeing that the residuals of the regressions yield far better results than the regular dependent variable (Y) would when used in the OPOSITE of the original logic. $$ residue < lower threshold (negative int), then, BUY $$ regular meta code would be: $$ \hat y > upper threshold (positive int), then, BUY $$

I tried this because of a mistake in the beginning of my coding and later found out it was a far better predictor in the out of sample analysis I ran. So I kept up testing and this kept showing up in the results.

The possible explanation I thought about this is that, if the residue is large enough, I should expect that the contrary of it to happen next. Like a mean reversal logic of the error. If the error is large enough, it will probably return to the mean, so I use it as the "- (Y hat)"

Is there some kind of explanation to this, is this some kind of statistical technique that I just don't know? Or I'm am I just delusional and this makes no sense at all, using the residuals as a predictor for my model?

Thank in advance

EDIT : What is the expected correlation between residual and the dependent variable? talks about this kind of comparison between residuals and Y.. Not much value in it.

I'll post the other scatter plots of residuals against fitted values and Y against fitted values.

Answer 1

Error measures are based on residuals. If you denote $y$ as predicted variable and $\hat y$ as your estimate for it, then we define residuals as

$$ r = y - \hat y $$

Error measures such as MAE, or RMSE are defined as $1/n \sum |r|$ and $1/n \sum r^2$. If you take $-r$ as your prediction, then what you are doing is you take

$$ y - y - \hat y = -\hat y $$

as your "residuals". So MAE becomes $1/n \sum |-\hat y|$ and RMSE becomes $1/n \sum -\hat y^2$, this applies also to other error measures. This means that you are not measuring error at all.

So by predicting all zeros, you would conclude that they perfectly fit to any data (MAE = RMSE = 0). Small predictions would lead to small "errors" and vice versa. You can easily extrapolate this example to other cases.

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If you say that this approach works for you in financial setting then this means that betting on random noise is better then being on predictions from your model. If regression assumptions are met then residuals are random around zero and there is no trend in them. So if this works then either you use regression for data that does not qualify for it, or your model is really poor.