# Regress predicted and realized values

I am trying to fit a model of the form

$$y=f(x)+u$$

where $x$ is a vector variable and $u$ an independent random noise. I tried to fit the unknown function $f$ by different machine learning methods (SVM, MARS, etc.)

I split my data into training and validation sets. After fitting the model on the training set, I test it on a validation set, and I plot $y$-realized versus $y$-predicted, and do a linear regression on that scatter plot. I obtain a very small intercept but a coefficient of $0.5$, while I was naively expecting a coefficient near to 1.

Does this regression coefficient tell me something about the quality of fit of my original model or about the original model itself?

Yes, it does. Indeed, a good forecast should be associated with a zero intercept and a unit slope in your setting. Deviations from that indicate forecast suboptimality. The idea is formalized by the Mincer-Zarnowitz test. Given forecasts $\hat y_i$ and realized values $y_i$, the regression
$$y_i = \beta_0 + \beta_1 \hat y_i + \varepsilon_i$$
should ideally (optimally) have zero intercept and unit slope, i.e. $\beta_0=0$ and $\beta_1=1$, which constitutes a testable hypothesis (the Mincer-Zarnowitz test); see Diebold "Forecasting in Economics, Business, Finance and Beyond", p. 337, for an introduction (the textbook is regularly updated so the page number may change over time; just search the pdf for "Mincer" and you will find it).
However, note that a zero intercept and a unit slope do not completely determine the goodness of your forecast. $\beta_0=0$ and $\beta_1=1$ indicates good accuracy but does not say anything about precision. You could have $\beta_0=0$ and $\beta_1=1$ combined with very high error variance $\text{Var}(\varepsilon_i)$ so that mean squared forecast error would still be high and hence the forecast not very satisfactory.