What is the interpretation of the "traditional" $R^2$? Suppose the following data correspond to observed responses and their predictions obtained from some model.
observed  <- runif(100)
predicted <- 1 + observed + rnorm(100, sd = 0.1)  # suppose they are obtained from some model instead

# predicted vs actual plot
plot(observed, predicted)


We can observe a linear trend and therefore expect a high $R^2$. $R^2$ can be interpreted as

*

*the degree of linear association between "predicted" and "observed", or

*the proportion of the total variation of the y-variable (here, "predicted") explained by the x-variable (here, "observed") using a straight line.

Question
Below are two versions of $R^2$ reported by caret. How do you interpret the "traditional" version, which gives -11 for the above example?

R2_default <- caret::R2(pred = predicted, obs = observed)
R2_tradi   <- caret::R2(pred = predicted, obs = observed, form = "traditional")
R2_tradi2  <- 1 - sum((observed - predicted)^2) / sum((observed - mean(observed))^2)



In ordinary least squares linear regression with just one feature and an intercept (so “simple” linear regression), there are a number of calculations that yield the same result.

*

*$(cor(x,y))^2$


*$(cor(\hat y, y))^2$


*$1 -\dfrac{\sum_{i=1}^n
\left(
y_i-\hat y_i
\right)^2
}{\sum_{i=1}^n
\left(
y_i-\bar y
\right)^2
}$
Consequently, any of these three can justifiably be called $R^2$.
Even moving to regression with multiple features, option #1 is not viable, but the other two remain viable calculations for any poring of truth values and predicted values.
However, when you obtain the predictions by a method other than ordinary least squares, options #2 and #3 need not be equal.
Your software is giving you the ability to pick which of those two calculations you want to perform. Option #2 is called the “corr” method, while option #3 is called the “traditional” method.
For reasons I discuss here, it is the third calculation (equivalent to #4 in the link) that makes sense to me as a comparison of your modeling of the conditional expected value compared to a baseline model of the conditional expected value that naïvely guesses the pooled/marginal expected value $\bar y$ every time; pay particular attention to the simulation that shows how correlation-based $R^2$ can mislead you into thinking your predictions are good. If you are in a setting where #2 and #3 are equivalent, then feel free to do the easier correlation-based calculation, but, in my view, that is only because of the equivalence, not because of inherent value to the correlation-based calculations.
(I can see some value to the correlation-based calculations in the sense that, a high correlation-based calculation and low (even negative) value of #3 could suggest a simple mapping between your poor predictions and good predictions, though I would wonder why I didn’t get good predictions at the beginning.)
