Which regression model to choose? [duplicate]

Question

This question already has an answer here:

• Removal of statistically significant intercept term increases $R^2$ in linear model 2 answers

I have two models, one lm(y ~ x1 + x2 + 0) which gives me a close to 0.90 something $R^2$ and another model lm(y ~ x1 + x2) which gives me a close to 0.0003 ish $R^2$. Based on $R^2$ one might choose the first model. I have domain (prior) knowledge that a 0 value of $x_1$ and a 0 value of $x_2$ should produce a 0 value of $y$. So, is my explicit setting of 0 a fair thing? Based on this knowledge and the obtained value of $R^2$ can I go with the first model?

Edit: From the threads Removal of statistically significant intercept term increases $R^2$ in linear model and When is it ok to remove the intercept in a linear regression model? I have learned that it is okay to do it as long as the domain knowledge above holds true. From Is $R^2$ useful or dangerous? I learned that $R^2$ alone is insufficient to make a decision about how good the model is. I need to look at AIC and BIC to make further decisions. However, because I my background is not in statistics, I am having trouble understanding the following in the first link:

1. Why does the first equality occur only when intercept is included?
2. why does $\bar{y}$ become 0 when intercept disappears?
3. Why in the second case, $R_0^2$ uses a reference model corresponding to noise only?
4. What is the $\bar{y_1}$ in the notation?
5. Why is $\|\ y \|_2^2 = \| y - \bar{y_1}\|_2^2 + n \bar y^2$ true?

Answer 1

In R formula +0 is the same as -1 and it means that you want a model with no intercept. Also, if regression model is:

$$y = \alpha + \beta_1 x_1 + \beta_2 x_2$$

so if $x_1 = x_2 = 0$, then:

$$y = \alpha + \beta_1 \times 0 + \beta_2 \times 0 = \alpha$$

so if you know that if $x_1 = x_2 = 0$ then $y=0$, it also means that in this case $\alpha$ has to be $0$.

On another hand, removing an intercept in many cases is not a good idea (as also gung suggeses in a comment). In your case model with no intercept will provide a perfect fit for values only in the case where $x_1 = x_2 = y = 0$, but could give worse fitting in other cases so you should consider if this idea is resonable.

However, you do not "explicitly set 0" in here, but remove an intercept (i.e. $\alpha$) from the model. $R^2$ also suggests that the model without intercept is better, however beware that $R^2$ values could be misleading and you should not judge a model only basing on $R^2$.