Suppose I have the true model that $y= 1+3x_1+4x_2+\xi$ where $\xi\sim\mathcal{N}(0,1)$. However, $x_2=5x_1+\eta$ where $\eta\sim\mathcal{N}(0,\sigma^2)$.
If I was to regress $y$ on $x_1,x_2$ I use the following code in R:
x1=1:20
sigma=0.1
x2=x1+sigma*rnorm(20,1)
y=y=1+3*x1+4*x2+rnorm(20,1)
data=data.frame(y,x1,x2)
summary(lm(y~.,data))
For $\sigma=0.1$ I get the following output:
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 2.3771 0.5735 4.145 0.000677 ***
x1 3.5004 2.3680 1.478 0.157643
x2 3.4736 2.3700 1.466 0.161002
---
However, if I change it to $\sigma=1$ I get:
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 2.5992 0.4652 5.588 3.27e-05 ***
x1 3.0217 0.3171 9.529 3.13e-08 ***
x2 3.9469 0.2978 13.252 2.17e-10 ***
---
Now when I try graphical lasso (glasso) using R's huge library I get the following graph: (This part is just to illustrate that some algorithms can pick up the correlation. Not necessarily related to question below.)
library(huge)
a=huge(as.matrix(data),method="glasso")
g=graph.adjacency(a$path[[10]],mode="undirected")
plot(g)
The question is:
When I know that some input variables are linearly correlated what do the p-values mean? I'm asking this in a general context and not in relation to the demonstrated problem.
What if $x_2$ was $f(x_1)$ where $f(\cdot)$ is non-linear? What is the interpretation of a p-value?
(optional) Can graphical Lasso replace stepwise regression to some extent? I understand that this algorithm is highly dependent on the regularising parameter.