Why does summary overestimate the R-squared with a "no-intercept" model formula

Question

I wanted to make a simple linear model (`lm()`) without intercept coefficient so I put `-1` in my model formula as in the following example. The problem is that the R-squared return by `summary(myModel)` seems to be overestimated. `lm()`, `summary()` and `-1` are among the very classic function/functionality in R. Hence I am a bit surprised and I wonder if this is a bug or if there is any reason for this behaviour. Here is an example: ``` x <- rnorm(1000, 3, 1) mydf <- data.frame(x=x, y=1+x+rnorm(1000, 0, 1)) plot(y ~ x, mydf, xlim=c(-2, 10), ylim=c(-2, 10)) mylm1 <- lm(y ~ x, mydf) mylm2 <- lm(y ~ x - 1, mydf) abline(mylm1, col="blue") ; abline(mylm2, col="red") abline(h=0, lty=2) ; abline(v=0, lty=2) r2.1 <- 1 - var(residuals(mylm1))/var(mydf$y) r2.2 <- 1 - var(residuals(mylm2))/var(mydf$y) r2 <- c(paste0("Intercept - r2: ", format(summary(mylm1)$r.squared, digits=4)), paste0("Intercept - manual r2: ", format(r2.1, digits=4)), paste0("No intercept - r2: ", format(summary(mylm2)$r.sq…

Accepted Answer

Your formula is wrong. Here is what I do in `fastLm.R` in RcppArmadillo when computing the summary:

```
## cf src/library/stats/R/lm.R and case with no weights and an intercept 
f <- object$fitted.values    
r <- object$residuals         
mss <- if (object$intercept) sum((f - mean(f))^2) else sum(f^2)     
rss <- sum(r^2)

r.squared <- mss/(mss + rss) 
df.int <- if (object$intercept) 1L else 0L

n <- length(f)     
rdf <- object$df     
adj.r.squared <- 1 - (1 - r.squared) * ((n - df.int)/rdf)      
```

There are two places where you need to keep track of whether there is an intercept or not.

Why does summary overestimate the R-squared with a "no-intercept" model formula

Solution

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