# On linear models with no constant and R2

In econometrics course we always say to our students that “if you fit a linear model with no constant, then you might have trouble. For instance, you might have a negative R-squared”. So I tried to find databases on the internet such that, when we compute a linear regression, we actually obtain a negative R squared. I have generated hundreds to random databases that should exhibit such a property, in R. With no success. Perhaps to be more specific, I should explain what might happen if we do not include a constant in a linear model. Consider the following dataset, where points are on a straight line, with a negative slope, far from the origin, symmetric with respect to the first diagonal.

```> x=1:3
> y=3:1
> plot(x,y)```

Points are on a straight line, so it is actually possible to get a perfect linear model. But only if we integrate a constant in our model. This is related to the fact that the correlation between our two variates is -1,

```> cor(x,y)
[1] -1```

The least-square program is here

i.e. the estimate of the slope is

Numerically, we obtain

```> sum(x*y)/sum(x^2)
[1] 0.7142857```

which is the actual slope on the illustration above. If we compute the sum of squares of errors (as a function of the slope), we have here

```> ssr=function(b){sum((y-b*x)^2)}
> SSR=Vectorize(ssr)
> B=seq(-1,3,by=.1)
> plot(B,SSR(B),ylim=c(0,ssr(3)),cex=.6,type="b")```

so the value we have computed is actually the minimum of the sum of squares of errors. But note that the sum of squares always exceeds the total sum of squares in red on the graph above

```> ssr(b)
[1] 6.857143
> sum((y-mean(y))^2)
[1] 2```

Recall that the total “coefficient of variation“, denoted , is defined as

i.e.

```> 1-ssr(b)/sum((y-mean(y))^2)
[1] -2.428571```

which is negative. It is also sometimes defined as “the square of the sample correlation coefficient between the outcomes and their predicted values“. Here it would be related to

```> cor(b*x,y)
[1] -1```

so we would have a unit . So obviously, using the in a model without a constant would give odd results. But the weird part is that if we run that regression with R, we get

```> summary(lm(y~0+x))

Call:
lm(formula = y ~ 0 + x)

Residuals:
1       2       3
2.2857  0.5714 -1.1429

Coefficients:
Estimate Std. Error t value Pr(>|t|)
x   0.7143     0.4949   1.443    0.286

Residual standard error: 1.852 on 2 degrees of freedom
Multiple R-squared: 0.5102,	Adjusted R-squared: 0.2653
F-statistic: 2.083 on 1 and 2 DF,  p-value: 0.2857```

Here, the estimation is correct. But the we obtain tells us that the model is not that bad… So if anyone knows what R computes, I’d be glad to know. The value given by R (thanks Vincent for asking me to look carefully at the R source code) is obtained using Pythagoras’s theorem to compute the total sum of square,

```> sum((b*x)^2)/(sum((b*x)^2)+sum((y-b*x)^2))
[1] 0.5102041```

So be careful, the might look good, but meaningless !

Cite this blog post
Arthur Charpentier (2012, February 2). On linear models with no constant and R2. Freakonometrics. Retrieved June 21, 2024, from https://doi.org/10.58079/ouk5

## 3 thoughts on “On linear models with no constant and R2”

1. Here is a book which mention the link between “negative R-square” and “no constant term“, It can also be found on wikipedia “Additionally, negative values of R2 may occur when fitting non-linear trends to data” with a reference to that paper by Cameron Colin and Frank Windmeijer (where I did not see any reference to negative R-square).

2. Pierre-Hugues Carmichael says:

Looking at the analysis of variance table in SAS gives the same observation:

SS(model): 7.15
SS(error): 6.85
SS(total): 14

But isn’t it clear that the model sum of squares can’t be negative ? By the fundamental partition identity the total sum of squares needs to exceed the error sum of squares.

I think that what is going on is that the when you remove the intercept you are basically constraining mean(y)=0. Your total sum of squares then becomes 1^2+2^2+3^3=14. Of course, if that constraint is inappropriate, your model will be meaningless. But it certainly won’t give you negative R-squared…

3. Pierre-Hugues Carmichael says:

Checked this out using SAS… Got the same result as obtained with R, so they must be doing similar things. I’ve never seen a negative r-squared, never even heard of that warning about no intercept terms in our stats courses. There’s probably something else going on here…

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