# Can Adjusted R squared be equal to 1?

I have a dataset with around 15 independent variables. I am using a multi-regression model to fit the dataset. For model selection, I am using a backward elimination procedure based on the p-values. The adjusted R^2 for the model with all predictors is exactly 1. At this point, I concluded that maybe the model is also picking up noise. But, based on the model selection I removed 5 predictor variables and still the adjusted R^2 is 1. I am not sure if this correct or I am just modeling noise. Can someone comment on this?

• That shouldn't happen. You should be able to find out if all the residuals are 0. That is the only way the unadjusted R$^2$ can be 1. – Michael Chernick Jan 31 at 1:08

## 2 Answers

Dan and Michael point out the relevant issues. Just for completeness, the relationship between adjusted $$R^2$$ and $$R^2$$ is given by (see, e.g., here)

$$R^2_{adjusted}=1-(1-R^2)\frac{n-1}{n-K},$$ (with $$K$$ the number of regressors, including the constant). This shows that $$R^2_{adjusted}=1$$ if $$R^2=1$$, unless (see below) $$K=n$$.

$$R^2=1$$ occurs when all residuals $$\hat u_i=y_i-\hat y_i$$ are zero, as $$R^2=1-\frac{\hat{u}'\hat{u}/n}{\tilde{y}'\tilde{y}/n}.$$ Here, $$\hat u$$ denotes the vector of residuals and $$\tilde y$$ the vector of demeaned observations on the dependent variable.

Dan discusses one reason to get an $$R^2$$ of 1. Another is to have as many regressors as observations, i.e., $$K=n$$.

Technically, this is because the $$n\times K$$ regressor matrix $$X$$ then is square. The OLS estimator $$\hat\beta=(X'X)^{-1}X'y$$ can then be written as (assuming no exact multicollinearity) $$\hat\beta=(X'X)^{-1}X'y=X^{-1}{X'}^{-1}X'y=X^{-1}y$$ so that the fitted values $$\hat y=X\hat\beta$$ are just $$\hat y=XX^{-1}y=y$$, so that all residuals are zero.

Here is an illustration using artificial data (code below), in which regressors are generated totally independently of $$y$$, and yet we achieve an $$R^2$$ of 1 once we have as many of them as we have observations. Code:

n <- 15
regressors <- n-1 # enough, as we'll also fit a constant
y <- rnorm(n)
X <- matrix(rnorm(regressors*n),ncol=regressors)

collectionR2s <- rep(NA,regressors)
for (i in 1:regressors){
collectionR2s[i] <- summary(lm(y~X[,1:i]))$r.squared } plot(1:regressors,collectionR2s,col="purple",pch=19,type="b",lwd=2) abline(h=1, lty=2)  When $$K=n$$, R however, correctly, does not report an adjusted $$R^2$$: > summary(lm(y~X)) Call: lm(formula = y ~ X) Residuals: ALL 15 residuals are 0: no residual degrees of freedom! Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2.36296 NA NA NA X1 -1.09003 NA NA NA X2 0.39177 NA NA NA X3 0.19273 NA NA NA X4 0.51528 NA NA NA X5 -0.04530 NA NA NA X6 -1.28539 NA NA NA X7 -0.72770 NA NA NA X8 -0.14604 NA NA NA X9 0.34385 NA NA NA X10 -0.93811 NA NA NA X11 2.23064 NA NA NA X12 0.06744 NA NA NA X13 0.21220 NA NA NA X14 -2.29134 NA NA NA Residual standard error: NaN on 0 degrees of freedom Multiple R-squared: 1, Adjusted R-squared: NaN F-statistic: NaN on 14 and 0 DF, p-value: NA  • Do you want to expand on what u hat and y tilde are? I wonder how many cases the OP has as too few cases can easily lead to perfect prediction. – mdewey Jan 31 at 15:37 • Right, I should have spelled out all my notation in the first place. I agree it would be helpful information to know the sample size OP has access to. – Christoph Hanck Jan 31 at 15:42 • By quoting an unnecessarily limited formula for the adjusted$R^2$(which applies only to the ordinary regression situation but not to multiple regression) you arrive at an incorrect conclusion: the adjusted$R^2$isn't even defined when there are as many regressors as observations. It's certainly not equal to unity in that case! – whuber Feb 1 at 17:54 • Ouch, I should have seen that this was not the right expression to connect the adjusted and standard$R^2\$. I hope that my edit fixes this. – Christoph Hanck Feb 1 at 19:39

An adjusted R squared equal to one implies perfect prediction and is an indication of a problem in your model. Adjusted R squared is a penalised version of R square, which is a way of describing the ratio of the residual sum of squares to the total sum of squares - as you approach 1 the implication is that there is no variation/deviation away from your model.

I would suggest you begin by looking at a correlation matrix, or put each predictor into your model individually to see which predictor is causing the issue.

In R, you will get a warning from lm: "essentially perfect fit..."

In the (single predictor) example below you will see that adjusted R square is less than 1 even when the correlation between y and x is greater than 0.99.

# create a data frame with some strongly correlated variables
myData<- data.frame(y = rnorm(n = 1000, mean = 0, sd = 1))
myData$$x1<- myData$$y
myData$$x2<- jitter(myData$$x1, factor = 10)
myData$$x3<- jitter(myData$$x1, factor = 1000)

# fit models
myModel1<- lm(y ~ x1, data = myData)
myModel2<- lm(y ~ x2, data = myData)
myModel3<- lm(y ~ x3, data = myData)

# output
summary(myModel1)
#> Warning in summary.lm(myModel1): essentially perfect fit: summary may be
#> unreliable
#>
#> Call:
#> lm(formula = y ~ x1, data = myData)
#>
#> Residuals:
#>        Min         1Q     Median         3Q        Max
#> -4.551e-15 -1.200e-17  6.000e-18  2.090e-17  3.455e-16
#>
#> Coefficients:
#>              Estimate Std. Error   t value Pr(>|t|)
#> (Intercept) 1.404e-17  4.924e-18 2.852e+00  0.00444 **
#> x1          1.000e+00  5.085e-18 1.966e+17  < 2e-16 ***
#> ---
#> Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#>
#> Residual standard error: 1.553e-16 on 998 degrees of freedom
#> Multiple R-squared:      1,  Adjusted R-squared:      1
#> F-statistic: 3.867e+34 on 1 and 998 DF,  p-value: < 2.2e-16
summary(myModel2)
#>
#> Call:
#> lm(formula = y ~ x2, data = myData)
#>
#> Residuals:
#>        Min         1Q     Median         3Q        Max
#> -1.996e-03 -9.643e-04 -1.996e-05  1.009e-03  2.034e-03
#>
#> Coefficients:
#>              Estimate Std. Error  t value Pr(>|t|)
#> (Intercept) 3.278e-06  3.647e-05     0.09    0.928
#> x2          1.000e+00  3.766e-05 26550.25   <2e-16 ***
#> ---
#> Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#>
#> Residual standard error: 0.00115 on 998 degrees of freedom
#> Multiple R-squared:      1,  Adjusted R-squared:      1
#> F-statistic: 7.049e+08 on 1 and 998 DF,  p-value: < 2.2e-16
summary(myModel3)
#>
#> Call:
#> lm(formula = y ~ x3, data = myData)
#>
#> Residuals:
#>       Min        1Q    Median        3Q       Max
#> -0.214135 -0.097828 -0.003721  0.099000  0.226453
#>
#> Coefficients:
#>              Estimate Std. Error t value Pr(>|t|)
#> (Intercept) -0.001598   0.003602  -0.444    0.657
#> x3           0.983900   0.003685 266.982   <2e-16 ***
#> ---
#> Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#>
#> Residual standard error: 0.1136 on 998 degrees of freedom
#> Multiple R-squared:  0.9862, Adjusted R-squared:  0.9862
#> F-statistic: 7.128e+04 on 1 and 998 DF,  p-value: < 2.2e-16

cor(myData$$x1, myData$$x2)
#>  0.9999993
cor(myData$$x1, myData$$x3)
#>  0.9930721