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First of all, we should understand what the R software is doing when no intercept is included in the model. Recall that the usual computation of $R^2$ when an intercept is present is $$ R^2 = \frac{\sum_i (\hat y_i - \bar y)^2}{\sum_i (y_i - \bar y)^2} = 1 - \frac{\sum_i (y_i - \hat y_i)^2}{\sum_i (y_i - \bar y)^2} \>. $$ The first equality only occurs because of the inclusion of the intercept in the model even though this is probably the more popular of the two ways of writing it. The second equality actually provides the more general interpretation! This point is also address in this related question.

First of all, we should understand what the R software is doing when no intercept is included in the model. Recall that the usual computation of $R^2$ when an intercept is present is $$ R^2 = \frac{\sum_i (\hat y_i - \bar y)^2}{\sum_i (y_i - \bar y)^2} = 1 - \frac{\sum_i (y_i - \hat y_i)^2}{\sum_i (y_i - \bar y)^2} \>. $$ The first equality only occurs because of the inclusion of the intercept in the model even though this is probably the more popular of the two ways of writing it. The second equality actually provides the more general interpretation! This point is also address in this related question.

set.seed(.Random.seed[1])

n <- 220
a <- 0.5
b <- 0.5
se <- 0.25

# Make sure x has a strong mean offset
x <- rnorm(n)/3 + a
y <- a + b*x + se*rnorm(x)

int.lm   <- lm(y~x)
noint.lm <- lm(y~x+0) 

# For comparison to summary(.) output
rsq.int <- cor(y,x)^2
rsq.noint <- 1-mean((y-noint.lm$fit)^2) / mean(y^2)

This gives the output:

> summary(int.lm)

Call:
lm(formula = y ~ x)

Residuals:
      Min        1Q    Median        3Q       Max
-0.656010 -0.161556 -0.005112  0.178008  0.621790

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept)  0.48521    0.02990   16.23   <2e-16 ***
x            0.54239    0.04929   11.00   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.2467 on 218 degrees of freedom
Multiple R-squared: 0.3571,     Adjusted R-squared: 0.3541
F-statistic: 121.1 on 1 and 218 DF,  p-value: < 2.2e-16

> summary(noint.lm)

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

Residuals:
     Min       1Q   Median       3Q      Max
-0.62108 -0.08006  0.16295  0.38258  1.02485

Coefficients:
  Estimate Std. Error t value Pr(>|t|)
x  1.20712    0.04066   29.69   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.3658 on 219 degrees of freedom
Multiple R-squared: 0.801,      Adjusted R-squared: 0.8001
F-statistic: 881.5 on 1 and 219 DF,  p-value: < 2.2e-16

set.seed(.Random.seed[1])

n <- 220
a <- 0.5
b <- 0.5
se <- 0.25

# Make sure x has a strong mean offset
x <- rnorm(n)/3 + a
y <- a + b*x + se*rnorm(x)

int.lm   <- lm(y~x)
noint.lm <- lm(y~x+0)  # Intercept be gone!

# For comparison to summary(.) output
rsq.int <- cor(y,x)^2
rsq.noint <- 1-mean((y-noint.lm$fit)^2) / mean(y^2)

This gives the following output. We begin with the model with intercept.

# Include an intercept!
> summary(int.lm)

Call:
lm(formula = y ~ x)

Residuals:
      Min        1Q    Median        3Q       Max
-0.656010 -0.161556 -0.005112  0.178008  0.621790

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept)  0.48521    0.02990   16.23   <2e-16 ***
x            0.54239    0.04929   11.00   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.2467 on 218 degrees of freedom
Multiple R-squared: 0.3571,     Adjusted R-squared: 0.3541
F-statistic: 121.1 on 1 and 218 DF,  p-value: < 2.2e-16

Then, see what happens when we exclude the intercept.

# No intercept!
> summary(noint.lm)

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

Residuals:
     Min       1Q   Median       3Q      Max
-0.62108 -0.08006  0.16295  0.38258  1.02485

Coefficients:
  Estimate Std. Error t value Pr(>|t|)
x  1.20712    0.04066   29.69   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.3658 on 219 degrees of freedom
Multiple R-squared: 0.801,      Adjusted R-squared: 0.8001
F-statistic: 881.5 on 1 and 219 DF,  p-value: < 2.2e-16

Well, in that case, R (silently!) uses the modified form $$ R_0^2 = 1 - \frac{\sum_i (y_i - \hat y_i)^2}{\sum_i y_i^2} \>. $$

It helps to recall what $R^2$ is trying to measure. In the former case, it is comparing your current model to the reference model that only includes an intercept (i.e., constant term). In the second case, there is no intercept, so it makes little sense to compare it to such a model. So, instead, $R_0^2$ is computed, which implicitly uses as a reference model corresponding to noise only.

But, how are they different, and when?

Let's take a brief digression into some linear algebra and see if we can figure out what is going on. First of all, let's call the fitted values from the model with intercept as $\newcommand{\yhat}{\hat {\mathbf y}}\newcommand{\ytilde}{\tilde {\mathbf y}}\yhat$ and the fitted values from the model without intercept $\ytilde$.

Ok, so how do we make the ratio on the left-hand side small. Well, $\newcommand{\P}{\mathbf P}\ytilde = \P_0 \y$ and $\yhat = \P_1 \y$ where $\P_0$ and $\P_1$ are projection matrices corresponding to subspaces $S_0$ and $S_1$ such that $S_0 \subset S_1$.

Here we try to generate an example close to the case in the question where an intercept is explicitly in the model. Below is some simple R code to demonstrate

set.seed(.Random.seed[1])

n <- 220
a <- 0.5
b <- 0.5
se <- 0.25

x <- rnorm(n)/3 + a
y <- a + b*x + se*rnorm(x)

int.lm   <- lm(y~x)
noint.lm <- lm(y~x+0) 

# For comparison to summary(.) output
rsq.int <- cor(y,x)^2
rsq.noint <- 1-mean((y-noint.lm$fit)^2) / mean(y^2)

Well, in that case, R (silently!) uses the modified form $$ R_0^2 = \frac{\sum_i \hat y_i^2}{\sum_i y_i^2} = 1 - \frac{\sum_i (y_i - \hat y_i)^2}{\sum_i y_i^2} \>. $$

It helps to recall what $R^2$ is trying to measure. In the former case, it is comparing your current model to the reference model that only includes an intercept (i.e., constant term). In the second case, there is no intercept, so it makes little sense to compare it to such a model. So, instead, $R_0^2$ is computed, which implicitly uses a reference model corresponding to noise only.

In what follows below, I focus on the second expression for both $R^2$ and $R_0^2$ since that expression generalizes to other contexts and it's generally more natural to think about things in terms of residuals.

But, how are they different, and when?

Let's take a brief digression into some linear algebra and see if we can figure out what is going on. First of all, let's call the fitted values from the model with intercept $\newcommand{\yhat}{\hat {\mathbf y}}\newcommand{\ytilde}{\tilde {\mathbf y}}\yhat$ and the fitted values from the model without intercept $\ytilde$.

Ok, so how do we make the ratio on the left-hand side small?

Recall that $\newcommand{\P}{\mathbf P}\ytilde = \P_0 \y$ and $\yhat = \P_1 \y$ where $\P_0$ and $\P_1$ are projection matrices corresponding to subspaces $S_0$ and $S_1$ such that $S_0 \subset S_1$.

Here we try to generate an example with an intercept explicitly in the model and which behaves close to the case in the question. Below is some simple R code to demonstrate.

set.seed(.Random.seed[1])

n <- 220
a <- 0.5
b <- 0.5
se <- 0.25 

# Make sure x has a strong mean offset
x <- rnorm(n)/3 + a
y <- a + b*x + se*rnorm(x)

int.lm   <- lm(y~x)
noint.lm <- lm(y~x+0) 

# For comparison to summary(.) output
rsq.int <- cor(y,x)^2
rsq.noint <- 1-mean((y-noint.lm$fit)^2) / mean(y^2)