The lm function in R retrieves an R^2 value.
The glm function, even if applied to a Gaussian family, does not retrieve an R^2 value.
What is/are the reason/reasons for this?
Thank you!
R-squared Value – Why the GLM Function Does Not Return an R^2 Value
r-squared
Related Solutions
In addition to the numerous (correct) comments by other users pointing out that the $p$-value for $r^2$ is identical to the $p$-value for the global $F$ test, note that you can also get the $p$-value associated with $r^2$ "directly" using the fact that $r^2$ under the null hypothesis is distributed as $\textrm{Beta}(\frac{v_n}{2},\frac{v_d}{2})$, where $v_n$ and $v_d$ are the numerator and denominator degrees of freedom, respectively, for the associated $F$-statistic.
The 3rd bullet point in the Derived from other distributions subsection of the Wikipedia entry on the beta distribution tells us that:
If $X \sim \chi^2(\alpha)$ and $Y \sim \chi^2(\beta)$ are independent, then $\frac{X}{X+Y} \sim \textrm{Beta}(\frac{\alpha}{2}, \frac{\beta}{2})$.
Well, we can write $r^2$ in that $\frac{X}{X+Y}$ form.
Let $SS_Y$ be the total sum of squares for a variable $Y$, $SS_E$ be the sum of squared errors for a regression of $Y$ on some other variables, and $SS_R$ be the "sum of squares reduced," that is, $SS_R=SS_Y-SS_E$. Then $$ r^2=1-\frac{SS_E}{SS_Y}=\frac{SS_Y-SS_E}{SS_Y}=\frac{SS_R}{SS_R+SS_E} $$ And of course, being sums of squares, $SS_R$ and $SS_E$ are both distributed as $\chi^2$ with $v_n$ and $v_d$ degrees of freedom, respectively. Therefore, $$ r^2 \sim \textrm{Beta}(\frac{v_n}{2},\frac{v_d}{2}) $$ (Of course, I didn't show that the two chi-squares are independent. Maybe a commentator can say something about that.)
Demonstration in R (borrowing code from @gung):
set.seed(111)
x = runif(20)
y = 5 + rnorm(20)
cor.test(x,y)
# Pearson's product-moment correlation
#
# data: x and y
# t = 1.151, df = 18, p-value = 0.2648
# alternative hypothesis: true correlation is not equal to 0
# 95 percent confidence interval:
# -0.2043606 0.6312210
# sample estimates:
# cor
# 0.2618393
summary(lm(y~x))
# Call:
# lm(formula = y ~ x)
#
# Residuals:
# Min 1Q Median 3Q Max
# -1.6399 -0.6246 0.1968 0.5168 2.0355
#
# Coefficients:
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) 4.6077 0.4534 10.163 6.96e-09 ***
# x 1.1121 0.9662 1.151 0.265
# ---
# Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
# Residual standard error: 1.061 on 18 degrees of freedom
# Multiple R-squared: 0.06856, Adjusted R-squared: 0.01681
# F-statistic: 1.325 on 1 and 18 DF, p-value: 0.2648
1 - pbeta(0.06856, 1/2, 18/2)
# [1] 0.2647731
McCullagh and Nelder 1989 (page 34) give for the deviance function $D$ for the Poisson distribution:
$$ D = 2 \sum\left(y \log\left(\frac{y}{\mu} \right) + (y-\mu)\right) $$
where y represents your data and $\mu$ your modelled output. I use this function to estimate the explained deviance $ED$ of a GLM with Poisson distribution like this:
$$ ED = 1 - \frac{D}{\text{total deviance}} $$
where total deviance is given by the same equation for $D$ but using the mean of $y$ (a single number, i.e., $\mathrm{mean}(y)$) instead of the array of modelled estimates $\mu$.
I do not know if this is 100% correct, it sounds logical for me and seems to work as you would expect an estimate of the explained deviance to work (it gives you 1 if you use $\mu = y$, etc).
Best Answer
The glm function uses a maximum likelihood estimator (or restricted maximum likelihood). Maximum likelihood does not minimize the squared error (this is called [ordinary] least squares). Sometimes both estimators give the same results (in the linear/ordinary case for normal distributed error terms, see here) but this does not hold in general. Since the coefficient of determination $R^2$ is calculated by ordinary least-squares regression and not by maximum likelihood, there is no reason to display this measure.
PS: Also regard Nick Cox very valid comment below: $R^2$ may be also well-definied and interesting for GLM. My personal experience is that (as so often) some people like/accept it, while others do not.