# Why report r-squared in Instrumental Variables Estimation?

I mean the the R-squared calculated such as in $R^2=1-\frac{RSS}{TSS}$ when you use the $RSS$ from the original structural model and not recalculation that you should do in order to do an F test. With said $R^2$, you will not have a proper interpretation for the $R^2$ statistic, as I understand. So why report it? I am familiar with Stata reporting it in commands such as ivreg2 and I think other software packages do it too.

On another note, why is it more popular to run the calculations of $R^2$ with $R^2=1-\frac{RSS}{TSS}$ instead of $R^2=\frac{MSS}{TSS}$?

It's true that $R^2$ in instrumental variables regressions is not useful. Since one of the explanatory variables $x$ is correlated with the error $\epsilon$ we can't decompose the variance of the outcome $y$ into $\beta^2 Var(x) + Var(\epsilon)$, so the obtained $R^2$ neither has a natural interpretation nor can it be used for computation of F-tests for joint rejection. Also $R^2$ in instrumental variables regression can be negative and for this point it makes not difference for whether you use $$R^2 = \frac{MSS}{TSS} \quad \text{or} \quad R^2 = 1- \frac{RSS}{TSS}$$ because when $RSS>TSS$, then we also have that $MSS = TSS - RSS < 0$. In general the two expressions are the same so there should be no reason for why one would be more popular than the other. The issue is discussed in more length on the Stata website resources and support FAQs (link).

 to address the additional question in the comment
When you instrument the endogenous variable $x$ with your instrument $z$ as $$x = \alpha + \pi z + \eta$$ you use the predicted values $\widehat{x}$ in the second stage $$y = a + \beta \widehat{x} + \epsilon$$ and if you do this procedure by hand in Stata like

reg x z
predict x_hat, xb
reg y x_hat


the standard errors will be calculated as $y - \widehat{x}\beta$ but these standard errors will be wrong. They are wrong because $\widehat{x}$ is an estimated quantity and not a random variable. The property of these standard errors though is that $RSS < TSS$ and there would be no negative $R^2$ and $\widehat{x}\beta$ is going to be a better predictor of $y$ than $\overline{y}$.

To calculate the corrected standard errors you use the actual values of the endogenous variable $x$ and not its fitted values when computing $e = y − x\beta$. The issue with this is that in this case you are computing the $RSS$ from a different set of regressors than those that are used to actually fit the model from which we take the $TSS$. For this reason it can happen that $x\beta$ is a worse predictor for $y$ than $\overline{y}$.

• The link you posted says that the residuals are calculated in way in which there is no nested constant-only model within the calculations of the 2SLS -- that just means that you run the calculations without the intercept (for the structural equation)? Mar 1 '15 at 16:39
• I don't think that's what it means. The issue is that for the calculation of the standard errors you use the variation of the original endogenous variable $x$ and not the predicted $\widehat{x}$ (which is not a random variable since it's estimated). But the intercept doesn't take this fact into account so it is not the intercept that would minimize the RSS when you use the actual values of $x$ rather than $\widehat{x}$.
– Andy
Mar 1 '15 at 16:47
• I don't think I follow. When you have a negative $R^2$, it means that the $\bar{y}$ is a better predictor than $X\beta$ -- a situation that might occur when you do an estimation without the intercept. If I were to run just a normal OLS (with the intercept) using the equation of the structural model, my $R^2$ could never be negative, as I understand. So if I am calculating an $R^2$ in which $R^2= 1- \frac{RSS}{TSS}$ and $RSS=sum_{i=1}^{n}(y_i-\hat{y_i})^2$ and $TSS=sum_{i=1}^{n}(y_i-\bar{y_i})^2$ (with an intercept) with residuals of the structural equation, how can it exist a negative $R^2$? Mar 2 '15 at 2:25
• @JohnDoe I expanded the answer, I hope it now addresses the additional question you raised in the comments.
– Andy
Mar 2 '15 at 10:37
• "The issue with this is that in this case you are computing the RSS from a different set of regressors than those that are used to actually fit the model from which we take the TSS". See if you follow me: we know that $TSS=sum_{i=1}^{n}(y_i−\bar{y_i})^2$, i.e. it does not change in case of estimating OLS or 2SLS. Now, if we use the residuals of the structural equation, a normal OLS estimation with the intercept, I do not see how can we have a negative $R^2$. I what I hope to understand is what it means to calculate the residuals without a nested constant-only model. Mar 2 '15 at 13:02