Can I ignore the negative R-squared value when I am using instrumental variable regression?

I am running an instrumental variable regression using 'ivreg' command in R program.

I find that all my validity tests related to endogeneity are satisfied only except the R-squared value which is negative.

May I know whether I can ignore this negative R-squared value without reporting?

If not, what is an alternative manner to resolve this issue? The code is as below:

    > Y_ivreg=ivreg(Y~x1+x2+x3+x4+x5+x6+x7|x2+x8+x9+x10+x5+x6+x7,data=DATA)
> summary(Y_ivreg,diagnostics=TRUE)

Call:
ivreg(formula = Y ~ x1 + x2 + x3 + x4 + x5 +
x6 + x7 | x2 + x8 + x9 + x10 +
x5 + x6 + x7, data = DATA)

Residuals:
Min        1Q    Median        3Q       Max
-0.747485 -0.053721 -0.009349  0.044285  1.085256

Coefficients:
Estimate  Std. Error  t value Pr(>|t|)
(Intercept)  0.0979178  0.0319244   3.067  0.00218 **
x1        0.0008438  0.0004927   1.712  0.08691 .
x2        0.0018515  0.0009135   2.027  0.04277 *
x3       -0.0130133  0.0073484  -1.771  0.07667 .
x4       -0.0018486  0.0009552  -1.935  0.05303 .
x5       -0.0000294  0.0000126  -2.333  0.01971 *
x6        0.0018214  0.0008908   2.045  0.04096 *
x7       -0.0024457  0.0005488  -4.456 8.61e-06 ***

Diagnostic tests:
df1  df2 statistic p-value
Weak instruments (x1)    3 3313   185.440  <2e-16 ***
Weak instruments (x3)    3 3313  3861.526  <2e-16 ***
Weak instruments (x4)    3 3313  3126.315  <2e-16 ***
Wu-Hausman               3 3310     1.943   0.121
Sargan                   0   NA        NA      NA
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.1142 on 3313 degrees of freedom
Multiple R-Squared: -0.009029,  Adjusted R-squared: -0.01116
Wald test: 4.231 on 7 and 3313 DF,  p-value: 0.0001168


There is a Stata post link related to this issue and ivregression for your reference: https://www.stata.com/support/faqs/statistics/two-stage-least-squares/

• Are the changes in Y very small? Your coefficients and residuals are all tiny, except lag(Y, 1) Commented Aug 4, 2019 at 0:16
• @Henry: Yes it is.
– Eric
Commented Aug 4, 2019 at 7:26
• Then your $R^2$ is not $1$, just so close to $1$ that you cannot see the difference. You might investigate how much better your model is than just predicting that Y is equal to lag(Y, 1) Commented Aug 4, 2019 at 9:32
• @Roland: My question is more general. Whether it is a requirement to report R-squared values or not in iv regression. If so, why. If not, why. That's it.
– Eric
Commented Aug 6, 2019 at 9:27
• I changed my entire question to avoid any uncertainty.
– Eric
Commented Aug 6, 2019 at 11:57

$$R^2$$ really has no statistical meaning in the context of 2SLS/IV.

How can $$R^2$$ be negative?

Wikipedia has a great visualization of $$R^2$$:

On the left, we see the $$\color{red}{\text{total sum of squares}}$$, obtained by using the mean ($$\bar{y}$$) as a prediction:

$${\text{total sum of squares}} = \sum_{i = 1}^n (y_i - \bar{y})^2$$

On the right, we see the $$\color{blue}{\text{residual sum of squares}}$$, obtained by using the model's predictions ($$\hat{y}$$):

$${\text{residual sum of squares}} = \sum_{i = 1}^n (y_i - \hat{y})^2 = \sum_{i = 1}^n \bigg(y_i - \Big( \hat{\beta}_0 + \sum_{j = 1}^p \hat{\beta}_j \cdot x_j \Big) \bigg)^2$$

Ordinarily, $$R^2 = 1 - \frac{\color{blue}{\text{residual sum of squares}}}{\color{red}{\text{total sum of squares}}} \geq 0$$, because any model with an intercept ($$\beta_0$$) should perform at least as well as the image on the left (the intercept could simply be the mean).

However, if you interpret instrumental variable regression as a two-stage linear regression, it is easy to show why it could end up being negative. Namely, suppose the endogenous variables ($$\mathbf{X}$$) are regressed on the exogenous variables ($$\mathbf{Z}$$), and the predicted values ($$\hat{\mathbf{X}}$$) are then used as covariates in the second stage:

$$\text{Stage 1:} \quad \mathbf{X} = \mathbf{Z\delta} + \text{error} \\ \text{Stage 2:} \quad \mathbf{y} = \hat{\mathbf{X}}\mathbf{\beta} + \text{error}$$

Since $$\hat{\mathbf{X}} \neq \mathbf{X}$$, the error that is minimized in the second stage is not the same as the error used to calculate the residual sum of squares. Consequently, the residual sum of squares need not be less than the total sum of squares anymore. (And more importantly, the $$R^2$$ has become meaningless.)