# Will excluding the intercept affect other variables' coefficients?

Is there any explanation if running without intercept caused the change in coefficients of other variables ?

I am talking about a panel data regression.

xtreg y x

and

xtreg y x, nocons

For simplicity, consider the case of one covariate. We have two regressions, the first with and the second without an intercept. \begin{align*} &y_i = \alpha + \beta x_{i} + \varepsilon_i, \tag{1}\\ &y_i = \gamma x_i + \delta_i \tag{2} \end{align*} For the first regression, you will estimate the coefficients by setting the following sample moment conditions: \begin{align*} &\frac{1}{n} \sum_i \varepsilon_i = 0,\\ &\frac{1}{n} \sum_i \varepsilon_i x_i = 0 \end{align*} So: \begin{align*} &\frac{1}{n} \sum_i y_i = \hat \alpha + \hat \beta \left(\frac{1}{n} \sum_i x_i\right),\\ &\frac{1}{n} \sum_i y_ix_i = \hat \alpha \left(\frac{1}{n} \sum_i x_i\right) + \hat \beta \left(\frac{1}{n} \sum_i (x_i)^2\right) \end{align*} Then: \begin{align*} &\hat \alpha = \left(\frac{1}{n} \sum_i y_i\right) - \hat \beta \left(\frac{1}{n} \sum_i x_i\right),\\ &\frac{1}{n} \sum_i y_i x_i = \left(\frac{1}{n} \sum_i y_i\right) \left( \frac{1}{n} \sum_i x_i\right) - \hat \beta \left(\frac{1}{n} \sum_i x_i\right)^2 + \hat \beta \left(\frac{1}{n} \sum_i (x_i)^2\right), \end{align*} So using $$\widehat{cov}(y_i, x_i) = \left(\frac{1}{n} \sum_i y_i x_i\right) - \left(\frac{1}{n} \sum_i y_i\right) \left( \frac{1}{n} \sum_i x_i\right)$$ and $$\widehat{var}(x_i) = \left(\frac{1}{n} \sum_i (x_i)^2\right) - \left(\frac{1}{n} \sum_i x_i\right)^2$$: $$\hat \beta = \frac{\widehat{cov}(y_i, x_i)}{\widehat{var}(x_i)}. \tag{3}$$ For the regression $$(2)$$ without intercept, we only have the second moment condition: \begin{align*} &\frac{1}{n} \sum_i \delta_i x_i = 0,\\ \to &\frac{1}{n} \sum_i y_i x_i = \hat \gamma \left(\frac{1}{n} \sum_i (x_i)^2\right) \end{align*} So: $$\hat \gamma = \frac{\frac{1}{n} \sum_i y_i x_i}{\frac{1}{n} \sum_i (x_i)^2} \tag{4}$$ Comparing $$(3)$$ and $$(4)$$, we see that they are the same if the mean of $$x$$ is zero: $$\frac{1}{n} \sum_i x_i = 0$$.
The figure below demonstrates. We have 5 data points with $$\frac{1}{n} \sum_i x_i = 0$$. We see that the two regression lines, the black with intercept and the red without intercept have the same slope.
• @Henry Indeed, the intercept will be biased (as you fix it to zero). However, the slope will still be unbiased. The whole derivation is in fact a special case of the Firsh-Waugh-Lovell theorem. First de-mean the covariates by regressing on the constant term and taking the residuals. Then regress $y$ on these de-meaned variables. – tdm Jun 17 at 8:49
• Hi @tdm, a novice question, I can explain and rpvoe that (3) equalling to (4) based on this condition. However, can I ask why for the (1) regression ,For the first regression, I can estimate the coefficients by setting the following sample moment conditions: \begin{align*} &\frac{1}{n} \sum_i \varepsilon_i = 0,\\ &\frac{1}{n} \sum_i \varepsilon_i x_i = 0 \end{align*} – BeautifulMindset Jun 17 at 23:20
• @BeautifulMindset One way to see this is to notice that OLS minimizes $\frac{1}{n} \sum_i (y_i - \alpha - \beta x)^2$ with respect to $\alpha$ and $\beta$. This gives the conditions $\frac{1}{n} \sum_i (y_i - \alpha - \beta x) = 0$ and $\frac{1}{n} \sum_i (y_i - \alpha - \beta x) x = 0$. Subsituting back $\varepsilon = (y - \alpha - \beta x)$ gives you the two moment conditions. Alternatively, you can also see the conditions as the method of moments estimator (see for example here). – tdm Jun 18 at 5:31
While the previous answer was very instructive, the correct answer for Stata is that xtreg y x, nocons will result in error r(198) option noconstant not allowed. I don't know why Stata programmers did not allow noconstant option for random effects model, but that's a different question.