# Constant term in linear regresion

Can someone give a mathematical proof as to why including a constant in a linear regression equivalent is to running a regression with demeaned data and zero constant?

More specifically, consider the linear regression $$Y = b_0 + b_1 X_1 + b _2 X_2 + ... b_k X_k + e$$ where $X$'s and $Y$ are vectors, and the same regression with demeaned regressors $$\bar{Y} = b_1 \bar{X}_1 + b _2 \bar{X}_2 + ... b_k \bar{X}_k + e,$$ where the $\bar{Y}$ and the $\bar{X}_i$ are demeaned. It turns out that you'll obtain the same coefficients $\{b_i\}_{1 \leq i \leq k}$ in both regressions, but I struggle to prove this.

Basically, if you think of your fitting procedure as following a method of moments then you will have equivalence. Take our model as $$Y = \alpha + \beta X + u$$
Say our model coefficients were chosen such that $E(u) \neq 0$, then the presumed symmetry of the distribution of $u$ would mean a "better" model is available by adding a further constant to $\alpha$. Therefore, we have a primary statistical condition that the first moment of $u$ be zero, and therefore
$$E(Y)=\alpha+ \beta E(X)$$
• Right, in OLS the error term $e$ has an expected value of 0. So if I think of the $X$ and the $Y$'s as random variables, then $E(Y) = b_0 + b_1 E(X_1) + \cdots + b_k E(X_k) + e$. Subtracting this from the original equation gives me the desired result. – elliptics Apr 26 '14 at 16:07