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Suppose that $\hat\beta$ is the OLS estimator of the multiple linear regression given by $$ \hat\beta =\underset{\beta}{\operatorname{arg\,min}}\Bigl\{\sum_{i=1}^n(y_i-\beta_0-\sum_{j=1}^px_{ij}\beta_j)^2\Bigr\} $$ and $\hat\beta^c$ is the OLS estimator of the multiple linear regression when the variables are centered, i.e. $$ \hat\beta^c =\underset{\beta^c}{\operatorname{arg\,min}}\Bigl\{\sum_{i=1}^n((y_i-\bar y)-\beta_0^c-\sum_{j=1}^p(x_{ij}-\bar x_j)\beta_j^c)^2\Bigr\}. $$ It seems that $\hat\beta_0^c$ should be equal to $0$. We have that \begin{align*} \hat\beta &=\underset{\beta}{\operatorname{arg\,min}}\Bigl\{\sum_{i=1}^n(y_i-\beta_0-\sum_{j=1}^px_{ij}\beta_j)^2\Bigr\}\\ &=\underset{\beta}{\operatorname{arg\,min}}\Bigl\{\sum_{i=1}^n((y_i-\bar y+\bar y)-\beta_0-\sum_{j=1}^p(x_{ij}-\bar x_j+\bar x_j)\beta_j)^2\Bigr\}\\ &=\underset{\beta}{\operatorname{arg\,min}}\Bigl\{\sum_{i=1}^n((y_i-\bar y)+\underbrace{\bar y-\beta_0-\sum_{j=1}^p\bar x_j\beta_j}_{=\beta_0^c}-\sum_{j=1}^p(x_{ij}-\bar x_j)\underbrace{\beta_j}_{=\beta_j^c})^2\Bigr\}. \end{align*}

It is straightforward to see that $\hat\beta_0^c=0$ in the simple linear regression. We have that $$ \bar y-\hat\beta_0-\bar x\hat\beta_1=\bar y-\bar y+\bar x\hat\beta_1-\bar x\hat\beta_1=0. $$

How can we show that this also holds for the multiple linear regression, i.e. how can we show that $\hat\beta_0^c$ is equal to $0$?

Any help is much appreciated!

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2 Answers 2

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Consider the partitioned regression: $y_i=X_{i1}^\top\beta_1+X_{i2}^\top\beta_2+\epsilon_i$. It is easy to show that $$ \hat{\beta}_1=(\mathbf{X}_1^\top\mathbf{X}_1)^{-1}\mathbf{X}_1^{\top}(\mathbf{y}-\mathbf{X}_2\hat{\beta}_2), $$ where $\mathbf{y}\equiv[y_1,\ldots,y_n]^\top$ and $\mathbf{X}_j\equiv[X_{ij},\ldots,X_{nj}]^{\top}$. If $X_{i1}=1$, the last equation reduces to $$ \hat{\beta}_1=\mathbf{i}^\top(\mathbf{y}-\mathbf{X}_2\hat{\beta}_2)/n. $$ When $y_i$ and $X_{i2}$ are measured as deviations from the corresponding sample means, $\mathbf{i}^\top\mathbf{y}=0$ and $\mathbf{i}^\top\mathbf{X}_2=\mathbf{0}$, which yields $\hat{\beta}_1=0$.

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Consider the following multiple regression: $$y_t^*=\beta_0+\beta_1x_{1t}^*+...+\beta_nx_{nt}^*+\epsilon_t,$$ where $y_t^*=y_t-\bar{y}; x_{it}^*=x_{it}-\bar{x_i}$ for $i=1,...,n$. By applting OLS we have: $$\sum_{t=1}^k(y_t^*-\beta_0-\beta_1x_{1t}^*-...-\beta_nx_{nt}^*)^2 \rightarrow min.$$ Since we are interested in determining $\beta_0$, we have $$\beta_0=\frac{1}{n}\sum_{t=1}^k(y_t^*-\beta_1x_{1t}^*-...-\beta_nx_{nt}^*)=\frac{1}{n}\sum_{t=1}^{k}(y_{t}-\bar{y})-\frac{\beta_1}{n}\sum_{t=1}^{k}(x_{1t}-\bar{x_1})-...\frac{\beta_n}{n}\sum_{t=1}^{k}(x_{nt}-\bar{x_n})=0$$ The above equality holds since $\frac{1}{n}\sum_{t=1}^{k}(y_{t}-\bar{y})=\bar{y}-\bar{y}=0;\frac{\beta_i}{n}\sum_{t=1}^{k}(x_{it}-\bar{x_i})=\beta_i(\bar{x_i}-\bar{x_i})=0$ for all $i=1,...,n$.

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