# Proof for “The sum of the observed values $Y_i$ equals the sum of the estimated / fitted values $\hat Y_i$” [duplicate]

I needed some help trying to understand why the sum of the observed values $$Y_i$$ equals the sum of the estimated values $$\hat{Y}_i$$.

• In linear least squares regression the sum of estimated does not equal the sum of observed values when there is no intercept term. E.g. for $y_i = a x_i + \epsilon_i$ with $y_i = \lbrace -1,1 \rbrace$ and $x_i = \lbrace 0, 1 \rbrace$ the estimate will be $\hat {a} =1$ with $\sum y_i = 0 \neq 1 =\sum \hat {y}_i$ – Sextus Empiricus Oct 9 '19 at 17:37

$$Y_i = \hat{Y}_i + \hat{\epsilon_i}$$ by definition.
Also, we know that $$\frac{1}{n}\sum_{i=1}^{n}{\hat{\epsilon}_i}=0$$ because the intercept of the model absorbs the mean of the residuals.
So, $$\frac{1}{n}\sum_{i=1}^{n}{Y_i} = \frac{1}{n}\sum_{i=1}^{n}{\hat{Y}_i} + \frac{1}{n}\sum_{i=1}^{n}{\hat{\epsilon_i}} = \frac{1}{n}\sum_{i=1}^{n}{\hat{Y}_i} + 0 = \frac{1}{n}\sum_{i=1}^{n}{\hat{Y}_i}$$
Let $$P$$ be the projection matrix on $$X$$, where one of the columns of $$X$$ is $$\mathbf{1}$$, where $$\mathbf{1}$$ is vector of ones, then \begin{align} \mathbf{1}'\hat{y}&=\mathbf{1}'Py\\ &=\left(P'\mathbf{1}\right)'y\\ &=\mathbf{1}'y. \end{align}
Minimizing the sum of squared residuals $$S=\sum_{r=1}^{n}(y_{r}-a-b x_{r})^{2}$$ w.r.t. $$a$$ and $$b$$ gives the "normal equations" (first order conditions) $$\sum_{r=1}^{n}y_{r}=na+b\sum_{r=1}^{n}x_{r}\quad\text{and}\quad\sum_{r=1}^{n}y_{r}x_{r}=a\sum_{r=1}^{n}x_{r}+b\sum_{r=1}^{n}x_{r}^{2}.$$ The fitted values are $$\hat{y}_{r}=a+bx_{r}$$, with residuals $$\hat{u}_{r}=y_{r}-(a+bx_{r})$$ Consider $$\sum_{r=1}^{n}\hat{u}_{r}=\sum_{r=1}^{n}y_{r}-\sum_{r=1}^{n}\hat{y}_{r}=\sum_{r=1}^{n}y_{r}-\sum_{r=1}^{n}(a+bx_{r})=\sum_{r=1}^{n}y_{r}-na-b\sum_{r=1}^{n}x_{r}.$$ Now, substitute $$\sum_{r=1}^{n}y_{r}$$ for the first normal equation.