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If $\textbf{X}\in\textbf{R}^{n\times p}$ has full rank and $\textbf{Y}\in\textbf{R}^{n\times 1}$, prove that \begin{align*} \sum_{i=1}^{n}(Y_{i} - \hat{Y}_{i}) = 0 \end{align*}

where $\hat{\textbf{Y}}$ is the fitted value of the linear model $\textbf{Y} = \textbf{X}\beta + \textbf{e}$, $\textbf{e}\sim\mathcal{N}_{n}(\textbf{0},\sigma^{2}\textbf{I}_{n})$. Precisely speaking, $\hat{\textbf{Y}} = \textbf{X}\hat{\beta}$ such that $\textbf{X}^{\prime}\textbf{X}\hat{\beta} = \textbf{X}^{\prime}\textbf{Y}$. Can someone help me? Thanks!

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closed as off-topic by Michael Chernick, kjetil b halvorsen, Peter Flom Mar 31 at 12:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

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  • $\begingroup$ Hint: the first column of $\mathbf{X}$ is all $1$'s $\endgroup$ – Francis Mar 31 at 4:11
  • $\begingroup$ Sorry, I still did not grasp the idea. Can you provide a full answer? $\endgroup$ – user1337 Mar 31 at 4:40
  • $\begingroup$ Possible duplicate of Do these residuals sum to zero? $\endgroup$ – kjetil b halvorsen Mar 31 at 12:43
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$\newcommand{\X}{\mathbf{X}}\newcommand{\Y}{\mathbf{Y}}\newcommand{\bhat}{\hat{\beta}}\newcommand{\yhat}{\hat{\Y}}\newcommand{\0}{\mathbf{0}}$We know that $$\X'\Y =\X'\X\bhat,$$ so rearranging this, we have $$\X'\left(\Y - \X\bhat\right) = \0,\quad \text{i.e.}\quad \X'\left(\Y - \yhat\right) =\0.$$ Looking at the first component of both sides, we get (since the top row of $\X'$ is all $1$'s) using the definition of matrix multiplication that $$\sum\limits_{i=1}^{n}1\cdot \left(Y_i - \hat{Y}_i\right)=0,$$ which is the desired result.

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