<p><span class="math-container">$\sum_{i=1}^n(X_i -\bar{X})(Y_i -\bar{Y}) =\sum_{i=1}^n(X_i -\bar{X})(Y_i) -\sum_{i=1}^n(X_i -\bar{X})\bar{Y}  $</span></p>
<p>The key is that <span class="math-container">$\bar{Y}$</span> is a constant, so we can bring it in front of the summation.</p>
<p><span class="math-container">$=\sum_{i=1}^n(X_i -\bar{X})(Y_i) -\bar{Y}\sum_{i=1}^n(X_i -\bar{X})  $</span></p>
<p>Then apply, <span class="math-container">$\sum_{i=1}^n(X_i -\bar{X})=0$</span>.</p>
<p>It's an analogous argument for the other equality.</p>

CC BY-SA 4.0
2021-08-16T17:12:48.443Z
2021-08-16T22:37:48.980Z
2021-08-16T22:37:48.980Z
Michael Gmeiner
<p>I was reading a book on econometrics and I found this algebraic transformation and I don't know how I can get there.I would appreciate some help.<br />
<span class="math-container">$\sum_{i=1}^n (X_i-\bar{X})\cdot (Y_i-\bar{Y})=\sum_{i=1}^n (X_i-\bar{X})\cdot Y_i=\sum_{i=1}^n X_i\cdot (Y_i-\bar{Y}) $</span></p>

<econometrics><mathematical-economics>
How can I prove this?
MatiasNougues
# How can I prove this?

(Asked by: username_0 on 2021-08-16)

I was reading a book on econometrics and I found this algebraic transformation and I don't know how I can get there.I would appreciate some help.  
$\sum\_{i=1}^n (X\_i-\bar{X})\cdot (Y\_i-\bar{Y})=\sum\_{i=1}^n (X\_i-\bar{X})\cdot Y\_i=\sum\_{i=1}^n X\_i\cdot (Y\_i-\bar{Y}) $

**Tags:** econometrics, mathematical-economics


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## 1 answers

### Answer by username_1 on 2021-08-16. Score: 7

$\sum\_{i=1}^n(X\_i -\bar{X})(Y\_i -\bar{Y}) =\sum\_{i=1}^n(X\_i -\bar{X})(Y\_i) -\sum\_{i=1}^n(X\_i -\bar{X})\bar{Y} $

The key is that $\bar{Y}$ is a constant, so we can bring it in front of the summation.

$=\sum\_{i=1}^n(X\_i -\bar{X})(Y\_i) -\bar{Y}\sum\_{i=1}^n(X\_i -\bar{X}) $

Then apply, $\sum\_{i=1}^n(X\_i -\bar{X})=0$.

It's an analogous argument for the other equality.


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