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Does $\sum\frac{y_i}{n}-\sum(\frac{x_i-\bar x}{(x_i-\bar x)^2}\cdot y_i)\sum\frac{x_i}{n}$ reduce to:
$\sum[y_i(\frac{1}{n}-\frac{x_i-\bar x}{(x_i-\bar x)^2}\sum\frac{x_i}{n})]$
Basically, what does $\sum a-\sum b\sum c$ reduce to?
edit: it should be noted I'm trying to rewrite the first expression in terms of $\sum y_ik_i$ for some constant $k$. See comment for original equation.

cap
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  • Where did this formula pop up from? Looks like something from statistics – WaveX Feb 12 '20 at 00:53
  • Simple Linear Regression. More specifically, formula comes from: $\hat\beta_0=\bar y-\hat\beta_1\bar x$ – cap Feb 12 '20 at 00:56

1 Answers1

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Yes it does.

Let $a=y_{i}, b=\frac{1}{n}, c=\frac{x_i-\bar x}{(x_i-\bar x)^2}, d=\frac{_{}}{n}$.

Then you have:

$\sum\frac{y_i}{n}-\sum(\frac{x_i-\bar x}{(x_i-\bar x)^2}\cdot y_i)\sum\frac{x_i}{n} =$

$\sum ab - \sum ca \sum d$ =

$\sum ab - \sum ac \sum d$ =

$\sum ab - \sum a(c \sum d)$ =

$\sum a(b - c \sum d)$ =

$\sum[y_i(\frac{1}{n}-\frac{x_i-\bar x}{(x_i-\bar x)^2}\sum\frac{x_i}{n})]$

Mosquite
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