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Above is the capture from the Econometrics slide from "Hill,Griths and Lim (2018) Principles of Econometrics".

I have no problem the see the blue font sentence separately, it's all about the numerator and denominator issues. But, if I think deeply about its intuitions, it bugs me.

The first blue sentence could interpret as more coverage of $x_i$ can reduce the variance

then I expand the equation, larger $\sum(x_i-\bar{x})^2=\sum x_i^2-n\bar{x}^2$ means Larger the value of$\sum x_i^2$. It's reduce variance, and this contradict to second blue sentence Clearly second blue is right, so what's go wrong with my reasoning about the larger $\sum x_i^2$ .

Please enlighten me, Thank you so much

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    $\begingroup$ Welcome to economics.se, please next time do not include picture of texts or equations you should rewrite them so the whole question is searchable $\endgroup$
    – 1muflon1
    Oct 11 at 13:27
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You should be able to confirm the claim after dividing the numerator and denominator by $\sum x_i^2$. Then, ceteris paribus, a greater $\sum x_i^2$ increases the denominator and reduces the variance of the random variable $b_1$.

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I don't like the blue sentences. More rigorously, it should state that, if denominator is fixed, an increase in numerator gives a larger variance.

That is, if $\sum_{i=1}^{n}x_{i}^{2}$ increases, meanwhile, $\bar{x}$ increases too so that the denominator remains the same, then variance becomes smaller. Without any statement about the denominator (more precisely, about $\bar{x}$), you don't know whether the variance becomes larger or smaller. What if $\bar{x}=0$, will increase in $\sum_{i=1}^{n}x_{i}^{2}$ affect the variance?

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