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I'm trying to simplify this equation...: $$ t~=~\frac{\left( \sum_{i=1}^{n} x_iy_i \right) \left( \sqrt{ n - 1} \right) } {\left( \sqrt{ \left( \sum_{i'=1}^{n} x^2_{i'} \right) \left( \sum_{i=1}^{n} y^2_i \right) - \sum_{i=1}^{n} \left( 2x_iy_i\hat{\beta} + x_i^2\hat{\beta}^2 \right)} \right)} $$ ...to this: $$ t~=~\frac{\left( \sum_{i=1}^{n} x_iy_i \right) \left( \sqrt{ n - 1} \right) } {\sqrt{ \left( \sum_{i'=1}^{n} x^2_{i'} \right) \left( \sum_{i=1}^{n} y^2_i \right) - \left( \sum_{i=1}^{n} x_iy_i \right)^2}} $$ which, as I see it, comes down to simplifying...: $$ \sum_{i=1}^{n} \left( 2x_iy_i\hat{\beta} + x_i^2\hat{\beta}^2 \right) $$ ...to: $$ \left( \sum_{i=1}^{n} x_iy_i \right)^2 $$

I've looked here for multiple summation identities, however the one I can't find (and feel as though is key to answering this problem) is the simplification of: $$ \sum_{i=1}^{n} x_iy_i $$ Is there a simplification? Something along the lines of $ \sum_{i=1}^{n}x_i\sum_{i=1}^{n}y_i $?

Note: Problem Context - I'm attempting to show that the t-statistic for coefficient $\hat{\beta}$ when Y is regressed onto X without an intercept can be re-written as the second equation. The information given in the problem is: $$ \hat{\beta}~=~\frac{\sum_{i=1}^{n} x_iy_i}{\sum_{i'=1}^n x^2_{i'}}~~~~~and~~~~~SE(\beta)~=~\sqrt{\frac{ \sum_{i=1}^{n} \left( y_i - x_i\hat{\beta} \right)^2 }{\left( n - 1 \right) \sum_{i'=1}^{n} x^2_{i'}}}~~~~~and~~~~~t~=~\frac{\hat{\beta}} {SE\left( \hat{\beta} \right)} $$

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First, simplifying the numerator of $\,SE(\hat\beta)\,$ gives the following, where all sums are for $i=1$ to $n\,$:

$$\require{cancel} \begin{align} \sum \left( y_i - x_i\hat{\beta} \right)^2 &= \sum y_i^2 - 2 \hat\beta \sum x_iy_i + \hat\beta^2\sum x_i^2 \\ &= \sum y_i^2 - 2 \frac{\sum x_iy_i}{\sum x_i^2} \sum x_iy_i + \frac{\left(\sum x_iy_i\right)^2}{\left(\sum x_i^2\right)^\bcancel{2}}\bcancel{\sum x_i^2} \\ &= \frac{\left(\sum x_i^2\right)\left(\sum y_i^2\right)-\left(\sum x_iy_i\right)^2}{\sum x_i^2} \end{align} $$

Then:

$$ \begin{align} t~=~\frac{\hat{\beta}} {SE\left( \hat{\beta} \right)} = \frac{\sum x_iy_i}{\cancel{\sum x_i^2}} \cdot \sqrt{\frac{(n-1) \cancel{\sum x_i^2 \cdot \sum x_i^2}}{\left(\sum x_i^2\right)\left(\sum y_i^2\right)-\left(\sum x_iy_i\right)^2}} = \frac{\sqrt{n-1} \sum x_iy_i}{\sqrt{\left(\sum x_i^2\right)\left(\sum y_i^2\right)-\left(\sum x_iy_i\right)^2}} \end{align} $$

The above matches the second expression in the OP.

dxiv
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