1
$\begingroup$

What happened to $x^2$

[1]: http://i.stack.imgur.com/HfhdW.gif

My question is what happened to the yellow part of the equation, why does it disappear in the new line. I understand that if you take the square root of $x^2$ then it should be $x,$ but in the second line of equation we only have the sum of $xy$ with square-root of $n-1.$ Please provide a dummy explanation. Also if you have time, please explain the red part of the equation?

Please be aware there is a picture attached. Please click on "What happened to $x^2$".

$\endgroup$
  • 1
    $\begingroup$ The answer to the question in the title is no, the square root of $x^2$ is not $X$. Capital $X$ and $x$ are different things. $\endgroup$ – Ross Millikan Sep 1 '16 at 23:29
  • $\begingroup$ It's easier than it looks. Let $X =\sum x_i^2; S = \sum_x_iy_i $ and $V = \sum (y_i-x_i\beta)^2$. Then this statement is: $\frac SX\sqrt {\frac {(n-1)X}{V}}=\frac {\sqrt {n-1}S}{\sqrt {XV}} $. That's not surprising. $\endgroup$ – fleablood Sep 1 '16 at 23:54
3
$\begingroup$

$$\frac{\sum x_iy_i}{\sum x_i^2}\sqrt{\frac{(n-1)\sum x_i^2}{\sum(y_i-x_i \beta )^2}} = (\sum x_iy_i)\sqrt{\frac{(n-1)\sum x_i^2}{(\sum x_i^2)^2\sum(y_i-x_i \beta )^2}} = (\sum x_iy_i)\frac{\sqrt{(n-1)}}{\sqrt{(\sum x_i^2)\sum(y_i-x_i \beta )^2}}$$

The sum in the denominator was brought under the square root, then appropriate terms were cancelled in the fraction.

$\endgroup$
  • $\begingroup$ it says Math Processing Error, please edit $\endgroup$ – Boro Dega Sep 1 '16 at 23:34
  • $\begingroup$ @Boro Dega I'm not exactly sure how to handle this; it comes up fine on my screen. Is anyone else getting this error? $\endgroup$ – Christian Sep 1 '16 at 23:35
  • $\begingroup$ @Christian: I'm not getting an error. $\endgroup$ – Will R Sep 1 '16 at 23:38
  • $\begingroup$ @Christian I believe it is a browser thing. Works fine with Firefox, but not in Chrome:) $\endgroup$ – Boro Dega Sep 1 '16 at 23:42
  • $\begingroup$ @Christian Thanks for a very simplistic answer. $\endgroup$ – Boro Dega Sep 1 '16 at 23:49
5
$\begingroup$

Here's what happened: $$\frac{A}{C}\sqrt{\frac{B\cdot C\strut}{D}}=\frac{A}{C}\cdot\frac{\sqrt{B\strut}\cdot \sqrt{C\strut}}{\sqrt{D\strut}}=\frac{\sqrt{B\strut}\cdot A}{\sqrt{C\strut}\cdot \sqrt{D\strut}}=\frac{\sqrt{B\strut}\cdot A}{\sqrt{C\cdot D\strut}}$$ where $$\textstyle A=\sum x_iy_i\qquad B=n-1 \qquad C=\sum x_i^2\qquad D=\sum (y_i-x_i\beta)^2$$ Note that these steps are permitted and make sense since $A$, $B$, $C$, and $D$ are all non-negative.

$\endgroup$
  • $\begingroup$ That is indeed what happened; what appears to be missing from this answer is why. $\endgroup$ – Will R Sep 1 '16 at 23:37
  • $\begingroup$ @WillR: The point of my answer is that it may be clearer to see that what happened is nothing more than a simple manipulation with some judicious choices of what to hide underneath a variable name. However, I can add some more steps I suppose. $\endgroup$ – Zev Chonoles Sep 1 '16 at 23:39
  • $\begingroup$ "Why?"? What do you mean why? Because. That's why. It's just math. Square roots and fractions and positive real numbers. $\endgroup$ – fleablood Sep 1 '16 at 23:59
3
$\begingroup$

In the equation you show we can define $\sum x_i^2=k$ The upper equation has $\sqrt k$ in the numerator and $k$ in the denominator. The lower expression, whic should be equal to $t$ as well, has $\sqrt k$ in the denominator. As long as $k \gt 0, \frac {\sqrt k}k=\frac 1{\sqrt k}$ It was just brought into the square root sign.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.