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I'm working out some multivariable linear regression equations on paper for a class I'm taking, and I'm getting an erroneous factor of N in my solution according to the class. I'm sure it is my error but I can't figure out where.

Starting with these equations:

$ a = \frac{N\sum(xy)-\sum(x)\sum(y)}{N\sum(x^2)-(\sum(x))^2} $ $ b = \frac{\sum(y)\sum(x^2)-\sum(x)\sum(yx)}{N\sum(x^2)-(\sum(x)^2)}$

And using the identities:

$\bar{x} = \frac{1}{N}\sum(x) $

$\bar{xy} = \frac{1}{N}\sum(xy) $

I'm told that we can simplify the expressions by dividing both numberators and denominators by $N^2$.

They should yield these identities:

$ a = \frac{\bar{xy} - \bar{x}\bar{y}}{\bar{x^2}-\bar{x}^2}$ $ b = \frac{\bar{y}\bar{x^2}-{\bar{x}\bar{xy}}}{\bar{x^2}-\bar{x}^2}$

I'm attempting to simplify as suggested by dividing both the numerator and denominator by $N^2$

$ \frac{\frac{N\sum(xy)-\sum(x)*\sum(y)}{N^2}}{\frac{N\sum(x^2)-(\sum(x)^2)/}{N^2}} $

For $a$ I'm getting an extra factor of $N$ in both the numerator and denominator:

$ a = \frac{N\bar{xy}-\bar{x}\bar{y}}{N\bar{x^2}-\bar{x}^2}$

I haven't yet solved b, because I want to get this first. Can someone please confirm or point out the mistake? I've done it on paper twice and gotten the same answer, and can't tell when I go wrong.

Note: I hope my MathJax accurately portrayed all elements, as I haven't written many complex expressions in MathJax before.

EDIT: To show how I got to this solution, I had the following after dividing by $N^2$:

$ a = \frac{\bar{xy}-\bar{x}\bar{y}\frac{1}{N}}{\bar{x^2}-\bar{x}^2\frac{1}{N}} $

In order to get rid of the factor of $\frac{1}{N}$ on both numerator and denominator, I multiplied numerator and denominator by $N$, which yielded my solution above.

Here are screenshots from the class slides that show where we begin, and what we should yield (in case I misinterpreted something): Begin End

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  • $\begingroup$ why you get this expression for a? if you divide $N\sum\limits_{i=1}^N x_iy_i$ by $N^2$ you get $\overline{xy}$, for instance. Similar for $N\sum\limits_{i=1}^N x_i^2$ $\endgroup$ – callculus Jul 5 '18 at 17:18
  • $\begingroup$ @callculus True, which is what I got, but then I end up with an extra factor of $\frac{1}{N}$ on the second term of both the numerator and denominator. In order to get rid of that common factor for the second term of the numerator and denominator, I multiplied both numerator and denominator by $N$, which left me with what I have above. Does that make sense? I'm sorry if it's unclear in my comment. I've edited the question to include that information as well. $\endgroup$ – Hanzy Jul 5 '18 at 17:27
  • $\begingroup$ I´ve posted an answer for $\sum\limits_{i=1}^Nx_i\sum\limits_{i=1}^Ny_i$. I hope it is comprehensible. $\endgroup$ – callculus Jul 5 '18 at 17:35
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There is no extra factor. We have $\sum\limits_{i=1}^Nx_i\sum\limits_{i=1}^Ny_i$. Now we divide it by $N$.

$\underbrace{\frac1N\sum\limits_{i=1}^Nx_i}_{=\overline x}\sum\limits_{i=1}^Ny_i$

$\overline x\sum\limits_{i=1}^Ny_i$

Dividing the term by $N$ again

$\overline x\cdot \underbrace{ \frac1N\cdot \sum\limits_{i=1}^Ny_i}_{\overline y}=\overline x \ \overline y$

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  • $\begingroup$ Thank you, that is where I made my mistake. I combined the summations which is a no-no. I appreciate the help, you got me back on track. $\endgroup$ – Hanzy Jul 5 '18 at 17:38
  • $\begingroup$ @Hanzy You´re welcome. Nice to here that´s all clear now. $\endgroup$ – callculus Jul 5 '18 at 17:39

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