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I have a regression model in which the intercept is 0 and the slope is 3. Now I multiply the dependent variable by 10 and independent by 2. I wish to find out what happens to the intercept and slope. I ran a calculation and found out that the slope is now 15, but I am not sure how to prove it. If I could understand why the intercept is without change, I could prove it easily.

$10y=a+b\cdot 2x$

$30x=a+2xb$

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  • $\begingroup$ Drawing yourself a picture may help. You could also use the formulas for slope and intercept, see stats.stackexchange.com/questions/199575/… $\endgroup$
    – Frank
    May 2, 2019 at 19:35
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    $\begingroup$ I am aware of the formulas and of the fact that both averages will be multiplied by the same constant, but how do I tackle the sum of xy, for instance ? $\endgroup$ May 2, 2019 at 19:37
  • $\begingroup$ $\sum_i (c_1 x_i) (c_2 y_i) = ?$ $\endgroup$
    – Frank
    May 2, 2019 at 19:46

1 Answer 1

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For a classic linear regression model, $y=bx+a+\epsilon$, we have the following formulas: $$b=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}, \ \ \ a=\frac{\sum y \sum x^2-\sum x\sum xy}{n\sum x^2-(\sum x)^2}$$

Now, if we change the IV and DV as : $x\rightarrow cx, y\rightarrow dy$ $$b_{new}=\frac{ncd\sum xy-cd\sum x\sum y}{nc^2\sum x^2-c^2(\sum x)^2}=b\frac{d}{c}, \ \ a_{new}=\frac{c^2d\sum y \sum x^2-c^2d\sum x\sum xy}{nc^2\sum x^2-c^2(\sum x)^2}=ad$$

Intercept normally changes, but since it was $0$, new coefficients don't have any effects.

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  • $\begingroup$ This follows more directly from just focusing on the units and dimensions of the systematic part of the model. So the first part of the argument is to note that the intercept has the same units as $y$ and that the slope has units of $y$ / units of $x$. The second part of the argument is then easy to fill in. Tacitly you are assuming that errors have mean zero so variability around the line is immaterial. $\endgroup$
    – Nick Cox
    May 2, 2019 at 20:33
  • $\begingroup$ @NickCox ; I was just going to add it as an alternate solution (which doesn't narrow down the concept to OLS), but why don't you add it as another answer? $\endgroup$
    – gunes
    May 4, 2019 at 6:33
  • $\begingroup$ Feel free to build the idea into yours. $\endgroup$
    – Nick Cox
    May 4, 2019 at 6:36

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