# Regression line on origin

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1. May 23, 2017

### Faiq

1. The problem statement, all variables and given/known data

A random sample of size $n$ from a bivariate distribution is denoted by $(x_r,y_r), r=1,2,3,...,n$. Show that if the regression line of $y$ on $x$ passes through the origin of its scatter diagram then

$$\bar y \sum^n_{r=1} x_r^2=\bar x\sum^n_{r=1} x_r y_r$$ where $(\bar x,\bar y)$ is the mean point of the sample.

I don't really know how to begin. I am aware the line equation is $$b=\frac{y}{x}=\frac{\sum xy-\frac{\sum x\sum y}{n}}{\sum x^2-\frac{(\sum x)^2}{n}}$$

Not sure what to do next.

Last edited by a moderator: May 24, 2017
2. May 23, 2017

### Ray Vickson

(1) Do not write $\xbar$, write $\bar{x}$. Right-click on the formula to see its TeX commands.
Mod note: Fixed the TeX in the original post and above.
(2) What can you say about the data if the least-squares line has intercept $a$ equal to zero?

Last edited by a moderator: May 24, 2017
3. May 23, 2017

### Faiq

Y is proportional to X

4. May 23, 2017

### Ray Vickson

That answer is not useful. Take the formula for $a$, in terms of $(x_i,y_i)$, and set it to zero. What do you get?

5. May 24, 2017

### SammyS

Staff Emeritus
I suppose the that the linear model you are working with is: $\ \displaystyle y=a+bx\$.

You have the correct expression for finding the linear coefficient, $\ b\,.\$( Leave out the $\ \displaystyle \frac yx\$ ).

It seems to me that you must also consider the expression for $\ a\,.\$ Then show that if $\ a=0\,,\$ then you obtain the desired result:
$\displaystyle \bar y \sum^n_{r=1} x_r^2=\bar x\sum^n_{r=1} x_r y_r$​
.

Last edited: May 25, 2017