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Linear regression help

Help for my brother.

Hi, currently I'm revising the first statistics module and seem to be having a few problems. I'm a first year conomics student if that help.

stats.png

I was wondering, firstly what exactly the constancy of β is. And secondly why you do each step which leads to B-bar(xy)/B-bar(x^2) from xiyi = x^2iβ + xiεi

Thanks
(edited 9 years ago)
Original post by Liamnut
Help for my brother.

Hi, currently I'm revising the first statistics module and seem to be having a few problems. I'm a first year conomics student if that help.

stats.png

I was wondering, firstly what exactly the constancy of β is. And secondly why you do each step which leads to B-bar(xy)/B-bar(x^2) from xiyi = x^2iβ + xiεi

Thanks

"Constancy of b" means "because b is constant".
First line is by expanding xiyix_i y_i as given at the very top of the screenshot. Second line is by rearranging the first line for the constant β\beta. Third line is by multiplying the top and bottom of each term in the second line by 1N\frac{1}{N}. Fourth line is assuming x,ϵx, \epsilon are uncorrelated, and estimating ∑(xiϵi)=0\sum(x_i \epsilon_i) = 0 because the errors "on average" cancel out.
(edited 9 years ago)
Reply 2
Avatar for Liamnut
Liamnut
OP
13
Original post by Smaug123
"Constancy of b" means "because b is constant".
First line is by expanding xiyix_i y_i as given at the very top of the screenshot. Second line is by rearranging the first line for the constant β\beta. Third line is by multiplying the top and bottom of each term in the second line by 1N\frac{1}{N}. Fourth line is assuming x,ϵx, \epsilon are uncorrelated, and estimating ∑(xiϵi)=0\sum(x_i \epsilon_i) = 0 because the errors "on average" cancel out.


Why do you multiply by 1/N? Is it to get E?
Original post by Liamnut
Why do you multiply by 1/N? Is it to get E?

Yes. (You could multiply by anything, and it would still work - it's just that multiplying by 1/N gives us something we recognise.)
Reply 4
Avatar for Liamnut
Liamnut
OP
13
Original post by Smaug123
Yes. (You could multiply by anything, and it would still work - it's just that multiplying by 1/N gives us something we recognise.)


Thanks. It's just multiplying by 1 as 1/n divided by 1/n I guess.
Original post by Liamnut
Thanks. It's just multiplying by 1 as 1/n divided by 1/n I guess.

Exactly.

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