# Effect on beta one by changing x variables

• May 15th 2011, 02:58 AM
anonymous_maths
Effect on beta one by changing x variables
Hi,

I had to do a linear regression of y on x, where the x variables are 1851 to 1900 inclusive.
I was asked to change x varaibles to 1 to 50 inclusive and whether that would change beta one.

i know beta one would stay the same, but how would I be able to prove it?

• May 15th 2011, 03:48 AM
SpringFan25
do you mean x values instead of x variables? or did you really have 50 x variables in your first model?

why do you think it would stay the same?
• May 15th 2011, 04:30 AM
anonymous_maths
Yeah, the x-values for the x variable, i know coz I tried it out on excel....
• May 15th 2011, 04:32 AM
SpringFan25
write down the formula defining your model (ie, say what beta 1 is).

I wouldn't expect that to happen except by chance, unless your model has no random error term.
• May 15th 2011, 04:36 AM
anonymous_maths
nah it kinda make sense coz gradient equals rise/run yeah, and the run is still 50.... but that's a dodgy proof.... trying to find a better one lolz
• May 15th 2011, 04:59 AM
SpringFan25
oh wait i mininterpreted your question, my bad. if you're asking about the transformation:

x' = x - 1850

You know (i hope!) the formula for beta 1, so just show that Var(X) and Cov(X,Y) dont change.

ie, show that
Cov(X',Y) = Cov(X,Y)
Var(X') = Var(X)
• May 15th 2011, 09:08 PM
anonymous_maths
Hi,

I checked on the formula sheet and that

β_1= \frac{∑(X-\bar{X})(Y-\bar{Y}}{∑(X-\bar{X})^2}
• May 16th 2011, 12:07 PM
SpringFan25
i'll use "b" instead of beta 1 to save typing

$b = \frac{Cov(X,Y)}{Var(X)} = \frac{ \sum (X - \bar{X})(Y - \bar{Y})}{\sum (X - \bar{X})(X - \bar{X})}$

Define X' = X - 1850

If we run a regression using X', we will get

$b' = \frac{ \sum (X' - \bar{X'})(Y - \bar{Y})}{\sum (X' - \bar{X'})(X' - \bar{X'})}$

substitute X' = X - 1850

$b' = \frac{\sum (X - 1850 - \bar{X'})(Y - \bar{Y})}{\sum (X - 1850 - \bar{X'})(X - 1850 - \bar{X'})}$

You should know or be able to prove that $\bar{X'} = \bar{X} - 1850$. Substitute that in:

$b' = \frac{\sum (X - 1850 - \left[\bar{X} - 1850 \right])(Y - \bar{Y})}{\sum (X - 1850 - \left[\bar{X} - 1850 \right])(X - 1850 - \left[\bar{X} - 1850 \right])}$

$b' = \frac{\sum (X - 1850 + 1850 - \bar{X})(Y - \bar{Y})}{\sum (X - 1850 +1850 - \bar{X} )(X - 1850 +1850 - \bar{X} )}$

$b' = \frac{\sum (X - \bar{X})(Y - \bar{Y})}{\sum (X - \bar{X} )(X - \bar{X} )}$

$b' = b$

Which is the required result.
• May 17th 2011, 07:36 AM
logisc
thank you. however, how could you write up a proof to show that beta zero will change when x values have changed?
• May 17th 2011, 08:20 AM
SpringFan25
start from the formula for b0'

$b'_0 = \bar{y} - b_1 \bar{X'}$

and substitute $\bar{X'}=\bar{X} - 1850$

compare the result to the formula for $b0$.
• May 17th 2011, 10:31 AM
logisc
thank you very much :)