Thread: Calculating Variances and Regression Coefficients

This first part I am just looking for confirmation that I am working it correctly:
I have 30 observations and sumX=30, sumX²=150, sumY=120, sumY²=1830 and sumXY=480.

I found the variance of X to be V[X]= E[X²] - (E[X])²= (150/30) - (1²) = 5 - 1 = 4
the variance of X to be V[Y]= E[Y²] - (E[Y])²= (1830/30) - (4²) = 61 - 16 = 45
and the covariance of X,Y to be Cov[X,Y] = E[XY] - E[X]E[Y] = (480/30) - (1)(4) = 16 - 4 = 12

Then I was asked to fit a regression line and I found beta to be Cov[X,Y]/V[X] = 12/4 = 3.
Alpha = E[Y] - beta*E[X] = 4 - 3 * 1 = 1 and the regression equation is Y = 1 + 3X

Now the part I am stuck:
Suppose I am told that the scale of the X variable is was cut in half for this data (i.e. the original variable was 2X that given above). What would we find out about our regression coefficients?

In this case my sumX=60 and sumX²=300 ? But how would I know the effect on sumXY as I don't know the actual X inputs, only that they are twice what I was originally told. And I need sumXY to calculate new regression coefficients to determine a change.

I appreciate any help figuring this part out...thanks!

Hey drabbie.

Hint: Use the fact that E[aZ] = a*E[Z] (Let Z = XY).

Thank you chiro!

That's what I did for X and X². Wonder why it is so obvious for those, but not XY?

Was I on track with the others?

For E[(2X)^2] just be aware that this is 4*E[X^2] not 2*E[X^2].

Otherwise, the idea is correct.

Just to make sure I understand: E[aX]^2 = a^2 * E[X]^2. Because the whole term aX is squared rather than just the X, such as in E[X^2].

Thank you for all your help! I really appreciate it.

Yes that's right, but I was also thinking of E[(aX)^2] = E[a^2*X^2] = a^2*E[X^2].