regressions

btnh · #1 $Old$ June 4th, 2009, 11:06 PM

Suppose that you wish to use a linear regression to predict the dependent variable Y using the dependent variable X1. You collect a scatter plot of points (X1,i ,Yi) and notice that the plot seems to be well represented by a piecewise continuous linear function that satisfies the following conditions: For X1,i<=xo, the appropriate linear model appears to be Y1=a+B1X1,i+E1. However, when X1,1>xo the slope of the applicable linear relationship appears to change to B1+B2. Hint: First, determine the applicable form of the linear model that represents the plot of points (X1,i ,Yi) when X1,i>xo and note that both the slope and Y-intercept of the model will differ from the model applicable to the set of points (X1,i ,Yi) when X1,i<=xo. Now see if you can devise a way to combine these two linear relationships into a single multiple regression model by using the indicator variable X2,i where X2,i= +1 if X1,i >xo and X2,i=0 otherwise

a. There is no way to combine these two linear relationships into a single multiple regression. You should instead run each regression separately as simple regressions.
b. The combined model is: Y1= a + B1 X1,i + B2 X1,i X2,i + Ei
c. The combined model is: Y1= a + B1 X1,i + B2 (xo - X1,i) X2,i + Ei
d. The combined model is: Y1= a + B1 X1,i + B2 (a - xo - X1,i) X2,i + Ei
e. The combined model is: Y1= a + B1 X1,i + B2 (X1,i - xo) X2,i + Ei

matheagle · #2 $Old$ June 5th, 2009, 06:28 PM

I started to read this yesterday but I immediately saw y and x both dependent?
and don't you mean X1,i>xo not X1,1>xo?
You need to type this more carefully.

This is just curve fitting, not stats.
You want to incorporate both models using an indicator function $x_2$ .

You want $y=a+b_1x_1$ , when $x\le x_0$

and you want $y=a+(b_1+b_2)x_1$ , when $x>x_0$ , don't worry about the $\epsilon$ .
But, I'm concerned about using the same y-intercept here, a.

BUT it is the same a.
When $x_1=0$ , we have the same intercept.
So I would go with (b).

btnh · #3 $Old$ June 5th, 2009, 07:27 PM

Thank you, and sorry about the typos. You have been a big help.

adrian019 · #4 $Old$ June 5th, 2009, 09:53 PM

Hey guys, I am actually trying to find an answer to this exact same question as well. I messaged matheagle earlier, sorry for the confusion. It appears as if btnh and I are doing a similar assignment... Any help on this problem is appreciated.

matheagle · #5 $Old$ June 6th, 2009, 12:34 AM

B2 can be anything that means that the two slopes are completely different say m=B1+B2, but the a is a problem.

we have y=a+B1x for x<x0 and y=a+mx for x>x0 and these two different a's should not be the same.
However if x0>0 then the intercept is in the first region.

I feel that.... b. The combined model is: Y1= a + B1 X1,i + B2 X1,i X2,i + Ei
will estimate both B1 and B2 properly but I doubt it will estimate a correctly in both cases.

There is one real way to do this.
I can run a regression on b-e, but that means I need to this each equation via matrices $\hat\beta=(X^tX)^{-1}X^tY$ .

adrian019 · #6 $Old$ June 6th, 2009, 02:01 PM

Yes that is true. However, does the model's ability to estimate alpha properly affect the solutions? I don't know if that matters and would cause a difference? I guess I am torn between 'A' and 'B'. However, 'A' simply says that the relationships can't be combined, but I'm not sure that this is true. 'A' or 'B'...that is the question...

adrian019 · #7 $Old$ June 6th, 2009, 03:49 PM

Also, it seems that 'E' could be an answer as well. If you contruct a basic math model and plug in values, 'E' gives numbers that could very well be correct. What do you think matheagle?

matheagle · #8 $Old$ June 9th, 2009, 12:24 AM

I'm still leaning towards
b. The combined model is: Y= a + B1 X1,i + B2 X1,i X2,i + Ei
When X1 is less than x0, then X2 is zero and the model reduces to
Y= a + B1 X1,i + Ei
and when X2 is 1, we have
Y= a + B1 X1,i + B2 X1,i + Ei =a + (B1 + B2) X1,i + Ei
and what I'm not sure, but my guts say that (b) is right since
a is the same intercept in either case.
FOR if X1=0 we get the SAME intercept, a.