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Math Help - regressions

  1. #1
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    regressions

    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
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  2. #2
    MHF Contributor matheagle's Avatar
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    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).
    Last edited by matheagle; June 9th 2009 at 01:24 AM.
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  3. #3
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    Thank you, and sorry about the typos. You have been a big help.
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  4. #4
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    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.
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  5. #5
    MHF Contributor matheagle's Avatar
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    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.
    Last edited by matheagle; June 6th 2009 at 08:44 AM.
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  6. #6
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    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...
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  7. #7
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    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?
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  8. #8
    MHF Contributor matheagle's Avatar
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    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.
    Last edited by matheagle; June 9th 2009 at 01:14 AM.
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