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Least squares regression

  1. Oct 21, 2015 #1
    1. The problem statement, all variables and given/known data
    note a linear regression model with the response variable Y=(Y1..Yn) on a predictor variable X=(X1..Xn). the least squares estimates of the intercept and slope a(hat) and B(hat) are the values that minimize the function: (see attached image)

    and the problem reads on further -

    further predicted values equal y(hat)(x)=a(hat)+b(hat)x (note y(hat) is meant to be read as a function of x)

    i have been asked to find y(hat)(Xbar), where X(bar) is the average of the Xi's. (note y(hat) is meant to be read as a function of Xbar).

    i'm not sure where to start with this question. advice as to whether i'm on the right track is all i need for now.

    so i was thinking that i could use the fact that

    a(hat) = Y(bar)-B(hat)X(bar) and B(hat) = (sum) (Xi-X(bar))(Yi-Y(bar)) /(sum) (Xi-X(bar))^2

    but i'm not exactly sure how to solve for y(hat)X(bar) -(yhat as a function of Xbar)

    should i be trying to get a equation with only a(hat) , b(hat) , and Xbar?


    Thanks for the help - apologies for poor notation
     

    Attached Files:

  2. jcsd
  3. Oct 21, 2015 #2

    Ray Vickson

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    If you have set up and solved the least-squares equations (so that you know the parameters ##\hat{a}, \hat{b}## in terms of the ##\{x_i, y_i\}##, you can just substitute ##x = \bar{x}## into the equation ##\hat{y} = \hat{a} + \hat{b} x##, and carry out algebraic simplification.

    BTW: it is easy to employ good notation---just use LaTeX. To see how I did it, just right-click on an expression and to to the menu item to 'display math as tex commands'. This site has a brief LaTeX tutorial on the use of LaTeX, but I cannot say exactly where/how to find it; others may know.
     
  4. Oct 22, 2015 #3
    so if i do this should i get
    \hat{y} =Y(bar)?
     
  5. Oct 22, 2015 #4
    y(hat)(Xbar) = Y(bar)? still hopeless at notation :(
     
  6. Oct 22, 2015 #5
    or should i get,
    Y(bar) = a(hat) +b(hat)X(bar)
     
  7. Oct 22, 2015 #6

    Ray Vickson

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    You need to tell the system "LateX starts here" ... and "Latex ends here", with your mathematical expressions in between.

    For displayed equations use

    [t ex] ...your expressions ... [/t ex]

    with no spaces between the 't' and the 'ex', and not in a red-colored font (which I used just for emphasis). Doing that on your expression above gives
    [tex] \bar{Y} = \hat{a} + \hat{b} \bar{x} [/tex]
    Note that we write \bar{Y}, not Y(bar), and we write \hat{a}, not a(hat). Some people prefer the look of \overline{...} instead of \bar{...}, and using that instead gives
    [tex] \overline{Y} = \hat{a} + \hat{b} \overline{x} [/tex]

    For in-line equations or expressions, use

    # # ... expression... # #

    with no space between the two #s at the start and at the end, and not in a red font. Doing that with your expression above gives ##\bar{Y} = \hat{a} + \hat{b} \bar{x}##, as wanted.

    Anyway, you cannot just wrote that ##\bar{Y} = \hat{a} + \hat{b} \bar{x}##, because the right-hand-side is ##Y_{\text{fitted}}(\bar{x})##, but how do you know that ##Y_{\text{fitted}}(\bar{x}) = \bar{Y}##? Can you even be sure it is true?
     
    Last edited: Oct 22, 2015
  8. Oct 22, 2015 #7
    should i get y(Xbar) = a(hat)+b(hat)X(bar)
     
  9. Oct 22, 2015 #8

    Ray Vickson

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    Yes, but that is not the end of the story. You ought to be able to simplify it a lot, using either the explicit expressions for ##\hat{a}## and ##\hat{b}##, or by exploiting the fact that ##\hat{a}, \hat{b}## satisfy some particular equations obtained by minimization of the total squared error.
     
  10. Oct 22, 2015 #9
    ahhh... so after doing some algebra i get yhat(Xbar) = Y(bar)
     
  11. Oct 22, 2015 #10
    let me try in notation [tex] \hat{y}(\bar{x})=\bar{Y} [/tex]
     
  12. Oct 22, 2015 #11

    Ray Vickson

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    Yes, exactly.
     
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