Regression by dfhdhdhdhjr

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									   A regression line, a+bX, is interpreted as E[Y|X],
    the conditional expectation of Y given X.

   The slope of a regression line in general can be
    estimated as
    b=Cov(X,Y)/Var(X).

   The intercept of a regression line in general can
    be estimated as
    a=E[Y]-b*E[X].
   For a discrete PDF with only two possible
    outcomes for Y (Y1,Y2) and 2 possible
    outcomes for X (X1,X2), we can estimate the
    regression line in a very simple intuitive
    manner:
    1) Estimate E[Y|X=X1]
    2) Estimate E[Y|X=X2]
    3) Set up the two equations for two unknowns and
       solve for a and b:
                 E[Y | X  X 1 ]  a  bX1
                 E[Y | X  X 2 ]  a  bX 2
                                         rA
                                   18         -10
                            12    0.4         0.05
                       rI
                            -5    0.25        0.3



   P(A=18|I=12)=.4/.45=0.889
   P(A=-10|I=12)=.05/.45=0.111
   E[A|I=12]=18*0.889+(-10)*.111=14.89

   P(A=18|I=-5)=.25/.55=0.455
   P(A=-10|I=-5)=.30/.55=0.545
   E[A|I=-5]=18*0.455+(-10)*.545=2.74
   So we want an a and b such that
         14 .89  a  b  12
          2.74  a  b  (5)
First Solve for b
14.89  a  b  12         Then Solve for a
 2.74   a  b  (5)     14 .89  a  0.71  12
12.15  17b                a  14 .89  8.52  6.4
b  0.71
   How can we solve for the intercept and slope
    of a regression line in general?
    b=Cov(X,Y)/Var(X)
    a=E[Y]-b*E[X]

   In general, as long as we have a sample of
    historical outcomes, we know how to
    estimate Cov(X,Y), Var(X), E[Y], and E[X].
     We therefore know how to estimate a and b from
     a sample of historical data.
   How do we know these formulas work?
   Do they work in the case of a joint PDF?
   From the joint PDF:
       Cov(rA,rI)=51.17
       Var(rI)=71.53
       E[rA]=8.2
       E[rI]=2.65
   Let’s check our regression formulas:
     And 51.17/71.53=0.71=b
     And 8.2-2.65*(0.715)=6.3=a

   Yes, they work.

								
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