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here's a challenging one
Message posted by Neil (via 194.82.103.75) on February 28, 2002 at 9:47 AM (ET)
How do I solve the cofficients in the equation of type
y=a+b+cX+dX. by least squares where X has the same value. Ta. NH.
READERS RESPOND:
(In chronological order. Most recent at the bottom.)
Re: here's a challenging one
Message posted by Darius (via 200.23.217.10) on February 28, 2002 at 10:58 AM (ET)
y=a+b+cX+dX. by least squares where X has the same value. ?????????I tink you made a mistake, the equation:
y=a+b+cX+dX
being a+b=A and c+d = B >>> y= A+BX
A traditional univariate least squares
Re: here's a challenging one
Message posted by Jack Tomsky (via 12.144.103.66) on February 28, 2002 at 1:46 PM (ET)
If X has the same value, then the design matrix has a rank of one. What this means is that you cannot solve for a, b, c, and d separately. The least squares estimate for the linear function, a+b+cX+dX, is obtained by setting it equal to ybar.
Re: here's a challenging one
Message posted by Neil (via 194.82.103.76) on March 1, 2002 at 6:50 AM (ET)
Thanks to both of you, this problem is not mentioned in any book I can find. I did not understand Jack's response entirely but I assume it is the same as Darius's.
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