The suggestion below is the right idea, but I think is off in a detail. Doing least squares we take the partial derivatives of the sum of (y-b1-b2*x)^2 with respect to b1 and b2. I think the derivative w.r.t x is given below by mistake.
The easier proof may be to plug ybar and xbar into the equation for y, using the formulae for the least squares coefficients and show that. everything cancels. -Dick Startz BTW, it's common in statistics to order the pairs (y,x), so th original poster isn't necessarily violating convention. On 1 Mar 2004 08:46:47 GMT, Eric Bohlman <[EMAIL PROTECTED]> wrote: >"Maja" <[EMAIL PROTECTED]> wrote in >news:[EMAIL PROTECTED]: > >> Hi everyone, >> >> I was just doing some questions for practice b/c I have a midterm >> coming up, and there is a question that I'm having lots of difficulty >> with, it is not really covered in the textbook. >> Here it goes: >> Show that the least squares line y^=b1 + b2x passes through the point >> (ybar,xbar), where ybar is the mean of y and xbar the mean of x. > >That should be (xbar,ybar) since convention is to write points with X >first. > >The least squares line minimizes the sum of (y-b1-b2x)^2, which means that >the first derivative of the sum (which is the same as the sum of the >derivatives) will be zero. The first derivative of (y-b1-b2*x)^2 >is 2*b2*(b1-y+b2*x); setting it to zero and summing results in >ybar=b1+b2*xbar. ---------------------- Richard Startz [EMAIL PROTECTED] Lundberg Startz Associates . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
