Standard least squares measures the *vertical* distance to the line from each point and minimizes the sum of the squares. Interchanging x and y is equivalent to switching to measuring the *horizontal* distance. You can instead measure the perpendicular distance, which is invariant under interchanging x and y (or any other isometry for that matter) but the solution is correspondingly hairier. On Thu, Oct 4, 2018 at 9:24 AM Henry Baker <hbaker1@pipeline.com> wrote:
In common discussions of least squares, the parameters (m,b) are estimated for the equation y = m*x+b using as data various datapoints [x1,y1], [x2,y2], [x3,y3], etc.
For example, in Wikipedia (where m=beta2 and b=beta1):
https://en.wikipedia.org/wiki/Linear_least_squares#Example
So far, so good.
Now, if I merely exchange x and y, then my equation is x = m'*y+b', where should be m' = 1/m and b' = -b/m. (Let's ignore the case where the best m=0.)
However, if I then estimate (m',b') using the same least squares method, I don't get (1/m,-b/m) !
So either I'm doing something wrong, or perhaps there is a more symmetric least squares method that treats x and y symmetrically ??
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