ross <[EMAIL PROTECTED]> wrote in message 
news:<[EMAIL PROTECTED]>...
> Assume some 'ordinary' data - a set of (x,y) data points, ordered on x, 
>  with x & y real.  Assume we draw a boundary somewhere in the middle of 
> the x domain
> (x = k) to divide it into 2 adjacent subdomains.  I want to fit 2 
> regression lines of the form y = ax + b, one to each of the subdomains, 
> with a continuity constraint.
> 
> In other words I want to satisfy these criteria:
> (C1) the total sum of  squared deviates over both intervals is minimised
> (C2) the two fitted lines intersect the boundary x = k at the same 
> point.
> 
> The sums of squared deviates on the two intervals independently would 
> be:
>    s1 = S1(y - a1*x - b1)^2            [S1 = sum over n1 points in 
> subdomain #1]
>    s2 = S2(y - a2*x - b2)^2            [S2 = sum over n2 points in 
> subdomain #2]
> requiring the 4 parameters a1, b1, a2, b2.
> 
> but these fits are not independent because of criterion #2, which says 
> that a1*k + b1 = a2*k + b2.   This constraint means that we can replace 
> one of the parameters, eg,  b2 = a1*k + b1 - a2*k.
> 
> The total sum to minimise is:
>  s = s1 + s2
>    = S1[ (y - a1*x - b1)^2 ] + S2[ (y - a2*x - b2)^2 ]
>    = S1[ (y - a1*x - b1)^2 ] + S2[ (y - a2*x - (a1*k + b1 - a2*k))^2 ]
> 
> Then the solution to these 3 simultaneous partial deriv equations:
>      ds/d(a1) = ds/d(b1) = ds/d(a2) = 0            ..... eq.1
> is the solution to the problem, for a given k.
> 
> Expanding eq.1, I get:
> 
>      S1(x^2) +k^2.n2     S1(x) +k.n2      k.S2(x) - k^2.n2
> A =  S1(x) + k.n2        n1 + n2          S2(x) - k.n2
>      k.S2(x) - k^2.n2    S2(x) - k.n2     S2(x^2) - 2k.S2(x) + k^2.n2
> 
>     S1(xy) + k.S2(y)
> C = S1(y)  + S2(y)
>     S2(xy) - k.S2(y)
> 
> and solution B = A \ C      where B = column (a1,  b1,  a2)
> 
> When I program it and plot the results, the fit is clearly not correct. 
>  Is my expansion of eq.1 wrong?   Am I introducing the constraint in 
> the wrong way?

Your expansion seems ok. I programmed it, and the plots don't look bad
to me. What is it that looks wrong to you?
.
.
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