Re: leasqr problem - covp is 'NA'!!!

[Przemek Klosowski]

On 04/15/2011 12:09 PM, neo7891 wrote:
> hey guys,
>
> I'm try to solve exponential fitting with leasqr.
> The fitting itself is really good - and I want to measure the estimation
> error of parameters.
...
> but the problem is... my 'covp' is only...
> ========================================
> covp =
>
>      NA    NA    NA    NA
...
> data = load("data2.txt");

so , covariance matrix is invalid, and we don't know why. It probably 
has to do with what particular data you're using, so we'd need your 
data2.txt. It could also be a bug in the leasqr() implementation so 
you'd need to specify which version of Octave you're using.

I tried my Octave 3.2.4, using synthetic data that should match your 
function:

x=[0:.1:3]'; y=1-exp(-x)+0.1*(rand(length(x),1)-1); pin=[0,1,1,0];
[f,p,kvg,iter,corp,covp,covr,stdresid,Z,r2] = leasqr(x,y,pin,F);
plot(x, y, "r+", x, f, "b-");
p

and got reasonable fit, p=0.061542   0.915680   0.931379   0.104827

The covariance matrix results in sqrt(diag(covp))) values of 77.468751 
 77.468741    0.061083   90.835584. These values look strange, given 
the goodness of the fit; I haven't dwelved into it but the covp matrix 
looks strange--- approximately

1000 * [1 -1 0 1; -1 1 0 -1; 0 0 0 0; 1 -1 0 1]

which doesn't make sense to me but I haven't thought about it long 
enough to convince myself one way or the other. Practitioners of leasqr, 
what do you think about the reliability of covp values?



> x = data(1:length(data),1);
> y = data(1:length(data),2);
>
> orgFx = @(x, p) p(1) + p(2) * ( 1 - exp( -p(3) * ( x - p(4))));
>
> pin = ones(1,4);
> pin(1) = median(y(1:5));
> pin(2) = max(y) - min(y);
> pin(3) = abs(log(max(y) - min(y))) / (max(x)-min(x));
> pin(4) = min(x);
>
> F = orgFx;
>
> [f,p,kvg,iter,corp,covp,covr,stdresid,Z,r2] = leasqr(x,y,pin,F);
>
> plot(x, y, "r+", x, f, "b-");
>
> fprintf("%f     %f     %f     %f\n", p(1), p(2), p(3), p(4));