Re: Covariance with bounds in leasqr

[Olaf Till]

On Thu, Mar 22, 2012 at 11:26:32AM -0300, Pablo wrote:
> 
> 
> On 03/22/2012 03:41 AM, Olaf Till wrote:
> >On Wed, Mar 21, 2012 at 06:31:51PM -0300, Pablo wrote:
> >>Hi everyone,
> >>   I've been intrigued by a detail in leasqr (from Octave-Forge) for
> >>a while. In the function's documentation it's stated that "If
> >>constraints (or bounds) are set, returned guesses of corp, covp, and
> >>Z are generally invalid, even if no constraints are active for the
> >>final parameters".
> >>
> >>   I wonder why is that, particularly for covp (it's usually the most
> >>useful for me). Unfortunately, I don't seem to have access to the
> >>reference given (Bard) so I can't check if that's mentioned there.
> >>Also, a few web searches didn't give me any clue.
> >>   I'm familiar with the way the covariance of the parameters is
> >>calculated for a linear fit, and as far as I can see it's
> >>essentially the same for leasqr.
> >>
> >>   Could anyone give me some pointers as to why that statement is
> >>true? Any references or links to help me understand it would also be
> >>highly appreciated.
> >>
> >>   Best Regards,
> >>     Arnoques
> >
> >This is not specific to leasqr. Roughly speaking, the covariance
> >matrix of estimated parameters is a guess on how estimated parameters
> >change due to random in the input data. But if they change, they might
> >hit (and so are influenced by) a constraint, even if the constraint is
> >not active for the current estimation of parameters. This prevents an
> >easy estimation of the covariance matrix of parameters from (possibly
> >estimated) (co)variance of input data.
> >
> >Olaf
> >
> 
> Ok, so the covariance estimation is only affected by the bounds if
> the probability distribution for the parameters is not negligible at
> the bounds.
> 
> Just to check if I got that right, if I set [0,1] as the bounds for
> the parameter, and the fit gives me 0.3 with a variance of 0.1, the
> variance is fine. If, on the other hand, the fit gives me a variance
> of 1, this value is meaningless.

Probably right. Olaf

> Thank you,
>   Arnoques

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