Re: Minuit Parabolic Errors Smaller than both Minos Errors?

Hi,

what I am trying to do exactly is to float some parameters x,y,z to minimize the following function:

1/2*[ (x-x0)^2/sx(x)^2a + (y-y0)^2/sy(y)^2 + (z-z0)^2/sz(z)^2]

where

sx(x)= sxn when x<x0
and sx(x)= sxp when y>=y0

and where sy(y) and sz(z) are defined similarly.

I would expect the parabolic errors returned by Minuit to represent somehow an average between the asymmetric errors returned by Minos. Minos returns exactly the right asymmetrical errors, i.e. sxm, sxp, sym, syp, szm, szp. However, sometimes the parabolic error for x can be smaller than both sxm and sxp. This is what I find unexpected...

Thanks

Lorenzo Moneta wrote:
> Hi,
>
> The parabolic errors are correct only if your chi2 or log-likelihood function is a parabola around the minimum.
> I have not fully understood which function you minimize in Minuit, but it is clear, if you have an asymmetric one, the function is not anymore a parabola.
>
> Minos will give you correct errors, in the asymptotic approximation. See also http://seal.web.cern.ch/seal/documents/minuit/mnerror.pdf or if you have it , the F. James book (Statistical methods in experimental physics, chapter 9.3 )
>
> Best Regards
>
> Lorenzo
>
> On Dec 3, 2009, at 11:57 PM, Pierre-Luc Drouin wrote:
>
>
>> Hi,
>>
>> I was doing some tests to understand exactly how Minuit computes the parabolic errors vs the Minos errors and I observed a strange result. Using my test, Minuit finds parabolic errors that are smaller than both Minos errors for all the fit parameters:
>>
>> FCN=1.10038e-14 FROM MINOS STATUS=SUCCESSFUL 275 CALLS 363 TOTAL
>> EDM=8.9213e-11 STRATEGY= 1 ERROR MATRIX ACCURATE
>> EXT PARAMETER PARABOLIC MINOS ERRORS NO. NAME VALUE ERROR NEGATIVE POSITIVE 1 x 1.23000e+02 7.26365e+00 -1.30000e+01 2.60000e+01
>> 2 y 5.73000e+02 4.78688e+01 -4.50000e+01 7.50000e+01
>> 3 z 2.50000e+01 1.53528e+00 -3.00000e+00 5.00000e+00
>> ERR DEF= 0.5
>>
>> What I do for this test is to simply ask Minuit to find the maximum of a multivariate Gaussian function with correlated variables and asymmetrical widths (I simply use a different width depending if the variables are smaller or greater than their expected fit values). I would expect Minuit to find the following correlation matrix:
>>
>> 1.000000000
>> 2.338779e-01 1.000000000
>> 8.988122e-01 4.169567e-01 1.000000000
>>
>> but the covariance matrix it returns,
>> 5.276059e+01 -2.782138e+01 4.362987e+00
>> -2.782138e+01 2.291426e+03 1.611347e+01
>> 4.362987e+00 1.611347e+01 2.357071e+00
>>
>> ,gives the following correlations:
>> 1.000000e+00 -8.000878e-02 3.911778e-01
>> -8.000878e-02 1.000000e+00 2.194311e-01
>> 3.911778e-01 2.194311e-01 1.000000e+00
>>
>> Could someone explain me how such results from Minuit are possible? Is it simply because the 2nd derivative of my function is not continuous at the maximum (note that the first derivative is)? Is there another way I can use Minuit to evaluate the covariance matrix properly?
>>
>> Thanks!
>>
>>
>
>
Received on Fri Dec 04 2009 - 16:19:41 CET