| LINFIT CHISQ and SIGMA values are correct?? [message #91593] |
Tue, 04 August 2015 12:54  |
Krishnakumar M.A
Messages: 19
Registered: March 2013
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Junior Member |
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Hi,
I was trying to do a linfit in the following data (I'm using IDL 6.3).
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x = [150.0, 235.0, 325.0, 410.0, 610.0]
y = [200.0, 35.0, 8.4, 3.0, 0.6]
err = [25.0, 5.0, 2.1, 0.8, 0.2]
result = linfit(alog10(x),alog10(y),MEASURE_ERRORS=alog10(err), CHISQ=chi, COVAR=covmatrix, SIGMA=error, YFIT=fit)
---------------------------------------------------
It gave me surprisingly odd values for CHISQ and SIGMA. The values are given below.
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result
11.6899 -4.29070
chisq
0.00799291
sigma
4.32894 1.66352
---------------------------------------------------
Fit result looked good, but the values of sigma and chisq are way off. I believe that the chisq it gives is reduced chisq. Is there anything went wrong in the fitting procedure, or are there any issues with the linfit algorithm?
Please let me know whether I'm doing it right or not.
Thanks,
Krishnakumar
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| Re: LINFIT CHISQ and SIGMA values are correct?? [message #91598 is a reply to message #91593] |
Tue, 04 August 2015 19:19  |
wlandsman
Messages: 743
Registered: June 2000
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Senior Member |
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You are giving linfit negative errors -- alog10(0.2) = -0.69897
If you use the absolute value of alog10(err) you will get consistent results.
But probably it is better to do your logarithmic transformation correctly
if z = alog10(y) then dz = 0.434*dy/y (I think)
where dy is your original err and dz is your transformed err .
On Tuesday, August 4, 2015 at 3:54:39 PM UTC-4, Krishnakumar M.A wrote:
> Hi,
>
> I was trying to do a linfit in the following data (I'm using IDL 6.3).
>
> ---------------------------------------------------
>
> x = [150.0, 235.0, 325.0, 410.0, 610.0]
> y = [200.0, 35.0, 8.4, 3.0, 0.6]
> err = [25.0, 5.0, 2.1, 0.8, 0.2]
>
> result = linfit(alog10(x),alog10(y),MEASURE_ERRORS=alog10(err), CHISQ=chi, COVAR=covmatrix, SIGMA=error, YFIT=fit)
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| Re: LINFIT CHISQ and SIGMA values are correct?? [message #91603 is a reply to message #91598] |
Wed, 05 August 2015 00:07 |
Krishnakumar M.A
Messages: 19
Registered: March 2013
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Junior Member |
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On Wednesday, August 5, 2015 at 7:49:24 AM UTC+5:30, wlandsman wrote:
> You are giving linfit negative errors -- alog10(0.2) = -0.69897
>
> If you use the absolute value of alog10(err) you will get consistent results.
>
> But probably it is better to do your logarithmic transformation correctly
>
> if z = alog10(y) then dz = 0.434*dy/y (I think)
>
> where dy is your original err and dz is your transformed err .
>
> On Tuesday, August 4, 2015 at 3:54:39 PM UTC-4, Krishnakumar M.A wrote:
>> Hi,
>>
>> I was trying to do a linfit in the following data (I'm using IDL 6.3).
>>
>> ---------------------------------------------------
>>
>> x = [150.0, 235.0, 325.0, 410.0, 610.0]
>> y = [200.0, 35.0, 8.4, 3.0, 0.6]
>> err = [25.0, 5.0, 2.1, 0.8, 0.2]
>>
>> result = linfit(alog10(x),alog10(y),MEASURE_ERRORS=alog10(err), CHISQ=chi, COVAR=covmatrix, SIGMA=error, YFIT=fit)
Thanks for the reply. I did not get any difference by giving abs(alog10(err)).
But I got better values for chisq and sigma when I used dz = 0.434*dy/y. Could you please tell me why a factor of 0.434?
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| Re: LINFIT CHISQ and SIGMA values are correct?? [message #91604 is a reply to message #91603] |
Wed, 05 August 2015 00:23 |
Helder Marchetto
Messages: 520
Registered: November 2011
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Senior Member |
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I think that what he's saying is:
if f(x) = log_b(x) then f'(x) = 1/(x ln(b)) where b is the base of the logarithm. In your case, you're using IDL's alog10(). So the derivative of the function f(x) = alog10(x) is f'(x) = 1/(x*alog(10)) and can be rewritten as:
f'(x) = 0.434/x
So 0.434 = 1/alog(10)
I hope it helps.
Cheers,
Helder
PS: In case you're unsure what or why the step from z=alog10(y) to dz=0.434*dy/y was taken, then you should look at error propagation and differentials. Here are some google result I found: http://tutorial.math.lamar.edu/Classes/CalcI/Differentials.a spx
http://www.rit.edu/cos/uphysics/uncertainties/Uncertaintiesp art2.html
On Wednesday, August 5, 2015 at 9:08:05 AM UTC+2, Krishnakumar M.A wrote:
> On Wednesday, August 5, 2015 at 7:49:24 AM UTC+5:30, wlandsman wrote:
>> You are giving linfit negative errors -- alog10(0.2) = -0.69897
>>
>> If you use the absolute value of alog10(err) you will get consistent results.
>>
>> But probably it is better to do your logarithmic transformation correctly
>>
>> if z = alog10(y) then dz = 0.434*dy/y (I think)
>>
>> where dy is your original err and dz is your transformed err .
>>
>> On Tuesday, August 4, 2015 at 3:54:39 PM UTC-4, Krishnakumar M.A wrote:
>>> Hi,
>>>
>>> I was trying to do a linfit in the following data (I'm using IDL 6.3).
>>>
>>> ---------------------------------------------------
>>>
>>> x = [150.0, 235.0, 325.0, 410.0, 610.0]
>>> y = [200.0, 35.0, 8.4, 3.0, 0.6]
>>> err = [25.0, 5.0, 2.1, 0.8, 0.2]
>>>
>>> result = linfit(alog10(x),alog10(y),MEASURE_ERRORS=alog10(err), CHISQ=chi, COVAR=covmatrix, SIGMA=error, YFIT=fit)
>
>
> Thanks for the reply. I did not get any difference by giving abs(alog10(err)).
>
> But I got better values for chisq and sigma when I used dz = 0.434*dy/y. Could you please tell me why a factor of 0.434?
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