forum.alglib.net

Hello,

I am trying to use alglib.invstudenttdistribution() in the same way as Excel's TINV() function or
.NET's Chart.DataManipulator.Statistics.InverseTDistribution() function.

However, I am getting different results.

double x = alglib.invstudenttdistribution(3, 0.05); // gives me -2.3533634348018238

x = chrt.DataManipulator.Statistics.InverseTDistribution(0.05, 3); // .NET gives me 3.1824463049985425

The InverseTDistribution() result is correct and it is close to the Excel value, but invstudenttdistribution's result looks not OK.

I checked your docs, but there is not a lot of description for usage of invstudenttdistribution().

Is this a bug, my improper usage of your function, or what else is happening?

I'd appreciate any help you can offer on this puzzle.

Thank you.

I think that ALGLIB result is correct. ALGLIB invstudenttdistribution(k,p) returns such t that integral(student_PDF(x)dx : x=-INF...t)=p.

If p=0.5, t=0.
If p=1.0, t=+INF.
If p=0.0, t=-INF.

ALGLIB returns inverse CDF. DataManipulator and Excel's TINV() just use another definition of inverse distribution. Read http://office.microsoft.com/en-gb/excel ... 09317.aspx to find out what definition they use.

Thank you Sergey!

I looked at that link and played around with the relationship between the probability used by alglib.invstudenttdistribution() versus Microsoft's versions and I found this relationship:

p_alglib = 1.0 - (0.5 * p_Excel)

With this conversion, alglib.invstudenttdistribution() appears to work like a charm in my software.

Thanks for your excellent help!

I think there is something wrong with this relation , p_alglib = 1.0 - (0.5 * p_Excel) , please can anybody show the exact relation, I tried the above figures but it is not matching

I need excel to give me the same results as alglib

I reconciled both excel tinv and invstudenttdistribution and as a reference for this in the forum I'll mention it

the reconciliation will be according to the one tail result and two tails result

for one tail result

-1*InvStudentTDistribution(3,.05) = TINV(.05*2,3)

for two tails result

-1*InvStudentTDistribution(3,.05/2) = TINV(.05,3)