
Re: A guess of the Probability density function from percentile values
Posted:
Dec 27, 2010 7:41 AM


On 27 déc, 13:19, Allamarein <matteo.diplom...@gmail.com> wrote: > On 27 Dic, 12:23, Valeri Astanoff <astan...@gmail.com> wrote: > > > > > > > On 27 déc, 11:17, Allamarein <matteo.diplom...@gmail.com> wrote: > > > > I have just posted a similar thread in another group. > > > I hope to be more lucky here. > > > > I know three percentile values. > > > Let's say they are: > > > 95% 82.1 > > > 50% 80.3 > > > 5% 77.8 > > > > I would to get a Probability density function. > > > I presume I would guess the shape of this curve. > > > Since these data refer to a scientific measuring, I would find a t > > > Student or a Gaussian distribution that is consistent with the > > > previous percentiles. > > > > Any suggestions? > > > Good day, > > > A suggestion (for a normal distribution) : > > > FindMinimum[ > > (Quantile[NormalDistribution[m, s], 5/100]  77.8)^2 + > > (Quantile[NormalDistribution[m, s], 50/100]  80.3)^2 + > > (Quantile[NormalDistribution[m, s], 95/100]  82.1)^2, > > {m, 1}, {s, 1}] > > > {0.0816667, {m > 80.0667, s > 1.30711}} > > >  > > Valeri > > It seems you write a code in Mathematica. > I am not so familiar with this language. > May you give me more details? Masquer le texte des messages précédents  > >  Afficher le texte des messages précédents 
I'm not a specialist. I just use Mathematica to solve such problems. All I can say is that you get the same result using derivatives wrt m and s, instead of FindMinimum :
q[x_] = Quantile[NormalDistribution[m, s], x]
m + Sqrt[2]*s*InverseErf[2*x1]
Solve[ D[(q[0.05]  77.8)^2 + (q[0.50]  80.3)^2 + (q[0.95]  82.1)^2, m] == 0 && D[(q[0.05]  77.8)^2 + (q[0.50]  80.3)^2 + (q[0.95]  82.1)^2, s] == 0, {m, s}]
{{m > 80.0667, s > 1.30711}}
 Valeri

