[SciPy-User] solving integration, density function
Johannes Radinger
JRadinger@gmx...
Tue Dec 21 08:33:16 CST 2010
-------- Original-Nachricht --------
> Datum: Tue, 21 Dec 2010 09:18:15 -0500
> Von: Skipper Seabold <jsseabold@gmail.com>
> An: SciPy Users List <scipy-user@scipy.org>
> Betreff: Re: [SciPy-User] solving integration, density function
> On Tue, Dec 21, 2010 at 7:48 AM, Johannes Radinger <JRadinger@gmx.at>
> wrote:
> >
> > -------- Original-Nachricht --------
> >> Datum: Tue, 21 Dec 2010 13:20:47 +0100
> >> Von: Gregor Thalhammer <Gregor.Thalhammer@gmail.com>
> >> An: SciPy Users List <scipy-user@scipy.org>
> >> Betreff: Re: [SciPy-User] solving integration, density function
> >
> >>
> >> Am 21.12.2010 um 12:06 schrieb Johannes Radinger:
> >>
> >> > Hello,
> >> >
> >> > I am really new to python and Scipy.
> >> > I want to solve a integrated function with a python script
> >> > and I think Scipy should do that :)
> >> >
> >> > My task:
> >> >
> >> > I do have some variables (s, m, K,) which are now absolutely set, but
> in
> >> future I'll get the values via another process of pyhton.
> >> >
> >> > s = 400
> >> > m = 0
> >> > K = 1
> >> >
> >> > And have have following function:
> >> > (1/((s*K)*sqrt(2*pi)))*exp((-1/2*(((x-m)/s*K))^2) which is the
> density
> >> function of the normal distribution a symetrical curve with the mean
> (m) of
> >> 0.
> >> >
> >> > The total area under the curve is 1 (100%) which is for an
> integration
> >> from -inf to +inf.
> >> > I want to know x in the case of 99%: meaning that the integral (-x to
> >> +x) of the function is 0.99. Due to the symetry of the curve you can
> also set
> >> the integral from 0 to +x equal to (0.99/2):
> >> >
> >> > 0.99 = integral((1/((s*K)*sqrt(2*pi)))*exp((-1/2*(((x-m)/s*K))^2)),
> -x,
> >> x)
> >> > resp.
> >> > (0.99/2) =
> integral((1/((s*K)*sqrt(2*pi)))*exp((-1/2*(((x-m)/s*K))^2)),
> >> 0, x)
> >> >
> >> > How can I solve that question in Scipy/python
> >> > so that I get x in the end. I don't know how to write
> >> > the code...
> >>
> >>
> >> --->
> >> erf(x[, out])
> >>
> >> y=erf(z) returns the error function of complex argument defined
> as
> >> as 2/sqrt(pi)*integral(exp(-t**2),t=0..z)
> >> ---
> >>
> >> from scipy.special import erf, erfinv
> >> erfinv(0.99)*sqrt(2)
> >>
> >>
> >> Gregor
> >>
> >
> >
> > Thank you Gregor,
> > I only understand a part of your answer... I know that the integral of
> the density function is a error function and I know that the argument "from
> scipy.special import erf, erfinv" is to load the module.
> >
> > But how do I write the code including my orignial function so that I can
> modify it (I have also another function I want to integrate). how do i
> start? I want to save the whole code to a python-script I can then load e.g.
> into ArcGIS where I want to use the value of x for further calculations.
> >
>
> Are you always integrating densities? If so, you don't want to use
> integrals probably, but you could use scipy.stats
>
> erfinv(.99)*np.sqrt(2)
> 2.5758293035489004
>
> from scipy import stats
>
> stats.norm.ppf(.995)
> 2.5758293035489004
> Skipper
The second function I want to integrate is different, it is a combination of two normal distributions like:
0.99 = integrate(0.6*((1/((s1*K)*sqrt(2*pi)))*exp((-1/2*(((x-m)/s1*K))^2))+0,4*((1/((s2*K)*sqrt(2*pi)))*exp((-1/2*(((x-m)/s2*K))^2)))
and here again I know s1, s2, m and K and want to get x in the case when the integral is 0.99. What do I write into the script I want create?
I think it is better if I can explain it with a graph but I don't know if I can just attach pictures to the mail-list-mail.
/j
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