[SciPy-User] solving integration, density function

[josef.pktd at gmail.com]

On Thu, Jan 6, 2011 at 7:01 AM, Johannes Radinger <JRadinger at gmx.at> wrote:
> Thank you for the simplification of the formula,
> but I still get a different result in the case
> when x<s1 (eg. x=2, s1=3).
>
> here a code to try:
>
> ********************
> import math
> from scipy import stats
>
> s1 = 3
> m = 0
> p = 1
> x = 2
>
> func = stats.norm.pdf(x, loc=m, scale=(s1))
> func2 = (1/(s1*math.sqrt(2*math.pi)) * math.exp(-0.5*((x-m)/s1)**2))
>
> print func
> print func2
> ********************************

use floats, I think you just run into integer division

(x-m)/s1

Josef

>
> /j
>
> -------- Original-Nachricht --------
>> Datum: Thu, 6 Jan 2011 05:48:25 -0600
>> Von: Warren Weckesser <warren.weckesser at enthought.com>
>> An: SciPy Users List <scipy-user at scipy.org>
>> Betreff: Re: [SciPy-User] solving integration, density function
>
>> On Thu, Jan 6, 2011 at 5:27 AM, Johannes Radinger <JRadinger at gmx.at>
>> wrote:
>>
>> > Hey
>> >
>> > Last time you helped me a lot with my normal
>> > probabilty density function. My problem now is
>> > quite simple, I think it's just a problem with
>> > the syntax (brackets):
>> >
>> > There are two ways to calculate the pdf, with the
>> > stats-function and with pure mathematically, but
>> > the give different results and I can't find the
>> > where I make the mistake:
>> >
>> >
>> > func1 = stats.norm.pdf(x, loc=m, scale=(s1))
>> > func2 =
>> > 1/((s1)*(math.sqrt(2*math.pi))))*(math.exp(((-0.5)*((x-m)/(s1)))**2)
>> >
>> > Where is the problem
>> >
>>
>>
>> func2 = 1/(s1*math.sqrt(2*math.pi)) * math.exp(-0.5*((x-m)/s1)**2)
>>
>>
>> Warren
>>
>>
>>
>> > thank you...
>> >
>> > /j
>> >
>> > -------- Original-Nachricht --------
>> > > Datum: Tue, 21 Dec 2010 09:18:15 -0500
>> > > Von: Skipper Seabold <jsseabold at gmail.com>
>> > > An: SciPy Users List <scipy-user at scipy.org>
>> > > Betreff: Re: [SciPy-User] solving integration, density function
>> >
>> > > On Tue, Dec 21, 2010 at 7:48 AM, Johannes Radinger <JRadinger at gmx.at>
>> > > wrote:
>> > > >
>> > > > -------- Original-Nachricht --------
>> > > >> Datum: Tue, 21 Dec 2010 13:20:47 +0100
>> > > >> Von: Gregor Thalhammer <Gregor.Thalhammer at gmail.com>
>> > > >> An: SciPy Users List <scipy-user at 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
>> > > _______________________________________________
>> > > SciPy-User mailing list
>> > > SciPy-User at scipy.org
>> > > http://mail.scipy.org/mailman/listinfo/scipy-user
>> >
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