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MathGroup Archive 1996

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Re: Mathematica: Weibull distribution fnc.

  • To: mathgroup at smc.vnet.net
  • Subject: [mg5331] Re: [mg5309] Mathematica: Weibull distribution fnc.
  • From: BobHanlon at aol.com
  • Date: Wed, 27 Nov 1996 01:47:46 -0500
  • Sender: owner-wri-mathgroup at wolfram.com

This shows how to work with Weibull distribution.

Bob Hanlon (Col, USAF (Retired))
______________________________________

In[1]:=
Needs["Statistics`ContinuousDistributions`"]
Needs["Statistics`DiscreteDistributions`"]
n /: Negative[n] = False;
n /: IntegerQ[n] = True;
Off[General::spell1, General::intinit]

Let the number of failures, X, be a Poisson process with mean (mu t) where mu

is given by

In[6]:=
mu = (1/beta) (t/beta)^(alpha - 1);

In[7]:=
PDF[PoissonDistribution[mu t], n] // Simplify

Out[7]=
    t   alpha n
 ((----)     )
   beta
-----------------
         alpha
 (t/beta)
E              n!

Let T be the time until the first failure.  Then

     Pr{T <= t} = 1 - Pr{T > t} = 1 - PDF[PoissonDistribution[mu t], 0]

In[8]:=
cdf = 1 - PDF[PoissonDistribution[mu t], 0]

Out[8]=
                  -1 + alpha
     -((t (t/beta)          )/beta)
1 - E

The PDF for T is then

In[9]:=
pdf = D[cdf, t] // Simplify

Out[9]=
        t   alpha
alpha (----)
       beta
-----------------
         alpha
 (t/beta)
E              t

In[10]:=
PDF[WeibullDistribution[alpha, beta], t] - pdf == 0 // 
 PowerExpand

Out[10]=
True

Consequently, T has a Weibull distribution.

In[11]:=
Domain[WeibullDistribution[]]

Out[11]=
{0, Infinity}

The exponential and Rayleigh distributions are special cases of the Weibull 
distribution with alpha equal to 1 and 2 respectively:

In[12]:=
PDF[WeibullDistribution[1, 1/lambda], x] == 
 PDF[ExponentialDistribution[lambda], x]

Out[12]=
True

In[13]:=
PDF[WeibullDistribution[2, Sqrt[2] sigma], x] == 
 PDF[RayleighDistribution[sigma], x]

Out[13]=
True

In[14]:=
Mean[WeibullDistribution[alpha, beta]]

Out[14]=
                 1
beta Gamma[1 + -----]
               alpha

In[15]:=
Variance[WeibullDistribution[alpha, beta]]

Out[15]=
    2               1   2               2
beta  (-Gamma[1 + -----]  + Gamma[1 + -----])
                  alpha               alpha

Generalized moment:

In[16]:=
Integrate[x^t PDF[WeibullDistribution[alpha, beta], x], 
 {x, 0, Infinity}] // PowerExpand

Out[16]=
    t             t
beta  Gamma[1 + -----]
                alpha


FORWARDED MESSAGE:

Subj:  [mg5309] Mathematica: Weibull distribution fnc.
From:  riglinbd at ml.wpafb.af.mil
To: mathgroup at smc.vnet.net
X-From: riglinbd at ml.wpafb.af.mil (Brian)
To: mathgroup at smc.vnet.net

I am trying to employ a Weibull distribution in one of the
notebooks I am working on.  However, Mathematica provides
scant if any information on how to use this function. Could
some one with more experience than I please enlighten me as
to how the Weibull distribution function found in the
Statistics'ContinuousDistribution package is used.


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