# Re: WEIBULL DISTRIBUTION

In a message dated 3/31/98 4:42:21 PM, Simona Bignami wrote:

>I've read the message you sent on 1 January 1997 about Weibull distribution.
>I'm working with Weibull distribution and I would like to fit data with a
>THREE-PARAMETERS Weibull distribution.
>How can accomplish this on Mathematica 3.0?

Needs["Statistics`ContinuousDistributions`"]

PDF[WeibullDistribution[alpha, beta], x]

(alpha*x^(-1 + alpha))/(beta^alpha*E^(x/beta)^alpha)

Domain[WeibullDistribution[alpha, beta]]

Interval[{0, Infinity}]

Mean[WeibullDistribution[alpha, beta]]

beta*Gamma[1 + alpha^(-1)]

I do not know what specific generalization you are interested in to
obtain a  three-parameter distribution.  However, one possible
generalization would be a probability density of the form

Exp[-(x/beta)^alpha]*x^(gamma - 1);

To find the normalization factor, the integral of the probability
density must
be one.

PowerExpand[Integrate[Exp[-(x/beta)^alpha]*x^(gamma - 1), {x, 0,
Infinity},
Assumptions -> {Re[alpha] > 0, Re[(1/beta)^alpha] > 0}]]

(beta^gamma*Gamma[gamma/alpha])/alpha

Therefor, the probability density is given by

f[alpha_, beta_, gamma_, x_] :=
(alpha*Exp[-(x/beta)^alpha]*x^(gamma -
1))/(beta^gamma*Gamma[gamma/alpha])

PowerExpand[Integrate[f[alpha, beta, gamma, x], {x, 0, Infinity},
Assumptions -> {Re[alpha] > 0, Re[(1/beta)^alpha] > 0}]]

1

This clearly reduces to the two-parameter form when gamma equals alpha.
The  mean is given by

PowerExpand[Integrate[x*f[alpha, beta, gamma, x], {x, 0, Infinity},
Assumptions -> {Re[alpha] > 0, Re[(1/beta)^alpha] > 0}]]

(beta*Gamma[(1 + gamma)/alpha])/Gamma[gamma/alpha]

To find the estimated values of the parameters (alpha, beta, gamma) for
a  given set of data, use the method of maximum likelihood (maximize
the  likelihood function -- see whatever books on advanced statistics
which you  have available).

Bob Hanlon

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