*To*: mathgroup@smc.vnet.net*Subject*: [mg11851] Re: WEIBULL DISTRIBUTION*From*: Bob Hanlon <BobHanlon@aol.com>*Date*: Wed, 1 Apr 1998 00:35:54 -0500

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