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MathGroup Archive 1998

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Re: WEIBULL DISTRIBUTION



In a message dated 4/1/98 5:05:01 PM, Simona Bignami wrote:

>In truth, when I spoke about three parameters Weibull distribution, I was
>thinking about the following generalization:
>
>f(x,a,b,m)= (b/m)*[((x-a)/m)^(b-1)]*exp[(-(x-a)/m)^b]     x>a, a>=0, b,m>0

Needs["Statistics`ContinuousDistributions`"]

The exponential term in your probability density is not written
correctly.   The correct probability density is

f[x_, a_, b_, m_] := 
	b/m*((x - a)/m)^(b - 1)*Exp[-((x - a)/m)^b]

This reduces to the two-parameter case for a equal to zero

PowerExpand[f[x, 0, b, m] == 
	PDF[WeibullDistribution[b, m], x]]

True

Integrate[f[x, a, b, m], {x, a, Infinity}]

(b*Integrate[((-a + x)/m)^(-1 + b)/E^((-a + x)/m)^b, 
     {x, a, Infinity}])/m

A change in variables is required to evaluate the integral.  Let  y =
(x-a)/m. Then the integral is

Integrate[m*f[x, a, b, m] /. x -> m*y + a, 
	{y, 0, Infinity}, Assumptions -> b > 0]

1

The mean is then

Integrate[m*x*f[x, a, b, m] /. x -> m*y + a, 
	{y, 0, Infinity}, Assumptions -> b > 0]

a + m*Gamma[1 + b^(-1)]

Mean[WeibullDistribution[b, m]]

m*Gamma[1 + b^(-1)]

Proceed in this direction.

Bob Hanlon



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