Re: WEIBULL DISTRIBUTION
- To: mathgroup@smc.vnet.net
- Subject: [mg11872] Re: WEIBULL DISTRIBUTION
- From: Bob Hanlon <BobHanlon@aol.com>
- Date: Fri, 3 Apr 1998 03:45:31 -0500
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