Re: WeibullDistribution
- To: mathgroup at smc.vnet.net
- Subject: [mg42491] Re: WeibullDistribution
- From: "Robert Nowak" <robert.nowak at ims.co.at>
- Date: Fri, 11 Jul 2003 02:57:52 -0400 (EDT)
- References: <bej6em$5p9$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
hello bill,
> data = Table[{x, a x^(a - 1) Exp[-(x^a)] + Random[Real, {-0.2, 0.2}] /.
{a ->7}}, {x, 0.1, 1.5, 0.1}];
"usualy" you dont have randomly noised PDF(x)-funktionsvalues at positions
x.
"usualy" you only have random values which are expected to obey a
distribution with a specific PDF.
in the "usual" case you therefore cant fit your data against the PDF.
i think you have to do some of bob hanlons or similar calculations.
altenativly you could partition the random values in n intervalls centerd
about x(i) (1>i>n), count c(i) the members of each interval, and fit this
{{x(1),c(1)}, ....{x(i),c(i)}, ......{x(n),c(n)} against the expected
PDF(x).
as i´m not a stats expert i could be completly wrong !? pleas correct me if
i am wrong.
regards robert
"Bill Rowe" <listuser at earthlink.net> wrote in message
news:bej6em$5p9$1 at smc.vnet.net...
> On 7/9/03 at 8:24 AM, gohtk at rocketmail.com (goh tat kean) wrote:
>
> > The formula for the probability density function of the general
> > Weibull distribution is given in:
>
> > http://www.itl.nist.gov/div898/handbook/eda/section3/eda3668.htm
>
> <snip>
>
> > Fit the dummy data by using NonlinearRegress to obtain a and b,
>
> > NonlinearRegress[data, PDF[WeibullDistribution[a, b], x], x, {a, b},
> > MaxIterations -> 1000000]
>
> Under ideal conditions this will work fine. But real world data is seldom
ideal and this method will generally not be either efficient or robust.
>
> For the Weibull distribution it is much better to do a linear regression
on Log[H] vs Log[x] where H is the cumulative hazard function since Log[H]
== a Log[x] - a Log[b]. Note for all distributions H = -Log[R] and R = 1-F
where R is the reliability function and F is the cumulative distribution
function
>