Gumbel distribution: Difference between revisions

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	A more practical way of using the distribution could be		A more practical way of using the distribution could be
	p=exp(-exp(-0.367*(A-x)/(A-M)) ;-.367=ln(-ln(.5))		:p=exp(-exp(-0.367*(A-x)/(A-M)) ;-.367=ln(-ln(.5))
	where M is the ]. To fit values one could get the median		where M is the ]. To fit values one could get the median
	straight away and then vary A until it fits the list of values.		straight away and then vary A until it fits the list of values.

	Its variates(ie to get a list of random values) can be given as ;		Its variates(ie to get a list of random values) can be given as ;
	x=A-B*ln(-ln(rnd))		:x=A-B*ln(-ln(rnd))

	Its percentiles can be given by ;		Its percentiles can be given by ;
	x=A-B*ln(-ln(p))		:x=A-B*ln(-ln(p))

	ie Q1=A-B*ln(-ln(.25))		ie Q1=A-B*ln(-ln(.25))
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	The median is A-B*ln(-ln(.5))		The median is A-B*ln(-ln(.5))

	Q3=A-B*ln(-ln(.75))		:Q3=A-B*ln(-ln(.75))

	The mean is A+g*B '~~g=Eulers~~ constant = .57721		The mean is A+g*B where g =] = .57721.

	The sd = B * Pi()* sqr(1/6)		The sd = B * Pi()* sqr(1/6)

Revision as of 13:04, 9 March 2004

In probability theory and statistics the Gumbel distribution is used to find the minimum (or the maximum) of a number of samples of various distributions. For example we would use it to find the maximum level of a river in a particular year if we had the list of maximum values for the past ten years. It is therefore useful in predicting the chance that an extreme earthquake, flood or other natural disaster will occur.

The distribution of the samples could be of the normal or exponential type. The Gumbel distribution, and similar distributions, are used in extreme value theory.

In particular, the Gumbel distribution is a special case of the Fisher-Tippett distribution, also known as the log-Weibull distribution, whose cumulative distribution function is

p=\exp(-\exp((\mu -x)/\beta ).

The Gumbel distribution is the case where μ = 0 and β = 1.

A more practical way of using the distribution could be

p=exp(-exp(-0.367*(A-x)/(A-M)) ;-.367=ln(-ln(.5))

where M is the median. To fit values one could get the median straight away and then vary A until it fits the list of values.

Its variates(ie to get a list of random values) can be given as ;

x=A-B*ln(-ln(rnd))

Its percentiles can be given by ;

x=A-B*ln(-ln(p))

ie Q1=A-B*ln(-ln(.25))

The median is A-B*ln(-ln(.5))

Q3=A-B*ln(-ln(.75))

The mean is A+g*B where g =Euler's constant = .57721.

The sd = B * Pi()* sqr(1/6)

Its mode is A