[Numpy-discussion] weibull distribution has only one parameter?

Ryan May rmay@ou....
Mon Nov 12 11:21:31 CST 2007


D.Hendriks (Dennis) wrote:
> Alan G Isaac wrote:
>> On Mon, 12 Nov 2007, "D.Hendriks (Dennis)" apparently wrote: 
>>   
>>> All of this makes me doubt the correctness of the formula 
>>> you proposed. 
>>>     
>> It is always a good idea to hesitate before doubting Robert.
>> <URL:http://en.wikipedia.org/wiki/Weibull_distribution#Generating_Weibull-distributed_random_variates>
>>
>> hth,
>> Alan Isaac
>>   
> So, you are saying that it was indeed correct? That still leaves the 
> question why I can't seem to confirm that in the figure I mentioned (red 
> and green lines). Also, if you refer to X = lambda*(-ln(U))^(1/k) as 
> 'proof' for the validity of the formula, I have to ask if 
> Weibull(a,Size) does actually correspond to (-ln(U))^(1/a)?
> 

Have you actually looked at a histogram of the random variates generated 
this way to see if they are wrong?

Multiplying the the individual random values by a number changes the 
distribution differently than multiplying the distribution/density 
function by a number.

Ryan

-- 
Ryan May
Graduate Research Assistant
School of Meteorology
University of Oklahoma


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