[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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