Re: [Numpy-discussion] weibull distribution has only one parameter?
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.
>> 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)?
double rk_standard_exponential(rk_state *state)
{
/* We use -log(1-U) since U is [0, 1) */
return -log(1.0 - rk_double(state));
}
double rk_weibull(rk_state *state, double a)
{
return pow(rk_standard_exponential(state), 1./a);
}
Like Ryan says, multiplying a random deviate by a number is different from
multiplying the PDF by a number. Multiplying the random deviate by lambda is
equivalent to transforming pdf(x) to pdf(x/lambda) not lambda*pdf(x).
--
Robert Kern
"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco
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Re: [Numpy-discussion] weibull distribution has only one parameter?
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. >> 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 ___ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] weibull distribution has only one parameter?
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. 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)? ___ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] weibull distribution has only one parameter?
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. http://en.wikipedia.org/wiki/Weibull_distribution#Generating_Weibull-distributed_random_variates> hth, Alan Isaac ___ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] weibull distribution has only one parameter?
Robert Kern wrote: D.Hendriks (Dennis) wrote: According to (for instance) http://en.wikipedia.org/wiki/Weibull_distribution the Weibull distribution has two parameters: lambda > 0 is the scale parameter (real) and k > 0 is the shape parameter (real). However, the numpy.random.weibull function has only a single 'a' parameter (except for the size parameter which indicates the size of the array to fill with values - this is NOT a parameter of the distribution itself). My question is how this 'a' parameter translates to the Weibull distribution as it 'normally' is and how to sample the distribution when I have the lambda and k parameters? lambda * numpy.random.weibull(k) Thanks for the quick replay. However, when I look at the image of the probability density function at http://en.wikipedia.org/wiki/Weibull_distribution I see a red line and a green line, both with k=2. The red line is for lambda=0.5 and the green for lambda=1.0. The green line is not only half the height of the red one (while double the lambda factor!), but also has its mean a bit more to the right. Looking at the formulas on the same page, this makes sense. All of this makes me doubt the correctness of the formula you proposed... ___ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] weibull distribution has only one parameter?
D.Hendriks (Dennis) wrote: > According to (for instance) > http://en.wikipedia.org/wiki/Weibull_distribution the Weibull > distribution has two parameters: lambda > 0 is the scale parameter > (real) and k > 0 is the shape parameter (real). However, the > numpy.random.weibull function has only a single 'a' parameter (except > for the size parameter which indicates the size of the array to fill > with values - this is NOT a parameter of the distribution itself). My > question is how this 'a' parameter translates to the Weibull > distribution as it 'normally' is and how to sample the distribution when > I have the lambda and k parameters? lambda * numpy.random.weibull(k) -- Robert Kern "I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth." -- Umberto Eco ___ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
[Numpy-discussion] weibull distribution has only one parameter?
According to (for instance) http://en.wikipedia.org/wiki/Weibull_distribution the Weibull distribution has two parameters: lambda > 0 is the scale parameter (real) and k > 0 is the shape parameter (real). However, the numpy.random.weibull function has only a single 'a' parameter (except for the size parameter which indicates the size of the array to fill with values - this is NOT a parameter of the distribution itself). My question is how this 'a' parameter translates to the Weibull distribution as it 'normally' is and how to sample the distribution when I have the lambda and k parameters? ___ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
