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
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?
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 ___ 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. 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)? ___ 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: 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 ___ 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:
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)?
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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