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 _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion