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-distrib...>
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