Re: More Inquiries

*To*: mathgroup at smc.vnet.net*Subject*: [mg91206] Re: More Inquiries*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>*Date*: Sat, 9 Aug 2008 07:46:56 -0400 (EDT)*References*: <g7h9o1$b8u$1@smc.vnet.net>

Hi, in your code below, you worte pdf = BurrDistribution[x1, c, k] so you *know* that the PDF is BurrDistribution[x1, c, k] PDF[BurrDistribution[c, k], t] and this is complete nonsense because Mathematica don't know the BurrDistribution[] but you know it .. and PDF[BurrDistribution[c, k], t]=BurrDistribution[t, c, k] So to plot it Plot[ BurrDistribution[t, 37.81151579009424, 0.01356141249769735], {t, 2, 21}] should work. Regards Jens Dewi Anggraini wrote: > Hi All > > I've tried this following formula to Gamma distribution, it estimates the unknown parameters very well > > n = 69; > x1 = {3, 5, 6, 14, 11, 7, 7, 10, 11, 5, 4, 19, 9, 2, 8, 5, 6, 6, 5, > 5, > 4, 5, 5, 6, 8, 5, 8, 6, 5, 16, 5, 18, 16, 21, 6, 3, 16, 8, 3, > 8, > 11, 2, 3, 4, 8, 7, 9, 10, 8, 11, 8, 10, 9, 12, 12, 9, 6, 12, > 3, 9, > 14, 7, 4, 13, 8, 14, 5, 8, 2}; > pdf = PDF[GammaDistribution[\[Lambda], \[Beta]], x1] > logl = Plus @@ Log[pdf] > maxlogl = FindMinimum[-logl, {\[Lambda], 1}, {\[Beta], 1}] > mle = maxlogl[[2]] > > {190.145, {\[Lambda] -> 3.72763, \[Beta] -> 2.16947}} > > {\[Lambda] -> 3.72763, \[Beta] -> 2.16947} > > > and also the program produces the algebraic form of Gamma distribution by the following command and it calculates the probability of non-conformance (above the specification limit). The program also can produce the pdf graph. > > > PDF[GammaDistribution[\[Lambda], \[Beta]], t] > > (\[ExponentialE]^-t/\[Beta] t^(-1 + \[Lambda]) \ > \[Beta]^-\[Lambda])/Gamma[\[Lambda]] > > Integrate[ > PDF[GammaDistribution[3.7276258602234646, 2.169465718015002], t], {t, > 8, Infinity}] > > 0.439226 > > Plot[PDF[GammaDistribution[3.7276258602234646, 2.169465718015002], > t], {t, 2, 21}] > > > However, when I apply the same command towards Burr distribution (the following program below), especially for the case of producing algebraic formula of the pdf, calculating the probability of non-conformance and drawing the pdf graph (the last three commands), the program does not work very well. Do I have done something wrong in the program? > > > > n = 69; > x1 = {3, 5, 6, 14, 11, 7, 7, 10, 11, 5, 4, 19, 9, 2, 8, 5, 6, 6, 5, > 5, > 4, 5, 5, 6, 8, 5, 8, 6, 5, 16, 5, 18, 16, 21, 6, 3, 16, 8, 3, > 8, > 11, 2, 3, 4, 8, 7, 9, 10, 8, 11, 8, 10, 9, 12, 12, 9, 6, 12, > 3, 9, > 14, 7, 4, 13, 8, 14, 5, 8, 2}; > BurrDistribution[x1_, c_, > k_] := (c*k)*(x1^(c - 1)/(1 + x1^c)^(k + 1)) > pdf = BurrDistribution[x1, c, k] > logl = Plus @@ Log[pdf] > maxlogl = FindMinimum[{-logl, c > 0 && k > 0}, {c, 1}, {k, 2}, > MaxIterations -> 1000] > mle = maxlogl[[2]] > > {249.647, {c -> 37.8115, k -> 0.0135614}} > > {c -> 37.8115, k -> 0.0135614} > > > PDF[BurrDistribution[c, k], t] > > Integrate[ > PDF[BurrDistribution[37.81151579009424, 0.01356141249769735], t] {t, > 21, Infinity}] > > Plot[PDF[BurrDistribution[37.81151579009424, 0.01356141249769735], > t], {t, 2, 21}] > > > I already got assistance from some of you about gaining MLE of Burr distribution for my data and as recommended now I can run it very well. > > Now, please assist me again to find where I got wrong in my second case here (to find the algebraic form of Burr, do the integration and draw the graph). Thank You. > > > Kindly Regards, > Dewi > > >