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
>
>   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
>
>
>

```

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