1
$\begingroup$

This question is an extension of this question, if we change the initial values FindDistributionParameters[data1, dist, {{a, 0.1}, {b, 0.1}, {c, 0.1}, {al, 0.1}, {be, 0.1}, {ph, 0.1}}] to FindDistributionParameters[data1, dist, {{a, 0.2}, {b, 0.1}, {c, 0.1}, {al, 0.1}, {be, 0.1}, {ph, 0.1}}] parameter estimates also change. So my question is if we have no idea about initial values then how we search best initial values or best MLE estimates which close to empirical distribution of data1. The empirical distribution with two parameters which are estimated with initial value's given above are plotted as

enter image description here

Not a best fit. The code of above plot is:

data1 = {275, 13, 147, 23, 181, 30, 65, 10, 300, 173, 106, 300, 300, 212, 300, 300, 300, 2, 261, 293, 88, 247, 28, 143, 300, 23, 300, 80, 245, 266}; g1 = NIntegrate[(2.51339*2.51952*3.21532)/Beta[2.87484, 0.0825351]* y^-(3.21532 + 1)*(1 + 0.0609138*2.51952*y^-3.21532)^-((2.87484*2.51339)/0.0609138 + 1)*(1 - (1 + 0.0609138*2.51952*y^-3.21532)^-(2.51339/ 0.0609138))^(0.0825351 - 1), {y, 0, x}];

g2 = NIntegrate[(1.98156*1.98026*3.2635)/Beta[2.64639, 0.090212]* y^-(3.2635 + 1)*(1 + 0.0328252*1.98026*y^-3.2635)^-((2.64639*1.98156)/0.0328252 + 1)*(1 - (1 + 0.0328252*1.98026*y^-3.2635)^-(1.98026/ 0.0328252))^(0.090212 - 1), {y, 0, x}];

Plot[{g1, g2, CDF[EmpiricalDistribution[data1], x]}, {x, 2, 400},
Exclusions -> None]

Improve this question
$\endgroup$
2
  • $\begingroup$ Do you have the generating functions of your plots? $\endgroup$
    – Rod
    Commented Jul 20, 2013 at 14:27
  • $\begingroup$ I add code in question $\endgroup$
    – SAAN
    Commented Jul 20, 2013 at 14:31

1 Answer 1

Reset to default
1
$\begingroup$

Simply compare the value of loglikelihood of your estimates....

A = FindDistributionParameters[data1, dist, {{a, 0.1}, {b, 0.1}, {c, 0.1}, {al, 0.1}, {be, 0.1}, {ph,  0.1}}]

B = FindDistributionParameters[data1,  dist, {{a, 0.2}, {b, 0.1}, {c, 0.1}, {al, 0.1}, {be, 0.1}, {ph, 0.1}}]

LogLikelihood[dist /. A, data1]
-211.801

LogLikelihood[dist /. B, data1]
-212.959

So A is slightly a better fit.

Improve this answer
$\endgroup$
2
  • $\begingroup$ LogLikelihood tell us in two parameter estimates which one is the best, but my question is still stand which fit is best among all range of parameters `a,b,c,al,be,ph>0'. $\endgroup$
    – SAAN
    Commented Jul 21, 2013 at 4:17
  • $\begingroup$ I would recommend you to run a table with multiple parameters as the starting values then compare the loglikelihood of all of them!! $\endgroup$
    – Morry
    Commented Jul 22, 2013 at 9:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.