1
$\begingroup$
data = {1.312, 1.314, 1.479, 1.552, 1.700, 1.803, 1.861, 1.865, 1.944,
    1.958, 1.966, 1.997, 2.006, 2.021, 2.027, 2.055, 2.063, 2.098, 
   2.14, 2.179, 2.224, 2.240, 2.253, 2.270, 2.272, 2.274, 2.301, 
   2.301, 2.359, 2.382, 2.382, 2.426, 2.434, 2.435, 2.478, 2.490, 
   2.511, 2.514, 2.535, 2.554, 2.566, 2.57, 2.586, 2.629, 2.633, 
   2.642, 2.648, 2.684, 2.697, 2.726, 2.770, 2.773, 2.800, 2.809, 
   2.818, 2.821, 2.848, 2.88, 2.954, 3.012, 3.067, 3.084, 3.090, 
   3.096, 3.128, 3.233, 3.433, 3.585, 3.585};

I constructed a basic Pareto distribution and fitted it to the data to obtain the best-fit value of the distribution parameter $\lambda$:

dist = ProbabilityDistribution[{"CDF", 1 - x^(-\[Lambda])}, {x, 0, \[Infinity]}, Assumptions -> {\[Lambda] > 0}]


mle = FindDistributionParameters[data, dist]

This returned {\[Lambda] -> 1.14273}

I then wanted to plot fitted vs predicted values so I tried ProbabilityScalePlot[data, dist] but it returned an error:

an error

I ultimately want to calculate this DistributionFitTest[data, dist, {"TestDataTable", "AndersonDarling"}]: enter image description here

I would like to use my custom distribution, rather than a built-in from Mathematica, and i want to plot its empirical CDF.

$\endgroup$
2
  • $\begingroup$ Your custom distribution is equivalent to the built-in ParetoDistribution[1, lambda]. Have you tried using that? $\endgroup$
    – MarcoB
    Mar 13, 2022 at 15:20
  • $\begingroup$ no i want to write my custom distribution in mathematica $\endgroup$
    – A Day
    Mar 13, 2022 at 15:56

1 Answer 1

1
$\begingroup$
data = {1.312, 1.314, 1.479, 1.552, 1.700, 1.803, 1.861, 
        1.865, 1.944, 1.958, 1.966, 1.997, 2.006, 2.021,
        2.027, 2.055, 2.063, 2.098, 2.14, 2.179, 2.224,
        2.240, 2.253, 2.270, 2.272, 2.274, 2.301, 2.301,
        2.359, 2.382, 2.382, 2.426, 2.434, 2.435, 2.478,
        2.490, 2.511, 2.514, 2.535, 2.554, 2.566, 2.57,
        2.586, 2.629, 2.633, 2.642, 2.648, 2.684, 2.697,
        2.726, 2.770, 2.773, 2.800, 2.809, 2.818, 2.821,
        2.848, 2.88, 2.954, 3.012, 3.067, 3.084, 3.090,
        3.096, 3.128, 3.233, 3.433, 3.585, 3.585};

dist[λ_] = ProbabilityDistribution[{"CDF", 1 - x^(-λ)}, {x, 1, ∞}];

Λ = λ /. FindDistributionParameters[data, dist[λ]]
(*    1.14273    *)

ProbabilityPlot[data, dist[Λ]]

enter image description here

DistributionFitTest[data, dist[Λ], {"TestDataTable", "AndersonDarling"}]

$$ \begin{array}{l|ll} \text{} & \text{Statistic} & \text{P-Value} \\ \hline \text{Anderson-Darling} & 18.343 & \text{8.698112014204895$\grave{ }$*${}^{\wedge}$-6} \\ \end{array} $$

Notes:

I've modified the domain of the given CDF to $[1,\infty)$ so that it stays within the required range $[0,1]$.

The empirical distribution is EmpiricalDistribution[data]:

datacdf[x_] = CDF[datadist][x];
Plot[datacdf[x], {x, 1, 4}, PlotPoints -> 100]

enter image description here

$\endgroup$

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.