Defining a new distribution on wolfram Mathematica

Questions
1
Open in Wolfram Cloud ↗

Hello everyone!!
How can i implement a new distribution in Mathematica …i tried using the command

Kiesdist[\[Lambda]_,\[Alpha]_]=ProbabilityDistribution[(\[Alpha] \[Lambda])/(x - 1)^2 (x/(
x - 1))^(\[Alpha] - 1) Exp[-\[Lambda] (x/(x - 1))^\[Alpha]], {x, 0,
1}, Assumptions -> {\[Lambda] > 0, \[Alpha] > 0}]

As i am trying to simulate estimates for this distribution using the following command:

FindDistributionParameters[data, 
 testDistribution[\[Lambda], \[Alpha]], 
 ParameterEstimator -> "MaximumLikelihood"]

but i keep on getting an error message that it isn’t a recognized distribution
anyone knows how can i define this distribution in a way that allows me to deal with it as the built in ones??

2

Would you edit the code so that it can be pasted into a notebook? Alternatively, attaching a notebook would be helpful. What you’ve given is missing a few backslashes. For example: [Alpha] instead of \[Alpha].

3

Perhaps you forgot the underscores in Kiesdist[\[Lambda]_, \[Alpha]_]. However, your function has divergent integral:

With[{\[Lambda] = 1, \[Alpha] = 1}, 
 Integrate[(\[Alpha]  \[Lambda])/(x - 1)^2  (x/(x - 1))^(\[Alpha] - 
      1)  Exp[-\[Lambda]  (x/(x - 1))^\[Alpha]], {x, 0, 1}]]

it cannot be a probability distribution.

4
Open in Wolfram Cloud ↗

here is the notebook ,will you please give it a look? :slight_smile:
the function i am working with is the pdf of the kies distribution …i got this form of the pdf from a past research

5

i am sorry for the inconvenience,i have attached the notebook :slight_smile:

6

S = Integrate[([Alpha] [Lambda])/(x - 1)^2 (x/(x - 1))^([Alpha] -
1) Exp[-[Lambda] (x/(x - 1))^[Alpha]], {x, 0, t},
Assumptions → {0 < t <= 1, [Alpha] > 0, [Lambda] > 0}]

(*Mathematica Can compute the CDF*)

(* -1 + E^(-E^(I \[Pi] \[Alpha]) (-(t/(-1 + t)))^\[Alpha] \[Lambda]) *)

(*Is not the same as: 1-\[ExponentialE]^(-\[Lambda]((t/(-1+t))^\[Alpha]) ) what you say ? *)

1 - E^(-\[Lambda] ((x/(-1 + x))^\[Alpha]) ) /. x -> 1/2 /. \[Alpha] ->1/3 /. \[Lambda] -> 2 // N
(*1.05907 + 0.363107 I*)

 (* Can there be complex numbers ?*)

Regards M.I.

7

When alpha is equal to 2, your probability density function is negative. When alpha is noninteger, your probability density function is complex-valued.

8

Your pdf is the opposite of the derivative of the cdf. Also your cdf as per the research

1 - E^(-\[Lambda] ((x/(-1 + x))^\[Alpha]) )

has some problems: it is complex-valued when alpha is noninteger. It is negative, decreasing and unbounded when alpha is an odd integer. It is only plausible when alpha is an even integer.

9

Wait a minute, with a little change in sign the problems go away:

kiesDistribution[\[Lambda]_, \[Alpha]_] :=
 ProbabilityDistribution[(\[Alpha] \[Lambda])/(x - 1)^2 (x/(-x + 
     1))^(\[Alpha] - 1)  Exp[-\[Lambda]  (x/(-x + 1))^\[Alpha]],
  {x, 0, 1},
  Assumptions -> {\[Lambda] > 0, \[Alpha] > 0}]
10

omgg thank you so much!!!