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Naouar El Khadiri

PDF of a Transformed discrete distribution

Naouar El Khadiri
Posted 9 months ago
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Hi all, This code returns results not matching the results of the book I am working on.

f[x_] := Piecewise[{{1/3, x == {1, 1}}, {2/3, x == {2, 0}}}];
\[ScriptCapitalD] := 
  ProbabilityDistribution[
   f[{x1, x2}], {x1, -Infinity, Infinity}, {x2, -Infinity, Infinity}]; 
PDF[TransformedDistribution[
   2 {x1, x2}, {x1, x2} \[Distributed] \[ScriptCapitalD]], {y1, 
   y2}] // FullSimplify

i should have a result like P(Y==y)=P(g^-1(y)) for g an increasing function which is not the result Mathematica proposes. Any suggestions? Thx.
POSTED BY: Naouar El Khadiri
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Jim Baldwin
Jim Baldwin
Posted 9 months ago
POSTED BY: Jim Baldwin
Answer

Thx for the feedback, I was really doubting myself. As per your comment "For discrete distributions there is a dx value that is needed", please share references since the book provided a satisfactory demonstration. Maybe you are referring to a continuous function f which is not the case in this problem.
POSTED BY: Naouar El Khadiri
Answer
Jim Baldwin
Jim Baldwin
Posted 9 months ago

The dx term is found in the online reference for ProbabilityDistribution.

POSTED BY: Jim Baldwin
Answer

thx, please help me improve and share the reference. here's mine https://www.statlect.com/fundamentals-of-probability/functions-of-random-variables-and-their-distribution#hid9
POSTED BY: Naouar El Khadiri
Answer
Jim Baldwin
Jim Baldwin, Retired
Posted 9 months ago

From the online reference:

ProbabilityDistribution for discrete distributions
POSTED BY: Jim Baldwin
Answer

I see, the dx here is the increment of x to move from xmin to xmax However, I am using the 1st definition in the case of multivariate. no need for dx. enter image description here Thank u for everything.
POSTED BY: Naouar El Khadiri
Answer
Jim Baldwin
Jim Baldwin, Retired
Posted 9 months ago

No. Read the online reference of Discrete Multivariate Distributions under Scope. You still need dx.
POSTED BY: Jim Baldwin
Answer
Jim Baldwin
Jim Baldwin, Retired
Posted 9 months ago

Your example I think just isn't the kind of probability distribution that ProbabilityDistribution and TransformedDistribution can handle. However, those two functions do handle Exercise 3 of the reference you posted:

dist = ProbabilityDistribution[x^2/30, {x, 1, 4, 1}]
PDF[dist, x]

Discrete distribution

Now the transformation part of the exercise:

dist2 = TransformedDistribution[x + 1, x \[Distributed] dist]
PDF[dist2, y]

Transformed distribution
POSTED BY: Jim Baldwin
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