# PDF of a Transformed discrete distribution

Posted 5 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. Answer
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Posted 5 months ago
 You're writing a book or going through an existing book?Everything you wrote looks good except for one thing. For discrete distributions there is a dx value that is needed. However, that doesn't seem fix things.The following seems to get a bit farther: f[{x1_, x2_}] := Piecewise[{{1/3, {x1, x2} == {1, 1}}, {2/3, {x1, x2} == {2, 0}}}]; \[ScriptCapitalD] = ProbabilityDistribution[f[{x1, x2}], {x1, -10, 10, 1}, {x2, -10, 10, 1}]; Mean[\[ScriptCapitalD]]  $$\left\{\frac{5}{3},\frac{1}{3}\right\}$$ Variance[\[ScriptCapitalD]]  $$\left\{2/9, 2/9\right\}$$ Covariance[\[ScriptCapitalD]]  $$\left\{\left\{2/9, -(2/9)\right\}, \left\{-(2/9), 2/9\right\}\right\}$$ \[ScriptCapitalD]2 = TransformedDistribution[2 {x1, x2}, {x1, x2} \[Distributed] \[ScriptCapitalD]] // FullSimplify; Mean[\[ScriptCapitalD]2]  $$\left\{\frac{10}{3},\frac{2}{3}\right\}$$ Variance[\[ScriptCapitalD]2]  $$\left\{\frac{8}{9},\frac{8}{9}\right\}$$ Covariance[\[ScriptCapitalD]2]  $$\left\{\left\{8/9, -(8/9)\right\}, \left\{-(8/9), 8/9\right\}\right\}$$But RandomVariate doesn't work with the above. For your particular example things can be rewritten so that RandomVariate works d = TransformedDistribution[2 {x + 1, 1 - x}, x \[Distributed] BernoulliDistribution[2/3]]; RandomVariate[d, 10] (* {{4, 0}, {2, 2}, {4, 0}, {4, 0}, {4, 0}, {4, 0}, {4, 0}, {2, 2}, {4, 0}, {4, 0}} *) I suspect that ProbabilityDistribution and/or TransformedDistribution just can't handle the structure you have. (I'd like to be wrong about that.) Answer
Posted 5 months ago
 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. Answer
Posted 5 months ago
 The dx term is found in the online reference for ProbabilityDistribution. Answer
Posted 5 months ago Answer
Posted 5 months ago
 From the online reference:  Answer
Posted 5 months ago
 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. Thank u for everything. Answer
Posted 5 months ago
 No. Read the online reference of Discrete Multivariate Distributions under Scope. You still need dx. Answer
Posted 5 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] Now the transformation part of the exercise: dist2 = TransformedDistribution[x + 1, x \[Distributed] dist] PDF[dist2, y]  Answer