PDF of a Transformed discrete distribution

Questions
1

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.

2

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.)

3

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.

4

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

5

thx, please help me improve and share the reference.
here’s mine

6

From the online reference:

ProbabilityDistribution.png
ProbabilityDistribution.png1836×129 74.4 KB

7

I see, the dx here is the increment of x to move from x_min to x_max
However, I am using the 1st definition in the case of multivariate. no need for dx.

mathematica.jpg
mathematica.jpg3375×309 104 KB

Thank u for everything.

8

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]

discretedistribution.png

Now the transformation part of the exercise:

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

discretedistribution2.png

9

No. Read the online reference of Discrete Multivariate Distributions under Scope. You still need dx.