       Re: Understanding TransformedDistribution

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• Subject: [mg131897] Re: Understanding TransformedDistribution
• From: Itai Seggev <itais at wolfram.com>
• Date: Mon, 28 Oct 2013 23:21:15 -0400 (EDT)
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```On Mon, Oct 28, 2013 at 12:40:08AM -0400, Ernst H.K. Stelzer wrote:
> Through a question on StackExchange I came to know TransformedDistribution, which allows me to sort of develop new distributions based on the ones defined in Mathematica.  Instead of using my own functions, which work perfectly well, I now tried to repeat a number of older calculations using e.g. UniformDistribution.
>
> When I combine two via
>
> TransformedDistribution[2 (x - 4/5 y), {x \[Distributed] UniformDistribution[{-1, 1}],   y \[Distributed] UniformDistribution[{-(4/5), 4/5}]}]
>
> to generate a kind of annular function and then look at the PDF, I get
>
> PDF[%, x]
>
> \[Piecewise]	1/4	-(18/25)<=x<=18/25
> 1/256 (82-25 x)	18/25<x<82/25
> 1/256 (82+25 x)	-(82/25)<x<-(18/25)
> 0	True
>
> However, I expect, what I get when I enter
>
> 2 (PDF[UniformDistribution[{-1, 1}], x] -
>    4/5 PDF[UniformDistribution[{-(4/5), 4/5}], x])
>
> 2 ((\[Piecewise]	1/2	-1<=x<=1
> 0	True
> )-4/5 (\[Piecewise]	5/8	-(4/5)<=x<=4/5
> 0	True
> ))
>
> Does anybody have a clue which mistake I made?

For one thing, your expectation isn't the PDF of anything--it's integral is
2/5, not 1.

PDF is not in any sense a linear operation, because it must always integrate to
1.  TransformedDistribution[expr, dists] represnets the distribution of the
values of expr, given that its components are distribution according to dists.
This means your first input represent a distribution ranging from -82/25 (when
x is -1 and y 4/5) to 82/5 (when x is 1 and y 4/5).  Since x and y are
independent of piecewise constant distributions, you get a constant expectation
in the "middle" going up from and back down to zero linearly as you go the
maximum negative and positive values.

--
Itai Seggev
Mathematica Algorithms R&D
217-398-0700

```

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