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Problem with TransformedDistribution

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1

$begingroup$

I am trying to use Mathematica to obtain the probability distribution of $frac12(A + B)$ where $A$ and $B$ are independent random variables each distributed according to the uniform distribution, with lower and upper bounds of $L$ and $H$ respectively.

I suspect the distribution is triangular with lower and upper bounds of $L$ and $H$ respectively and mode equal to $frac12(A + B)$. However, I am having difficulty using TransformedDistribution to show that.

My code is:

[ScriptCapitalD] = TransformedDistribution[1/2 (A + B), B [Distributed] UniformDistribution[L, H], A [Distributed] UniformDistribution[L, H]]

probability-or-statistics

share|improve this question

edited 1 hour ago

user120911

asked 1 hour ago

user120911user120911

71328

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That was a typo. But I am still not getting what I expect.
  $endgroup$
  – user120911
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you try PDF[[ScriptCapitalD], y]?
  $endgroup$
  – JimB
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  PDF[[ScriptCapitalD], y] produces one expression with a denominator that looks correct, but the triangular distribution is split at the mode. Mathematica is not showing that. At least not in a way that is easy to see.
  $endgroup$
  – user120911
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you aware TriangularDistribution[] is built-in?
  $endgroup$
  – J. M. is slightly pensive♦
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why not check the PDFs? Simplify[PDF[TransformedDistribution[(a + b)/2, a, b [Distributed] UniformDistribution[l, h, l, h]], t] == PDF[TriangularDistribution[l, h], t], l < t < h]
  $endgroup$
  – J. M. is slightly pensive♦
  41 mins ago
  

  
  
  
  

 | 
show 3 more comments

1

$begingroup$

I am trying to use Mathematica to obtain the probability distribution of $frac12(A + B)$ where $A$ and $B$ are independent random variables each distributed according to the uniform distribution, with lower and upper bounds of $L$ and $H$ respectively.

I suspect the distribution is triangular with lower and upper bounds of $L$ and $H$ respectively and mode equal to $frac12(A + B)$. However, I am having difficulty using TransformedDistribution to show that.

My code is:

[ScriptCapitalD] = TransformedDistribution[1/2 (A + B), B [Distributed] UniformDistribution[L, H], A [Distributed] UniformDistribution[L, H]]

probability-or-statistics

share|improve this question

edited 1 hour ago

user120911

asked 1 hour ago

user120911user120911

71328

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That was a typo. But I am still not getting what I expect.
  $endgroup$
  – user120911
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you try PDF[[ScriptCapitalD], y]?
  $endgroup$
  – JimB
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  PDF[[ScriptCapitalD], y] produces one expression with a denominator that looks correct, but the triangular distribution is split at the mode. Mathematica is not showing that. At least not in a way that is easy to see.
  $endgroup$
  – user120911
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you aware TriangularDistribution[] is built-in?
  $endgroup$
  – J. M. is slightly pensive♦
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why not check the PDFs? Simplify[PDF[TransformedDistribution[(a + b)/2, a, b [Distributed] UniformDistribution[l, h, l, h]], t] == PDF[TriangularDistribution[l, h], t], l < t < h]
  $endgroup$
  – J. M. is slightly pensive♦
  41 mins ago
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

I am trying to use Mathematica to obtain the probability distribution of $frac12(A + B)$ where $A$ and $B$ are independent random variables each distributed according to the uniform distribution, with lower and upper bounds of $L$ and $H$ respectively.

I suspect the distribution is triangular with lower and upper bounds of $L$ and $H$ respectively and mode equal to $frac12(A + B)$. However, I am having difficulty using TransformedDistribution to show that.

My code is:

[ScriptCapitalD] = TransformedDistribution[1/2 (A + B), B [Distributed] UniformDistribution[L, H], A [Distributed] UniformDistribution[L, H]]

probability-or-statistics

share|improve this question

edited 1 hour ago

user120911

asked 1 hour ago

user120911user120911

71328

$endgroup$

[ScriptCapitalD] = TransformedDistribution[1/2 (A + B), B [Distributed] UniformDistribution[L, H], A [Distributed] UniformDistribution[L, H]]

share|improve this question

edited 1 hour ago

user120911

asked 1 hour ago

user120911user120911

71328

share|improve this question

edited 1 hour ago

user120911

asked 1 hour ago

user120911user120911

71328

asked 1 hour ago

user120911user120911

71328

asked 1 hour ago

user120911user120911

71328

2 Answers
2

active

oldest

votes

2

$begingroup$

You get what you expect if you do it it in two steps

[ScriptCapitalD] = 
 TransformedDistribution[x/2, 
 x [Distributed] TransformedDistribution[(A + B), 
 B [Distributed] UniformDistribution[L, H],
 A [Distributed] UniformDistribution[L, H]]]

(* TriangularDistribution[L, H] *)

share|improve this answer

answered 33 mins ago

Bob HanlonBob Hanlon

60.9k33597

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That is very nice!
  $endgroup$
  – user120911
  20 mins ago
  

  
  
  
  

add a comment | 

0

$begingroup$

PDF[[ScriptCapitalD]][z]

(((-30 + z)Sign[-30 + z])/2 - (-20 + z)
Sign[-20 + z] + ((-10 + z)*Sign[-10 + z])/2)/100

For plotting, assign values to L and H:

L = 10; H = 30;
Plot[Evaluate@PDF[[ScriptCapitalD]][x], x, 10, 30]

pdF[l_, h_] := Module[L = l, H = h, Evaluate[PDF[[ScriptCapitalD]]]]
Plot[Evaluate @ Flatten@Table[pdF[l, h][x], l, 0, 5, h, 10, 15], x, 0, 15,
 PlotRange -> All, 
 PlotLegends -> (Flatten @ Table[ToString@l, h, l, 0, 5, h, 10, 15])]

share|improve this answer

answered 1 hour ago

kglrkglr

189k10206424

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That confirms my intution, but can you get Mathematica to output the PDF for the triangular distribution? That is what I am having trouble doing.
  $endgroup$
  – user120911
  1 hour ago
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

You get what you expect if you do it it in two steps

[ScriptCapitalD] = 
 TransformedDistribution[x/2, 
 x [Distributed] TransformedDistribution[(A + B), 
 B [Distributed] UniformDistribution[L, H],
 A [Distributed] UniformDistribution[L, H]]]

(* TriangularDistribution[L, H] *)

share|improve this answer

answered 33 mins ago

Bob HanlonBob Hanlon

60.9k33597

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That is very nice!
  $endgroup$
  – user120911
  20 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

You get what you expect if you do it it in two steps

[ScriptCapitalD] = 
 TransformedDistribution[x/2, 
 x [Distributed] TransformedDistribution[(A + B), 
 B [Distributed] UniformDistribution[L, H],
 A [Distributed] UniformDistribution[L, H]]]

(* TriangularDistribution[L, H] *)

share|improve this answer

answered 33 mins ago

Bob HanlonBob Hanlon

60.9k33597

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That is very nice!
  $endgroup$
  – user120911
  20 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

You get what you expect if you do it it in two steps

[ScriptCapitalD] = 
 TransformedDistribution[x/2, 
 x [Distributed] TransformedDistribution[(A + B), 
 B [Distributed] UniformDistribution[L, H],
 A [Distributed] UniformDistribution[L, H]]]

(* TriangularDistribution[L, H] *)

share|improve this answer

answered 33 mins ago

Bob HanlonBob Hanlon

60.9k33597

$endgroup$

[ScriptCapitalD] = 
 TransformedDistribution[x/2, 
 x [Distributed] TransformedDistribution[(A + B), 
 B [Distributed] UniformDistribution[L, H],
 A [Distributed] UniformDistribution[L, H]]]

(* TriangularDistribution[L, H] *)

share|improve this answer

answered 33 mins ago

Bob HanlonBob Hanlon

60.9k33597

answered 33 mins ago

Bob HanlonBob Hanlon

60.9k33597

answered 33 mins ago

Bob HanlonBob Hanlon

60.9k33597

0

$begingroup$

PDF[[ScriptCapitalD]][z]

(((-30 + z)Sign[-30 + z])/2 - (-20 + z)
Sign[-20 + z] + ((-10 + z)*Sign[-10 + z])/2)/100

For plotting, assign values to L and H:

L = 10; H = 30;
Plot[Evaluate@PDF[[ScriptCapitalD]][x], x, 10, 30]

pdF[l_, h_] := Module[L = l, H = h, Evaluate[PDF[[ScriptCapitalD]]]]
Plot[Evaluate @ Flatten@Table[pdF[l, h][x], l, 0, 5, h, 10, 15], x, 0, 15,
 PlotRange -> All, 
 PlotLegends -> (Flatten @ Table[ToString@l, h, l, 0, 5, h, 10, 15])]

share|improve this answer

answered 1 hour ago

kglrkglr

189k10206424

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That confirms my intution, but can you get Mathematica to output the PDF for the triangular distribution? That is what I am having trouble doing.
  $endgroup$
  – user120911
  1 hour ago
  
  

  
  
  
  

add a comment | 

0

$begingroup$

PDF[[ScriptCapitalD]][z]

(((-30 + z)Sign[-30 + z])/2 - (-20 + z)
Sign[-20 + z] + ((-10 + z)*Sign[-10 + z])/2)/100

For plotting, assign values to L and H:

L = 10; H = 30;
Plot[Evaluate@PDF[[ScriptCapitalD]][x], x, 10, 30]

pdF[l_, h_] := Module[L = l, H = h, Evaluate[PDF[[ScriptCapitalD]]]]
Plot[Evaluate @ Flatten@Table[pdF[l, h][x], l, 0, 5, h, 10, 15], x, 0, 15,
 PlotRange -> All, 
 PlotLegends -> (Flatten @ Table[ToString@l, h, l, 0, 5, h, 10, 15])]

share|improve this answer

answered 1 hour ago

kglrkglr

189k10206424

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That confirms my intution, but can you get Mathematica to output the PDF for the triangular distribution? That is what I am having trouble doing.
  $endgroup$
  – user120911
  1 hour ago
  
  

  
  
  
  

add a comment | 

$begingroup$

PDF[[ScriptCapitalD]][z]

(((-30 + z)Sign[-30 + z])/2 - (-20 + z)
Sign[-20 + z] + ((-10 + z)*Sign[-10 + z])/2)/100

For plotting, assign values to L and H:

L = 10; H = 30;
Plot[Evaluate@PDF[[ScriptCapitalD]][x], x, 10, 30]

pdF[l_, h_] := Module[L = l, H = h, Evaluate[PDF[[ScriptCapitalD]]]]
Plot[Evaluate @ Flatten@Table[pdF[l, h][x], l, 0, 5, h, 10, 15], x, 0, 15,
 PlotRange -> All, 
 PlotLegends -> (Flatten @ Table[ToString@l, h, l, 0, 5, h, 10, 15])]

share|improve this answer

answered 1 hour ago

kglrkglr

189k10206424

$endgroup$

PDF[[ScriptCapitalD]][z]

L = 10; H = 30;
Plot[Evaluate@PDF[[ScriptCapitalD]][x], x, 10, 30]

pdF[l_, h_] := Module[L = l, H = h, Evaluate[PDF[[ScriptCapitalD]]]]
Plot[Evaluate @ Flatten@Table[pdF[l, h][x], l, 0, 5, h, 10, 15], x, 0, 15,
 PlotRange -> All, 
 PlotLegends -> (Flatten @ Table[ToString@l, h, l, 0, 5, h, 10, 15])]

share|improve this answer

answered 1 hour ago

kglrkglr

189k10206424

answered 1 hour ago

kglrkglr

189k10206424

answered 1 hour ago

kglrkglr

189k10206424