Problem with TransformedDistribution












2












$begingroup$


I am trying to use Mathematica to obtain the probability distribution of $frac{1}{2}(A + B)$ 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 $frac{1}{2}(A + B)$. 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}]}]









share|improve this question











$endgroup$












  • $begingroup$
    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
    58 mins ago
















2












$begingroup$


I am trying to use Mathematica to obtain the probability distribution of $frac{1}{2}(A + B)$ 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 $frac{1}{2}(A + B)$. 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}]}]









share|improve this question











$endgroup$












  • $begingroup$
    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
    58 mins ago














2












2








2





$begingroup$


I am trying to use Mathematica to obtain the probability distribution of $frac{1}{2}(A + B)$ 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 $frac{1}{2}(A + B)$. 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}]}]









share|improve this question











$endgroup$




I am trying to use Mathematica to obtain the probability distribution of $frac{1}{2}(A + B)$ 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 $frac{1}{2}(A + B)$. 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















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago







user120911

















asked 1 hour ago









user120911user120911

72828




72828












  • $begingroup$
    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
    58 mins ago


















  • $begingroup$
    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
    58 mins ago
















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




$begingroup$
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




$begingroup$
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






$begingroup$
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




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




1




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
58 mins ago




$begingroup$
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
58 mins ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

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









$endgroup$













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



















1












$begingroup$

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}]


enter image description here



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}}])]


enter image description here






share|improve this answer









$endgroup$













  • $begingroup$
    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













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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









$endgroup$













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
















3












$begingroup$

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









$endgroup$













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














3












3








3





$begingroup$

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









$endgroup$



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












share|improve this answer



share|improve this answer










answered 50 mins ago









Bob HanlonBob Hanlon

60.9k33597




60.9k33597












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


















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
















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




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











1












$begingroup$

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}]


enter image description here



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}}])]


enter image description here






share|improve this answer









$endgroup$













  • $begingroup$
    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


















1












$begingroup$

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}]


enter image description here



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}}])]


enter image description here






share|improve this answer









$endgroup$













  • $begingroup$
    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
















1












1








1





$begingroup$

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}]


enter image description here



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}}])]


enter image description here






share|improve this answer









$endgroup$



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}]


enter image description here



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}}])]


enter image description here







share|improve this answer












share|improve this answer



share|improve this answer










answered 1 hour ago









kglrkglr

189k10206424




189k10206424












  • $begingroup$
    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




















  • $begingroup$
    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


















$begingroup$
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






$begingroup$
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




















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