Problem with TransformedDistribution
4
$begingroup$
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}]}]
probability-or-statistics distributions
edited Mar 24 at 12:59
J. M. is slightly pensive♦
98.6k10309466
asked Mar 23 at 19:06
user120911user120911
75128
$endgroup$
|
show 4 more comments
4
$begingroup$
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}]}]
probability-or-statistics distributions
edited Mar 24 at 12:59
J. M. is slightly pensive♦
98.6k10309466
asked Mar 23 at 19:06
user120911user120911
75128
$endgroup$
$begingroup$
That was a typo. But I am still not getting what I expect.
$endgroup$
– user120911
Mar 23 at 19:16
$begingroup$
Did you tryPDF[[ScriptCapitalD], y]?
$endgroup$
– JimB
Mar 23 at 19:18
$begingroup$
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
Mar 23 at 19:25
2
$begingroup$
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♦
Mar 23 at 20:04
1
$begingroup$
I think I have found a bug! Try the following two expressions. The only difference is that I use U in one and H in the other for the upper limit.Simplify[PDF[TransformedDistribution[(A + B)/2, {A, B} [Distributed] UniformDistribution[{{L, U}, {L, U}}]], t] == PDF[TriangularDistribution[{L, U}], t], L < t < U]andSimplify[PDF[TransformedDistribution[(A + B)/2, {A, B} [Distributed] UniformDistribution[{{L, H}, {L, H}}]], t] == PDF[TriangularDistribution[{L, H}], t], L < t < H]
$endgroup$
– user120911
Mar 23 at 20:35
|
show 4 more comments
4
4
4
$begingroup$
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}]}]
probability-or-statistics distributions
edited Mar 24 at 12:59
J. M. is slightly pensive♦
98.6k10309466
asked Mar 23 at 19:06
user120911user120911
75128
$endgroup$
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 distributions
probability-or-statistics distributions
edited Mar 24 at 12:59
J. M. is slightly pensive♦
98.6k10309466
asked Mar 23 at 19:06
user120911user120911
75128
edited Mar 24 at 12:59
J. M. is slightly pensive♦
98.6k10309466
asked Mar 23 at 19:06
user120911user120911
75128
edited Mar 24 at 12:59
J. M. is slightly pensive♦
98.6k10309466
edited Mar 24 at 12:59
J. M. is slightly pensive♦
98.6k10309466
edited Mar 24 at 12:59
J. M. is slightly pensive♦
98.6k10309466
98.6k10309466
asked Mar 23 at 19:06
user120911user120911
75128
asked Mar 23 at 19:06
user120911user120911
75128
asked Mar 23 at 19:06
user120911user120911
75128
75128
$begingroup$
That was a typo. But I am still not getting what I expect.
$endgroup$
– user120911
Mar 23 at 19:16
$begingroup$
Did you tryPDF[[ScriptCapitalD], y]?
$endgroup$
– JimB
Mar 23 at 19:18
$begingroup$
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
Mar 23 at 19:25
2
$begingroup$
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♦
Mar 23 at 20:04
1
$begingroup$
I think I have found a bug! Try the following two expressions. The only difference is that I use U in one and H in the other for the upper limit.Simplify[PDF[TransformedDistribution[(A + B)/2, {A, B} [Distributed] UniformDistribution[{{L, U}, {L, U}}]], t] == PDF[TriangularDistribution[{L, U}], t], L < t < U]andSimplify[PDF[TransformedDistribution[(A + B)/2, {A, B} [Distributed] UniformDistribution[{{L, H}, {L, H}}]], t] == PDF[TriangularDistribution[{L, H}], t], L < t < H]
$endgroup$
– user120911
Mar 23 at 20:35
|
show 4 more comments
$begingroup$
That was a typo. But I am still not getting what I expect.
$endgroup$
– user120911
Mar 23 at 19:16
$begingroup$
Did you tryPDF[[ScriptCapitalD], y]?
$endgroup$
– JimB
Mar 23 at 19:18
$begingroup$
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
Mar 23 at 19:25
2
$begingroup$
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♦
Mar 23 at 20:04
1
$begingroup$
I think I have found a bug! Try the following two expressions. The only difference is that I use U in one and H in the other for the upper limit.Simplify[PDF[TransformedDistribution[(A + B)/2, {A, B} [Distributed] UniformDistribution[{{L, U}, {L, U}}]], t] == PDF[TriangularDistribution[{L, U}], t], L < t < U]andSimplify[PDF[TransformedDistribution[(A + B)/2, {A, B} [Distributed] UniformDistribution[{{L, H}, {L, H}}]], t] == PDF[TriangularDistribution[{L, H}], t], L < t < H]
$endgroup$
– user120911
Mar 23 at 20:35
$begingroup$
That was a typo. But I am still not getting what I expect.
$endgroup$
– user120911
Mar 23 at 19:16
$begingroup$
That was a typo. But I am still not getting what I expect.
$endgroup$
– user120911
Mar 23 at 19:16
$begingroup$
Did you try
PDF[[ScriptCapitalD], y]?$endgroup$
– JimB
Mar 23 at 19:18
$begingroup$
Did you try
PDF[[ScriptCapitalD], y]?$endgroup$
– JimB
Mar 23 at 19:18
$begingroup$
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
Mar 23 at 19:25
$begingroup$
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
Mar 23 at 19:25
2
2
$begingroup$
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♦
Mar 23 at 20:04
$begingroup$
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♦
Mar 23 at 20:04
1
1
$begingroup$
I think I have found a bug! Try the following two expressions. The only difference is that I use U in one and H in the other for the upper limit.
Simplify[PDF[TransformedDistribution[(A + B)/2, {A, B} [Distributed] UniformDistribution[{{L, U}, {L, U}}]], t] == PDF[TriangularDistribution[{L, U}], t], L < t < U] and Simplify[PDF[TransformedDistribution[(A + B)/2, {A, B} [Distributed] UniformDistribution[{{L, H}, {L, H}}]], t] == PDF[TriangularDistribution[{L, H}], t], L < t < H]$endgroup$
– user120911
Mar 23 at 20:35
$begingroup$
I think I have found a bug! Try the following two expressions. The only difference is that I use U in one and H in the other for the upper limit.
Simplify[PDF[TransformedDistribution[(A + B)/2, {A, B} [Distributed] UniformDistribution[{{L, U}, {L, U}}]], t] == PDF[TriangularDistribution[{L, U}], t], L < t < U] and Simplify[PDF[TransformedDistribution[(A + B)/2, {A, B} [Distributed] UniformDistribution[{{L, H}, {L, H}}]], t] == PDF[TriangularDistribution[{L, H}], t], L < t < H]$endgroup$
– user120911
Mar 23 at 20:35
|
show 4 more comments
2 Answers
2
active
oldest
votes
7
$begingroup$
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}] *)
answered Mar 23 at 20:12
Bob HanlonBob Hanlon
61.1k33598
$endgroup$
$begingroup$
That is very nice!
$endgroup$
– user120911
Mar 23 at 20:24
add a comment |
3
$begingroup$
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}}])]
answered Mar 23 at 19:25
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
Mar 23 at 19:30
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
7
$begingroup$
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}] *)
answered Mar 23 at 20:12
Bob HanlonBob Hanlon
61.1k33598
$endgroup$
$begingroup$
That is very nice!
$endgroup$
– user120911
Mar 23 at 20:24
add a comment |
7
$begingroup$
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}] *)
answered Mar 23 at 20:12
Bob HanlonBob Hanlon
61.1k33598
$endgroup$
$begingroup$
That is very nice!
$endgroup$
– user120911
Mar 23 at 20:24
add a comment |
7
7
7
$begingroup$
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}] *)
answered Mar 23 at 20:12
Bob HanlonBob Hanlon
61.1k33598
$endgroup$
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}] *)
answered Mar 23 at 20:12
Bob HanlonBob Hanlon
61.1k33598
answered Mar 23 at 20:12
Bob HanlonBob Hanlon
61.1k33598
answered Mar 23 at 20:12
Bob HanlonBob Hanlon
61.1k33598
answered Mar 23 at 20:12
Bob HanlonBob Hanlon
61.1k33598
61.1k33598
$begingroup$
That is very nice!
$endgroup$
– user120911
Mar 23 at 20:24
add a comment |
$begingroup$
That is very nice!
$endgroup$
– user120911
Mar 23 at 20:24
$begingroup$
That is very nice!
$endgroup$
– user120911
Mar 23 at 20:24
$begingroup$
That is very nice!
$endgroup$
– user120911
Mar 23 at 20:24
add a comment |
3
$begingroup$
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}}])]
answered Mar 23 at 19:25
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
Mar 23 at 19:30
add a comment |
3
$begingroup$
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}}])]
answered Mar 23 at 19:25
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
Mar 23 at 19:30
add a comment |
3
3
3
$begingroup$
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}}])]
answered Mar 23 at 19:25
kglrkglr
189k10206424
$endgroup$
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}]
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}}])]
answered Mar 23 at 19:25
kglrkglr
189k10206424
answered Mar 23 at 19:25
kglrkglr
189k10206424
answered Mar 23 at 19:25
kglrkglr
189k10206424
answered Mar 23 at 19:25
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
Mar 23 at 19:30
add a comment |
$begingroup$
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
Mar 23 at 19:30
$begingroup$
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
Mar 23 at 19:30
$begingroup$
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
Mar 23 at 19:30
add a comment |
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$begingroup$
That was a typo. But I am still not getting what I expect.
$endgroup$
– user120911
Mar 23 at 19:16
$begingroup$
Did you try
PDF[[ScriptCapitalD], y]?$endgroup$
– JimB
Mar 23 at 19:18
$begingroup$
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
Mar 23 at 19:25
2
$begingroup$
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♦
Mar 23 at 20:04
1
$begingroup$
I think I have found a bug! Try the following two expressions. The only difference is that I use U in one and H in the other for the upper limit.
Simplify[PDF[TransformedDistribution[(A + B)/2, {A, B} [Distributed] UniformDistribution[{{L, U}, {L, U}}]], t] == PDF[TriangularDistribution[{L, U}], t], L < t < U]andSimplify[PDF[TransformedDistribution[(A + B)/2, {A, B} [Distributed] UniformDistribution[{{L, H}, {L, H}}]], t] == PDF[TriangularDistribution[{L, H}], t], L < t < H]$endgroup$
– user120911
Mar 23 at 20:35