Problem with TransformedDistributionCalculate probability functionWhen are `If`, `Piecewise`, `Switch`, and `Which` interchangeable and when are they not?Calculate PDF and CDF of a product of independent exponentially distributed random variablesConditional probabilityFullSimplify on TransformedDistributionNProbability not reliability analysis?TransformedDistribution using $k$ iid random variables, but $k$ not fixedConvolve discrete random variables efficientlyProbability distribution defined by partitioning an intervalDistribution of Function of Random Sum of Random Variables
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Problem with TransformedDistribution
Calculate probability functionWhen are `If`, `Piecewise`, `Switch`, and `Which` interchangeable and when are they not?Calculate PDF and CDF of a product of independent exponentially distributed random variablesConditional probabilityFullSimplify on TransformedDistributionNProbability not reliability analysis?TransformedDistribution using $k$ iid random variables, but $k$ not fixedConvolve discrete random variables efficientlyProbability distribution defined by partitioning an intervalDistribution of Function of Random Sum of Random Variables
$begingroup$
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
asked 2 hours ago
user120911user120911
72428
$endgroup$
|
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
asked 2 hours ago
user120911user120911
72428
$endgroup$
$begingroup$
That was a typo. But I am still not getting what I expect.
$endgroup$
– user120911
2 hours ago
$begingroup$
Did you tryPDF[[ScriptCapitalD], y]?
$endgroup$
– JimB
2 hours 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
2 hours ago
$begingroup$
Are you awareTriangularDistribution[]is built-in?
$endgroup$
– J. M. is slightly pensive♦
2 hours 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♦
1 hour 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
asked 2 hours ago
user120911user120911
72428
$endgroup$
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
probability-or-statistics
asked 2 hours ago
user120911user120911
72428
asked 2 hours ago
user120911user120911
72428
edited 2 hours ago
user120911
asked 2 hours ago
user120911user120911
72428
asked 2 hours ago
user120911user120911
72428
asked 2 hours ago
user120911user120911
72428
72428
$begingroup$
That was a typo. But I am still not getting what I expect.
$endgroup$
– user120911
2 hours ago
$begingroup$
Did you tryPDF[[ScriptCapitalD], y]?
$endgroup$
– JimB
2 hours 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
2 hours ago
$begingroup$
Are you awareTriangularDistribution[]is built-in?
$endgroup$
– J. M. is slightly pensive♦
2 hours 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♦
1 hour ago
|
show 3 more comments
$begingroup$
That was a typo. But I am still not getting what I expect.
$endgroup$
– user120911
2 hours ago
$begingroup$
Did you tryPDF[[ScriptCapitalD], y]?
$endgroup$
– JimB
2 hours 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
2 hours ago
$begingroup$
Are you awareTriangularDistribution[]is built-in?
$endgroup$
– J. M. is slightly pensive♦
2 hours 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♦
1 hour ago
$begingroup$
That was a typo. But I am still not getting what I expect.
$endgroup$
– user120911
2 hours ago
$begingroup$
That was a typo. But I am still not getting what I expect.
$endgroup$
– user120911
2 hours ago
$begingroup$
Did you try
PDF[[ScriptCapitalD], y]?$endgroup$
– JimB
2 hours ago
$begingroup$
Did you try
PDF[[ScriptCapitalD], y]?$endgroup$
– JimB
2 hours 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
2 hours 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
2 hours ago
$begingroup$
Are you aware
TriangularDistribution[] is built-in?$endgroup$
– J. M. is slightly pensive♦
2 hours ago
$begingroup$
Are you aware
TriangularDistribution[] is built-in?$endgroup$
– J. M. is slightly pensive♦
2 hours 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♦
1 hour 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♦
1 hour ago
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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 1 hour ago
Bob HanlonBob Hanlon
60.9k33597
$endgroup$
$begingroup$
That is very nice!
$endgroup$
– 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])]
answered 2 hours 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
2 hours ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 1 hour ago
Bob HanlonBob Hanlon
60.9k33597
$endgroup$
$begingroup$
That is very nice!
$endgroup$
– user120911
1 hour 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] *)
answered 1 hour ago
Bob HanlonBob Hanlon
60.9k33597
$endgroup$
$begingroup$
That is very nice!
$endgroup$
– user120911
1 hour 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] *)
answered 1 hour ago
Bob HanlonBob Hanlon
60.9k33597
$endgroup$
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 1 hour ago
Bob HanlonBob Hanlon
60.9k33597
answered 1 hour ago
Bob HanlonBob Hanlon
60.9k33597
answered 1 hour ago
Bob HanlonBob Hanlon
60.9k33597
answered 1 hour ago
Bob HanlonBob Hanlon
60.9k33597
60.9k33597
$begingroup$
That is very nice!
$endgroup$
– user120911
1 hour ago
add a comment |
$begingroup$
That is very nice!
$endgroup$
– user120911
1 hour ago
$begingroup$
That is very nice!
$endgroup$
– user120911
1 hour ago
$begingroup$
That is very nice!
$endgroup$
– 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])]
answered 2 hours 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
2 hours 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])]
answered 2 hours 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
2 hours 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])]
answered 2 hours ago
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 2 hours ago
kglrkglr
189k10206424
answered 2 hours ago
kglrkglr
189k10206424
answered 2 hours ago
kglrkglr
189k10206424
answered 2 hours 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
2 hours ago
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
2 hours 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
2 hours 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
2 hours ago
add a comment |
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$begingroup$
That was a typo. But I am still not getting what I expect.
$endgroup$
– user120911
2 hours ago
$begingroup$
Did you try
PDF[[ScriptCapitalD], y]?$endgroup$
– JimB
2 hours 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
2 hours ago
$begingroup$
Are you aware
TriangularDistribution[]is built-in?$endgroup$
– J. M. is slightly pensive♦
2 hours 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♦
1 hour ago