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
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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 1 hour ago
user120911user120911
71828
$endgroup$
|
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 1 hour ago
user120911user120911
71828
$endgroup$
$begingroup$
That was a typo. But I am still not getting what I expect.
$endgroup$
– user120911
1 hour ago
$begingroup$
Did you tryPDF[[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 awareTriangularDistribution[]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♦
49 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
asked 1 hour ago
user120911user120911
71828
$endgroup$
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 1 hour ago
user120911user120911
71828
asked 1 hour ago
user120911user120911
71828
edited 1 hour ago
user120911
asked 1 hour ago
user120911user120911
71828
asked 1 hour ago
user120911user120911
71828
asked 1 hour ago
user120911user120911
71828
71828
$begingroup$
That was a typo. But I am still not getting what I expect.
$endgroup$
– user120911
1 hour ago
$begingroup$
Did you tryPDF[[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 awareTriangularDistribution[]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♦
49 mins ago
|
show 3 more comments
$begingroup$
That was a typo. But I am still not getting what I expect.
$endgroup$
– user120911
1 hour ago
$begingroup$
Did you tryPDF[[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 awareTriangularDistribution[]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♦
49 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♦
49 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♦
49 mins 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 41 mins ago
Bob HanlonBob Hanlon
60.9k33597
$endgroup$
$begingroup$
That is very nice!
$endgroup$
– user120911
28 mins 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 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 |
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 41 mins ago
Bob HanlonBob Hanlon
60.9k33597
$endgroup$
$begingroup$
That is very nice!
$endgroup$
– user120911
28 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] *)
answered 41 mins ago
Bob HanlonBob Hanlon
60.9k33597
$endgroup$
$begingroup$
That is very nice!
$endgroup$
– user120911
28 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] *)
answered 41 mins 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 41 mins ago
Bob HanlonBob Hanlon
60.9k33597
answered 41 mins ago
Bob HanlonBob Hanlon
60.9k33597
answered 41 mins ago
Bob HanlonBob Hanlon
60.9k33597
answered 41 mins ago
Bob HanlonBob Hanlon
60.9k33597
60.9k33597
$begingroup$
That is very nice!
$endgroup$
– user120911
28 mins ago
add a comment |
$begingroup$
That is very nice!
$endgroup$
– user120911
28 mins ago
$begingroup$
That is very nice!
$endgroup$
– user120911
28 mins ago
$begingroup$
That is very nice!
$endgroup$
– user120911
28 mins 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 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])]
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])]
answered 1 hour 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 1 hour ago
kglrkglr
189k10206424
answered 1 hour ago
kglrkglr
189k10206424
answered 1 hour ago
kglrkglr
189k10206424
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
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
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
add a comment |
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$begingroup$
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♦
49 mins ago