Revisions to Visualizing density given by its Stieltjes transformation?

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edited Jun 25, 2023 at 1:21

Yaroslav Bulatov

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How can I use Mathematica to get $f(y)$, the PDF of a random variable given its Stieltjes transform $g(s)$?

$$g(s)=E\left[1/(s-X)\right]$$

I need to visualize probability density of a random variable $X\in(0,1]$ with the following Stieltjes transform:

stieltjes=(Sqrt[2] (\[Pi] - 2 ArcTan[Sqrt[s]/Sqrt[2]]))/(2 Sqrt[s] + Sqrt[2] s ArcTan[Sqrt[2]/Sqrt[s]])

Expecting this PDF to look similar to $\operatorname{Beta}\left(\frac{1}{2},1\right)$

Trying on some simpler distributions first:

Using Beta$(1/2,1)$ we have the following pdf $f(y)$ and $g(s)$

pdf = 1/2 y^(-1/2); (* BetaDistribution[1/2,1] *)
stieltjes = Integrate[pdf 1/(s - y), {y, 0, 1}]

$$f(y)=\frac{\frac{1}{\sqrt{y}}}{2} \leftrightarrow g(s)=\frac{\coth ^{-1}\left(\sqrt{s}\right)}{\sqrt{s}}$$

Also, using $1/x^2$ where $x$ is Zipf distributed as in the earlier question we have the following

distY = TransformedDistribution[1/z^2, 
   z \[Distributed] ZipfDistribution[1]];
Expectation[1/(s - i), i \[Distributed] distY]

$$f(y)=\left(1, \frac{6}{\pi ^2}\right),\left(\frac{1}{4}, \frac{3}{2 \pi ^2}\right), \left(\frac{1}{9} , \frac{2}{3 \pi ^2}\right),\ldots \leftrightarrow g(s)=-\frac{3 \left(\pi \cot \left(\frac{\pi }{\sqrt{s}}\right)-\sqrt{s}\right)}{\pi ^2 \sqrt{s}}$$

You can get Stieltjes Transforms by chaining two Laplace transforms together as seen below. Wondering if there's a way to use this to get symbolic or numeric inversion working for examples above.

dist = BetaDistribution[1/2, 1];
mgf = MomentGeneratingFunction[BetaDistribution[1/2, 1], t];
doubleLaplace = LaplaceTransform[mgf, t, s] // FullSimplify;
stieltjes = Expectation[1/(s - y), y \[Distributed] dist];
doubleLaplace == stieltjes (*True if s>1*)

(note you often need Feynman trick to do inverse Laplace transforms in Mathematica, documented here)

How can I use Mathematica to get $f(y)$, the PDF of a random variable given its Stieltjes transform $g(s)$?

$$g(s)=E\left[1/(s-X)\right]$$

I need to visualize probability density of a random variable with the following Stieltjes transform:

stieltjes=(Sqrt[2] (\[Pi] - 2 ArcTan[Sqrt[s]/Sqrt[2]]))/(2 Sqrt[s] + Sqrt[2] s ArcTan[Sqrt[2]/Sqrt[s]])

Trying on some simpler distributions first:

Using Beta$(1/2,1)$ we have the following pdf $f(y)$ and $g(s)$

pdf = 1/2 y^(-1/2); (* BetaDistribution[1/2,1] *)
stieltjes = Integrate[pdf 1/(s - y), {y, 0, 1}]

$$f(y)=\frac{\frac{1}{\sqrt{y}}}{2} \leftrightarrow g(s)=\frac{\coth ^{-1}\left(\sqrt{s}\right)}{\sqrt{s}}$$

Also, using $1/x^2$ where $x$ is Zipf distributed as in the earlier question we have the following

distY = TransformedDistribution[1/z^2, 
   z \[Distributed] ZipfDistribution[1]];
Expectation[1/(s - i), i \[Distributed] distY]

$$f(y)=\left(1, \frac{6}{\pi ^2}\right),\left(\frac{1}{4}, \frac{3}{2 \pi ^2}\right), \left(\frac{1}{9} , \frac{2}{3 \pi ^2}\right),\ldots \leftrightarrow g(s)=-\frac{3 \left(\pi \cot \left(\frac{\pi }{\sqrt{s}}\right)-\sqrt{s}\right)}{\pi ^2 \sqrt{s}}$$

You can get Stieltjes Transforms by chaining two Laplace transforms together as seen below. Wondering if there's a way to use this to get symbolic or numeric inversion working for examples above.

dist = BetaDistribution[1/2, 1];
mgf = MomentGeneratingFunction[BetaDistribution[1/2, 1], t];
doubleLaplace = LaplaceTransform[mgf, t, s] // FullSimplify;
stieltjes = Expectation[1/(s - y), y \[Distributed] dist];
doubleLaplace == stieltjes (*True if s>1*)

(note you often need Feynman trick to do inverse Laplace transforms in Mathematica, documented here)

How can I use Mathematica to get $f(y)$, the PDF of a random variable given its Stieltjes transform $g(s)$?

$$g(s)=E\left[1/(s-X)\right]$$

I need to visualize probability density of a random variable $X\in(0,1]$ with the following Stieltjes transform:

stieltjes=(Sqrt[2] (\[Pi] - 2 ArcTan[Sqrt[s]/Sqrt[2]]))/(2 Sqrt[s] + Sqrt[2] s ArcTan[Sqrt[2]/Sqrt[s]])

Expecting this PDF to look similar to $\operatorname{Beta}\left(\frac{1}{2},1\right)$

Trying on some simpler distributions first:

Using Beta$(1/2,1)$ we have the following pdf $f(y)$ and $g(s)$

pdf = 1/2 y^(-1/2); (* BetaDistribution[1/2,1] *)
stieltjes = Integrate[pdf 1/(s - y), {y, 0, 1}]

$$f(y)=\frac{\frac{1}{\sqrt{y}}}{2} \leftrightarrow g(s)=\frac{\coth ^{-1}\left(\sqrt{s}\right)}{\sqrt{s}}$$

Also, using $1/x^2$ where $x$ is Zipf distributed as in the earlier question we have the following

distY = TransformedDistribution[1/z^2, 
   z \[Distributed] ZipfDistribution[1]];
Expectation[1/(s - i), i \[Distributed] distY]

$$f(y)=\left(1, \frac{6}{\pi ^2}\right),\left(\frac{1}{4}, \frac{3}{2 \pi ^2}\right), \left(\frac{1}{9} , \frac{2}{3 \pi ^2}\right),\ldots \leftrightarrow g(s)=-\frac{3 \left(\pi \cot \left(\frac{\pi }{\sqrt{s}}\right)-\sqrt{s}\right)}{\pi ^2 \sqrt{s}}$$

You can get Stieltjes Transforms by chaining two Laplace transforms together as seen below. Wondering if there's a way to use this to get symbolic or numeric inversion working for examples above.

dist = BetaDistribution[1/2, 1];
mgf = MomentGeneratingFunction[BetaDistribution[1/2, 1], t];
doubleLaplace = LaplaceTransform[mgf, t, s] // FullSimplify;
stieltjes = Expectation[1/(s - y), y \[Distributed] dist];
doubleLaplace == stieltjes (*True if s>1*)

(note you often need Feynman trick to do inverse Laplace transforms in Mathematica, documented here)

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edited Jun 25, 2023 at 1:04

Yaroslav Bulatov

6.7k
1
21
47

How can I use Mathematica to get $f(y)$, the PDF of a random variable given its Stieltjes transform $g(s)$?

$$g(s)=E\left[1/(s-X)\right]$$

I'm tryingI need to visualize probability density function given by itsof a random variable with the following Stieltjes Transformation (Sqrt[2] (\[Pi] - 2 ArcTan[Sqrt[s]/Sqrt[2]]))/(2 Sqrt[s] + Sqrt[2] s ArcTan[Sqrt[2]/Sqrt[s]]) and getting stuck.transform:

stieltjes=(Sqrt[2] (\[Pi] - 2 ArcTan[Sqrt[s]/Sqrt[2]]))/(2 Sqrt[s] + Sqrt[2] s ArcTan[Sqrt[2]/Sqrt[s]])

Simpler knownTrying on some simpler distributions I tried to solve first:

Using Beta$(1/2,1)$ we have the following pdf $f(y)$ and $g(s)$

pdf = 1/2 y^(-1/2); (* BetaDistribution[1/2,1] *)
stieltjes = Integrate[pdf 1/(s - y), {y, 0, 1}]

$$f(y)=\frac{\frac{1}{\sqrt{y}}}{2} \leftrightarrow g(s)=\frac{\coth ^{-1}\left(\sqrt{s}\right)}{\sqrt{s}}$$

Also, using $1/x^2$ where $x$ is Zipf distributed as in the earlier question we have the following

distY = TransformedDistribution[1/z^2, 
   z \[Distributed] ZipfDistribution[1]];
Expectation[1/(s - i), i \[Distributed] distY]

$$f(y)=\left(1, \frac{6}{\pi ^2}\right),\left(\frac{1}{4}, \frac{3}{2 \pi ^2}\right), \left(\frac{1}{9} , \frac{2}{3 \pi ^2}\right),\ldots \leftrightarrow g(s)=-\frac{3 \left(\pi \cot \left(\frac{\pi }{\sqrt{s}}\right)-\sqrt{s}\right)}{\pi ^2 \sqrt{s}}$$

You can get Stieltjes Transforms by chaining two Laplace transforms together as seen below. Wondering if there's a way to use this to get symbolic or numeric inversion working for examples above.

dist = BetaDistribution[1/2, 1];
mgf = MomentGeneratingFunction[BetaDistribution[1/2, 1], t];
doubleLaplace = LaplaceTransform[mgf, t, s] // FullSimplify;
stieltjes = Expectation[1/(s - y), y \[Distributed] dist];
doubleLaplace == stieltjes (*True if s>1*)

(note you often need Feynman trick to do inverse Laplace transforms in Mathematica, documented here)

How can I use Mathematica to get $f(y)$, the PDF of a random variable given its Stieltjes transform $g(s)$?

$$g(s)=E\left[1/(s-X)\right]$$

I'm trying to visualize probability density function given by its Stieltjes Transformation (Sqrt[2] (\[Pi] - 2 ArcTan[Sqrt[s]/Sqrt[2]]))/(2 Sqrt[s] + Sqrt[2] s ArcTan[Sqrt[2]/Sqrt[s]]) and getting stuck.

Simpler known distributions I tried to solve first:

Using Beta$(1/2,1)$ we have the following pdf $f(y)$ and $g(s)$

pdf = 1/2 y^(-1/2); (* BetaDistribution[1/2,1] *)
stieltjes = Integrate[pdf 1/(s - y), {y, 0, 1}]

$$f(y)=\frac{\frac{1}{\sqrt{y}}}{2} \leftrightarrow g(s)=\frac{\coth ^{-1}\left(\sqrt{s}\right)}{\sqrt{s}}$$

Also, using $1/x^2$ where $x$ is Zipf distributed as in the earlier question we have the following

distY = TransformedDistribution[1/z^2, 
   z \[Distributed] ZipfDistribution[1]];
Expectation[1/(s - i), i \[Distributed] distY]

$$f(y)=\left(1, \frac{6}{\pi ^2}\right),\left(\frac{1}{4}, \frac{3}{2 \pi ^2}\right), \left(\frac{1}{9} , \frac{2}{3 \pi ^2}\right),\ldots \leftrightarrow g(s)=-\frac{3 \left(\pi \cot \left(\frac{\pi }{\sqrt{s}}\right)-\sqrt{s}\right)}{\pi ^2 \sqrt{s}}$$

You can get Stieltjes Transforms by chaining two Laplace transforms together as seen below. Wondering if there's a way to use this to get symbolic or numeric inversion working for examples above.

dist = BetaDistribution[1/2, 1];
mgf = MomentGeneratingFunction[BetaDistribution[1/2, 1], t];
doubleLaplace = LaplaceTransform[mgf, t, s] // FullSimplify;
stieltjes = Expectation[1/(s - y), y \[Distributed] dist];
doubleLaplace == stieltjes (*True if s>1*)

(note you often need Feynman trick to do inverse Laplace transforms in Mathematica, documented here)

How can I use Mathematica to get $f(y)$, the PDF of a random variable given its Stieltjes transform $g(s)$?

$$g(s)=E\left[1/(s-X)\right]$$

I need to visualize probability density of a random variable with the following Stieltjes transform:

stieltjes=(Sqrt[2] (\[Pi] - 2 ArcTan[Sqrt[s]/Sqrt[2]]))/(2 Sqrt[s] + Sqrt[2] s ArcTan[Sqrt[2]/Sqrt[s]])

Trying on some simpler distributions first:

Using Beta$(1/2,1)$ we have the following pdf $f(y)$ and $g(s)$

pdf = 1/2 y^(-1/2); (* BetaDistribution[1/2,1] *)
stieltjes = Integrate[pdf 1/(s - y), {y, 0, 1}]

$$f(y)=\frac{\frac{1}{\sqrt{y}}}{2} \leftrightarrow g(s)=\frac{\coth ^{-1}\left(\sqrt{s}\right)}{\sqrt{s}}$$

Also, using $1/x^2$ where $x$ is Zipf distributed as in the earlier question we have the following

distY = TransformedDistribution[1/z^2, 
   z \[Distributed] ZipfDistribution[1]];
Expectation[1/(s - i), i \[Distributed] distY]

$$f(y)=\left(1, \frac{6}{\pi ^2}\right),\left(\frac{1}{4}, \frac{3}{2 \pi ^2}\right), \left(\frac{1}{9} , \frac{2}{3 \pi ^2}\right),\ldots \leftrightarrow g(s)=-\frac{3 \left(\pi \cot \left(\frac{\pi }{\sqrt{s}}\right)-\sqrt{s}\right)}{\pi ^2 \sqrt{s}}$$

You can get Stieltjes Transforms by chaining two Laplace transforms together as seen below. Wondering if there's a way to use this to get symbolic or numeric inversion working for examples above.

dist = BetaDistribution[1/2, 1];
mgf = MomentGeneratingFunction[BetaDistribution[1/2, 1], t];
doubleLaplace = LaplaceTransform[mgf, t, s] // FullSimplify;
stieltjes = Expectation[1/(s - y), y \[Distributed] dist];
doubleLaplace == stieltjes (*True if s>1*)

(note you often need Feynman trick to do inverse Laplace transforms in Mathematica, documented here)

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edited Jun 25, 2023 at 0:38

Yaroslav Bulatov

6.7k
1
21
47

How can I use Mathematica to get $f(y)$, the PDF of a random variable given its Stieltjes transform $g(s)$?

$$g(s)=E\left[1/(s-X)\right]$$

I'm trying to visualize probability density function given by its Stieltjes Transformation (Sqrt[2] (\[Pi] - 2 ArcTan[Sqrt[s]/Sqrt[2]]))/(2 Sqrt[s] + Sqrt[2] s ArcTan[Sqrt[2]/Sqrt[s]]) and getting stuck.

Simpler known distributions I tried to solve first:

Using Beta$(1/2,1)$ we have the following pdf $f(y)$ and $g(s)$

pdf = 1/2 y^(-1/2); (* BetaDistribution[1/2,1] *)
stieltjes = Integrate[pdf 1/(s - y), {y, 0, 1}]

$$f(y)=\frac{\frac{1}{\sqrt{y}}}{2} \leftrightarrow g(s)=\frac{\coth ^{-1}\left(\sqrt{s}\right)}{\sqrt{s}}$$

Also, using $1/x^2$ where $x$ is Zipf distributed as in the earlier question we have the following

distY = TransformedDistribution[1/z^2, 
   z \[Distributed] ZipfDistribution[1]];
Expectation[1/(s - i), i \[Distributed] distY]

$$f(y)=\left(1, \frac{6}{\pi ^2}\right),\left(\frac{1}{4}, \frac{3}{2 \pi ^2}\right), \left(\frac{1}{9} , \frac{2}{3 \pi ^2}\right),\ldots \leftrightarrow g(s)=-\frac{3 \left(\pi \cot \left(\frac{\pi }{\sqrt{s}}\right)-\sqrt{s}\right)}{\pi ^2 \sqrt{s}}$$

I suspected one could make use of the following two facts to do the inversionYou can get Stieltjes Transforms by chaining two inverse Laplace Transformstransforms together, but somehow I can't get the "double Laplace transform" result as seen below. Wondering if there's a way to match the "Stieltjes transformation" resultuse this to get symbolic or numeric inversion working for the simple densitiesexamples above.

Density and $E[\exp(-t x)]$ are related through a Laplace transform.

$\exp(t x)$ and $\frac{1}{s-x}$ are related through a Laplace Transform

dist = BetaDistribution[1/2, 1];
mgf = MomentGeneratingFunction[BetaDistribution[1/2, 1], t];
doubleLaplace = LaplaceTransform[mgf, t, s] // FullSimplify;
stieltjes = Expectation[1/(s - y), y \[Distributed] dist];
doubleLaplace == stieltjes (*True if s>1*)

(note you often need Feynman trick to do inverse Laplace transforms in Mathematica, documented here)

How can I use Mathematica to get $f(y)$, the PDF of a random variable given its Stieltjes transform $g(s)$?

$$g(s)=E\left[1/(s-X)\right]$$

I'm trying to visualize probability density function given by its Stieltjes Transformation (Sqrt[2] (\[Pi] - 2 ArcTan[Sqrt[s]/Sqrt[2]]))/(2 Sqrt[s] + Sqrt[2] s ArcTan[Sqrt[2]/Sqrt[s]]) and getting stuck.

Simpler known distributions I tried to solve first:

Using Beta$(1/2,1)$ we have the following pdf $f(y)$ and $g(s)$

pdf = 1/2 y^(-1/2); (* BetaDistribution[1/2,1] *)
stieltjes = Integrate[pdf 1/(s - y), {y, 0, 1}]

$$f(y)=\frac{\frac{1}{\sqrt{y}}}{2} \leftrightarrow g(s)=\frac{\coth ^{-1}\left(\sqrt{s}\right)}{\sqrt{s}}$$

Also, using $1/x^2$ where $x$ is Zipf distributed as in the earlier question we have the following

distY = TransformedDistribution[1/z^2, 
   z \[Distributed] ZipfDistribution[1]];
Expectation[1/(s - i), i \[Distributed] distY]

$$f(y)=\left(1, \frac{6}{\pi ^2}\right),\left(\frac{1}{4}, \frac{3}{2 \pi ^2}\right), \left(\frac{1}{9} , \frac{2}{3 \pi ^2}\right),\ldots \leftrightarrow g(s)=-\frac{3 \left(\pi \cot \left(\frac{\pi }{\sqrt{s}}\right)-\sqrt{s}\right)}{\pi ^2 \sqrt{s}}$$

I suspected one could make use of the following two facts to do the inversion by chaining two inverse Laplace Transforms together, but somehow I can't get the "double Laplace transform" result to match the "Stieltjes transformation" result for the simple densities above.

Density and $E[\exp(-t x)]$ are related through a Laplace transform.

$\exp(t x)$ and $\frac{1}{s-x}$ are related through a Laplace Transform

(note you often need Feynman trick to do inverse Laplace transforms in Mathematica, documented here)

How can I use Mathematica to get $f(y)$, the PDF of a random variable given its Stieltjes transform $g(s)$?

$$g(s)=E\left[1/(s-X)\right]$$

I'm trying to visualize probability density function given by its Stieltjes Transformation (Sqrt[2] (\[Pi] - 2 ArcTan[Sqrt[s]/Sqrt[2]]))/(2 Sqrt[s] + Sqrt[2] s ArcTan[Sqrt[2]/Sqrt[s]]) and getting stuck.

Simpler known distributions I tried to solve first:

Using Beta$(1/2,1)$ we have the following pdf $f(y)$ and $g(s)$

pdf = 1/2 y^(-1/2); (* BetaDistribution[1/2,1] *)
stieltjes = Integrate[pdf 1/(s - y), {y, 0, 1}]

$$f(y)=\frac{\frac{1}{\sqrt{y}}}{2} \leftrightarrow g(s)=\frac{\coth ^{-1}\left(\sqrt{s}\right)}{\sqrt{s}}$$

Also, using $1/x^2$ where $x$ is Zipf distributed as in the earlier question we have the following

distY = TransformedDistribution[1/z^2, 
   z \[Distributed] ZipfDistribution[1]];
Expectation[1/(s - i), i \[Distributed] distY]

$$f(y)=\left(1, \frac{6}{\pi ^2}\right),\left(\frac{1}{4}, \frac{3}{2 \pi ^2}\right), \left(\frac{1}{9} , \frac{2}{3 \pi ^2}\right),\ldots \leftrightarrow g(s)=-\frac{3 \left(\pi \cot \left(\frac{\pi }{\sqrt{s}}\right)-\sqrt{s}\right)}{\pi ^2 \sqrt{s}}$$

You can get Stieltjes Transforms by chaining two Laplace transforms together as seen below. Wondering if there's a way to use this to get symbolic or numeric inversion working for examples above.

dist = BetaDistribution[1/2, 1];
mgf = MomentGeneratingFunction[BetaDistribution[1/2, 1], t];
doubleLaplace = LaplaceTransform[mgf, t, s] // FullSimplify;
stieltjes = Expectation[1/(s - y), y \[Distributed] dist];
doubleLaplace == stieltjes (*True if s>1*)

(note you often need Feynman trick to do inverse Laplace transforms in Mathematica, documented here)

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edited Jun 24, 2023 at 23:14

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edited Jun 24, 2023 at 22:43

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edited Jun 24, 2023 at 22:11

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edited Jun 24, 2023 at 22:05

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asked Jun 24, 2023 at 21:55

Yaroslav Bulatov

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