Annulus distribution in n dimensions: normalising constant, normed mean and variance

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I would like to know the normalising constant of a distribution which has the pdf,



$$f(x) propto sqrtfrac12pisigma^2textexp(-fracxsigma^2),$$



where $xinmathbbR^d$ and $|x|$ is the Euclidean norm of $x$. This distribution looks like an annulus in two-dimensions,



enter image description here



In higher dimensions, samples from it are normally-distributed centred on the surface of a hypersphere of radius $r_0$ (at least, I hope it is - please tell me if I am wrong).



Intuition (and a previous question), suggests to me that this distribution has an integral over space that equals the surface area of a hypersphere $mathbbS^n-1$, multiplied by $r_0^n-1$, where $n$ is the number of dimensions of $x$. Basically, this accounts for taking a normal density and smearing it over higher dimensional space.



The trouble is that I am having trouble confirming my results in Mathematica. The following seems to work ok,



volS[n_] := (2 (Pi^(n / 2)) ) / Gamma[n / 2]

fPDF[z_, r0_, sigma_] := 
 Block[n = Length@z, (1 / (volS[n] r0^(n - 1)))
 PDF[NormalDistribution[r0, sigma], Norm[z]]]

NIntegrate[fPDF[x, y, 10, 1], x, -Infinity, Infinity, y, -Infinity, Infinity]



1.




But when I vary $r_0$ and $sigma$ I get different answers. For example,



NIntegrate[fPDF[x, y, 20, 2], x, -Infinity, Infinity, y, -Infinity, Infinity]



0.772186




which seems to suggest I am wrong. However, I am not sure because I am a little fearful that the numerical integration is going awry.



Any ideas?



Also, I would have thought that the mean normed distance for this distribution should just be $r_0$ (as long as $r_0>>sigma$). When I evaluate this, however, I get a different answer,



NIntegrate[Norm[x,y] fPDF[x, y, 10, 0.5], x, -Infinity, Infinity, y, -Infinity, Infinity]



7.17138




Any idea as to how to calculate the mean normed distance and variance?



Whilst all these examples are in two-dimensions, note that I am after n-dimensional results.






calculus-and-analysis probability-or-statistics geometry







share|improve this question



























    up vote
    4
    down vote

    favorite












    I would like to know the normalising constant of a distribution which has the pdf,



    $$f(x) propto sqrtfrac12pisigma^2textexp(-fracxsigma^2),$$



    where $xinmathbbR^d$ and $|x|$ is the Euclidean norm of $x$. This distribution looks like an annulus in two-dimensions,



    enter image description here



    In higher dimensions, samples from it are normally-distributed centred on the surface of a hypersphere of radius $r_0$ (at least, I hope it is - please tell me if I am wrong).



    Intuition (and a previous question), suggests to me that this distribution has an integral over space that equals the surface area of a hypersphere $mathbbS^n-1$, multiplied by $r_0^n-1$, where $n$ is the number of dimensions of $x$. Basically, this accounts for taking a normal density and smearing it over higher dimensional space.



    The trouble is that I am having trouble confirming my results in Mathematica. The following seems to work ok,



    volS[n_] := (2 (Pi^(n / 2)) ) / Gamma[n / 2]

    fPDF[z_, r0_, sigma_] := 
     Block[n = Length@z, (1 / (volS[n] r0^(n - 1)))
     PDF[NormalDistribution[r0, sigma], Norm[z]]]

    NIntegrate[fPDF[x, y, 10, 1], x, -Infinity, Infinity, y, -Infinity, Infinity]



    1.




    But when I vary $r_0$ and $sigma$ I get different answers. For example,



    NIntegrate[fPDF[x, y, 20, 2], x, -Infinity, Infinity, y, -Infinity, Infinity]



    0.772186




    which seems to suggest I am wrong. However, I am not sure because I am a little fearful that the numerical integration is going awry.



    Any ideas?



    Also, I would have thought that the mean normed distance for this distribution should just be $r_0$ (as long as $r_0>>sigma$). When I evaluate this, however, I get a different answer,



    NIntegrate[Norm[x,y] fPDF[x, y, 10, 0.5], x, -Infinity, Infinity, y, -Infinity, Infinity]



    7.17138




    Any idea as to how to calculate the mean normed distance and variance?



    Whilst all these examples are in two-dimensions, note that I am after n-dimensional results.






    calculus-and-analysis probability-or-statistics geometry







    share|improve this question

























      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite











      I would like to know the normalising constant of a distribution which has the pdf,



      $$f(x) propto sqrtfrac12pisigma^2textexp(-fracxsigma^2),$$



      where $xinmathbbR^d$ and $|x|$ is the Euclidean norm of $x$. This distribution looks like an annulus in two-dimensions,



      enter image description here



      In higher dimensions, samples from it are normally-distributed centred on the surface of a hypersphere of radius $r_0$ (at least, I hope it is - please tell me if I am wrong).



      Intuition (and a previous question), suggests to me that this distribution has an integral over space that equals the surface area of a hypersphere $mathbbS^n-1$, multiplied by $r_0^n-1$, where $n$ is the number of dimensions of $x$. Basically, this accounts for taking a normal density and smearing it over higher dimensional space.



      The trouble is that I am having trouble confirming my results in Mathematica. The following seems to work ok,



      volS[n_] := (2 (Pi^(n / 2)) ) / Gamma[n / 2]

      fPDF[z_, r0_, sigma_] := 
       Block[n = Length@z, (1 / (volS[n] r0^(n - 1)))
       PDF[NormalDistribution[r0, sigma], Norm[z]]]

      NIntegrate[fPDF[x, y, 10, 1], x, -Infinity, Infinity, y, -Infinity, Infinity]



      1.




      But when I vary $r_0$ and $sigma$ I get different answers. For example,



      NIntegrate[fPDF[x, y, 20, 2], x, -Infinity, Infinity, y, -Infinity, Infinity]



      0.772186




      which seems to suggest I am wrong. However, I am not sure because I am a little fearful that the numerical integration is going awry.



      Any ideas?



      Also, I would have thought that the mean normed distance for this distribution should just be $r_0$ (as long as $r_0>>sigma$). When I evaluate this, however, I get a different answer,



      NIntegrate[Norm[x,y] fPDF[x, y, 10, 0.5], x, -Infinity, Infinity, y, -Infinity, Infinity]



      7.17138




      Any idea as to how to calculate the mean normed distance and variance?



      Whilst all these examples are in two-dimensions, note that I am after n-dimensional results.






      calculus-and-analysis probability-or-statistics geometry







      share|improve this question















      I would like to know the normalising constant of a distribution which has the pdf,



      $$f(x) propto sqrtfrac12pisigma^2textexp(-fracxsigma^2),$$



      where $xinmathbbR^d$ and $|x|$ is the Euclidean norm of $x$. This distribution looks like an annulus in two-dimensions,



      enter image description here



      In higher dimensions, samples from it are normally-distributed centred on the surface of a hypersphere of radius $r_0$ (at least, I hope it is - please tell me if I am wrong).



      Intuition (and a previous question), suggests to me that this distribution has an integral over space that equals the surface area of a hypersphere $mathbbS^n-1$, multiplied by $r_0^n-1$, where $n$ is the number of dimensions of $x$. Basically, this accounts for taking a normal density and smearing it over higher dimensional space.



      The trouble is that I am having trouble confirming my results in Mathematica. The following seems to work ok,



      volS[n_] := (2 (Pi^(n / 2)) ) / Gamma[n / 2]

      fPDF[z_, r0_, sigma_] := 
       Block[n = Length@z, (1 / (volS[n] r0^(n - 1)))
       PDF[NormalDistribution[r0, sigma], Norm[z]]]

      NIntegrate[fPDF[x, y, 10, 1], x, -Infinity, Infinity, y, -Infinity, Infinity]



      1.




      But when I vary $r_0$ and $sigma$ I get different answers. For example,



      NIntegrate[fPDF[x, y, 20, 2], x, -Infinity, Infinity, y, -Infinity, Infinity]



      0.772186




      which seems to suggest I am wrong. However, I am not sure because I am a little fearful that the numerical integration is going awry.



      Any ideas?



      Also, I would have thought that the mean normed distance for this distribution should just be $r_0$ (as long as $r_0>>sigma$). When I evaluate this, however, I get a different answer,



      NIntegrate[Norm[x,y] fPDF[x, y, 10, 0.5], x, -Infinity, Infinity, y, -Infinity, Infinity]



      7.17138




      Any idea as to how to calculate the mean normed distance and variance?



      Whilst all these examples are in two-dimensions, note that I am after n-dimensional results.






      calculus-and-analysis probability-or-statistics geometry




      calculus-and-analysis probability-or-statistics geometry






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 1 hour ago









      Henrik Schumacher

      39.5k254118




      39.5k254118










      asked 1 hour ago









      ben18785

      1,398622




      1,398622




















          1 Answer
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          up vote
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          The numerical integration is not the issue. You also have to calculate a radial integral in order to determine the normalization constant. So, let's do it! Again, we apply polar coordinates:



          ClearAll[r, r0, σ, n];
          f = r, r0, σ [Function] Evaluate[PDF[NormalDistribution[r0, σ], r]];
          assumptions = And @@ n ∈ Integers, n >= 1, r > 0, r0 > 0, σ > 0;
          volS[n_] = (2 (Pi^(n/2)))/Gamma[n/2];
          normalization[n_, r0_, σ_] = volS[n] Integrate[f[r, r0, σ] r^(n - 1), r, 0, ∞, 
           Assumptions -> assumptions];


          This defines a routine that creates a PDF for given dimension n, radius r0 and parameter σ:



          makePDF[n_, r0_, σ_] := Function[
           z,
           Evaluate[Simplify[
           f[Sqrt[Sum[Indexed[z, i]^2, i, 1, n]], r0, σ]/
           normalization[n, r0, σ]
           ]]
           ];


          For example, in dimension 2:



          ρ = makePDF[2, r0, σ]



          $$z mapsto frace^fractextr0^2-left(textr0-sqrtz_1^2+z_2^2right)^22 sigma
          ^2sqrt2 pi sigma left(sqrtpi sigma
          e^fractextr0^22 sigma ^2 sqrtfractextr0^2sigma
          ^2 texterfleft(fracsqrtfractextr0^2sigma
          ^2sqrt2right)+sqrtpi textr0 e^fractextr0^22
          sigma ^2+sqrt2 sigma right)$$




          Integrate[ρ[x, y], x, -∞, ∞, y, -∞, ∞, Assumptions -> assumptions]



          1







          share|improve this answer






















          • @ben18785 Good point. Thanks for the edit!
            – Henrik Schumacher
            46 mins ago










          • No problem. Thanks for your work here. My differential geometry isn't what it once (or ever) was!
            – ben18785
            38 mins ago










          • You're welcome!
            – Henrik Schumacher
            36 mins ago










          Your Answer




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          1 Answer
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          active

          oldest

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          active

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          up vote
          5
          down vote













          The numerical integration is not the issue. You also have to calculate a radial integral in order to determine the normalization constant. So, let's do it! Again, we apply polar coordinates:



          ClearAll[r, r0, σ, n];
          f = r, r0, σ [Function] Evaluate[PDF[NormalDistribution[r0, σ], r]];
          assumptions = And @@ n ∈ Integers, n >= 1, r > 0, r0 > 0, σ > 0;
          volS[n_] = (2 (Pi^(n/2)))/Gamma[n/2];
          normalization[n_, r0_, σ_] = volS[n] Integrate[f[r, r0, σ] r^(n - 1), r, 0, ∞, 
           Assumptions -> assumptions];


          This defines a routine that creates a PDF for given dimension n, radius r0 and parameter σ:



          makePDF[n_, r0_, σ_] := Function[
           z,
           Evaluate[Simplify[
           f[Sqrt[Sum[Indexed[z, i]^2, i, 1, n]], r0, σ]/
           normalization[n, r0, σ]
           ]]
           ];


          For example, in dimension 2:



          ρ = makePDF[2, r0, σ]



          $$z mapsto frace^fractextr0^2-left(textr0-sqrtz_1^2+z_2^2right)^22 sigma
          ^2sqrt2 pi sigma left(sqrtpi sigma
          e^fractextr0^22 sigma ^2 sqrtfractextr0^2sigma
          ^2 texterfleft(fracsqrtfractextr0^2sigma
          ^2sqrt2right)+sqrtpi textr0 e^fractextr0^22
          sigma ^2+sqrt2 sigma right)$$




          Integrate[ρ[x, y], x, -∞, ∞, y, -∞, ∞, Assumptions -> assumptions]



          1







          share|improve this answer






















          • @ben18785 Good point. Thanks for the edit!
            – Henrik Schumacher
            46 mins ago










          • No problem. Thanks for your work here. My differential geometry isn't what it once (or ever) was!
            – ben18785
            38 mins ago










          • You're welcome!
            – Henrik Schumacher
            36 mins ago














          up vote
          5
          down vote













          The numerical integration is not the issue. You also have to calculate a radial integral in order to determine the normalization constant. So, let's do it! Again, we apply polar coordinates:



          ClearAll[r, r0, σ, n];
          f = r, r0, σ [Function] Evaluate[PDF[NormalDistribution[r0, σ], r]];
          assumptions = And @@ n ∈ Integers, n >= 1, r > 0, r0 > 0, σ > 0;
          volS[n_] = (2 (Pi^(n/2)))/Gamma[n/2];
          normalization[n_, r0_, σ_] = volS[n] Integrate[f[r, r0, σ] r^(n - 1), r, 0, ∞, 
           Assumptions -> assumptions];


          This defines a routine that creates a PDF for given dimension n, radius r0 and parameter σ:



          makePDF[n_, r0_, σ_] := Function[
           z,
           Evaluate[Simplify[
           f[Sqrt[Sum[Indexed[z, i]^2, i, 1, n]], r0, σ]/
           normalization[n, r0, σ]
           ]]
           ];


          For example, in dimension 2:



          ρ = makePDF[2, r0, σ]



          $$z mapsto frace^fractextr0^2-left(textr0-sqrtz_1^2+z_2^2right)^22 sigma
          ^2sqrt2 pi sigma left(sqrtpi sigma
          e^fractextr0^22 sigma ^2 sqrtfractextr0^2sigma
          ^2 texterfleft(fracsqrtfractextr0^2sigma
          ^2sqrt2right)+sqrtpi textr0 e^fractextr0^22
          sigma ^2+sqrt2 sigma right)$$




          Integrate[ρ[x, y], x, -∞, ∞, y, -∞, ∞, Assumptions -> assumptions]



          1







          share|improve this answer






















          • @ben18785 Good point. Thanks for the edit!
            – Henrik Schumacher
            46 mins ago










          • No problem. Thanks for your work here. My differential geometry isn't what it once (or ever) was!
            – ben18785
            38 mins ago










          • You're welcome!
            – Henrik Schumacher
            36 mins ago












          up vote
          5
          down vote










          up vote
          5
          down vote









          The numerical integration is not the issue. You also have to calculate a radial integral in order to determine the normalization constant. So, let's do it! Again, we apply polar coordinates:



          ClearAll[r, r0, σ, n];
          f = r, r0, σ [Function] Evaluate[PDF[NormalDistribution[r0, σ], r]];
          assumptions = And @@ n ∈ Integers, n >= 1, r > 0, r0 > 0, σ > 0;
          volS[n_] = (2 (Pi^(n/2)))/Gamma[n/2];
          normalization[n_, r0_, σ_] = volS[n] Integrate[f[r, r0, σ] r^(n - 1), r, 0, ∞, 
           Assumptions -> assumptions];


          This defines a routine that creates a PDF for given dimension n, radius r0 and parameter σ:



          makePDF[n_, r0_, σ_] := Function[
           z,
           Evaluate[Simplify[
           f[Sqrt[Sum[Indexed[z, i]^2, i, 1, n]], r0, σ]/
           normalization[n, r0, σ]
           ]]
           ];


          For example, in dimension 2:



          ρ = makePDF[2, r0, σ]



          $$z mapsto frace^fractextr0^2-left(textr0-sqrtz_1^2+z_2^2right)^22 sigma
          ^2sqrt2 pi sigma left(sqrtpi sigma
          e^fractextr0^22 sigma ^2 sqrtfractextr0^2sigma
          ^2 texterfleft(fracsqrtfractextr0^2sigma
          ^2sqrt2right)+sqrtpi textr0 e^fractextr0^22
          sigma ^2+sqrt2 sigma right)$$




          Integrate[ρ[x, y], x, -∞, ∞, y, -∞, ∞, Assumptions -> assumptions]



          1







          share|improve this answer














          The numerical integration is not the issue. You also have to calculate a radial integral in order to determine the normalization constant. So, let's do it! Again, we apply polar coordinates:



          ClearAll[r, r0, σ, n];
          f = r, r0, σ [Function] Evaluate[PDF[NormalDistribution[r0, σ], r]];
          assumptions = And @@ n ∈ Integers, n >= 1, r > 0, r0 > 0, σ > 0;
          volS[n_] = (2 (Pi^(n/2)))/Gamma[n/2];
          normalization[n_, r0_, σ_] = volS[n] Integrate[f[r, r0, σ] r^(n - 1), r, 0, ∞, 
           Assumptions -> assumptions];


          This defines a routine that creates a PDF for given dimension n, radius r0 and parameter σ:



          makePDF[n_, r0_, σ_] := Function[
           z,
           Evaluate[Simplify[
           f[Sqrt[Sum[Indexed[z, i]^2, i, 1, n]], r0, σ]/
           normalization[n, r0, σ]
           ]]
           ];


          For example, in dimension 2:



          ρ = makePDF[2, r0, σ]



          $$z mapsto frace^fractextr0^2-left(textr0-sqrtz_1^2+z_2^2right)^22 sigma
          ^2sqrt2 pi sigma left(sqrtpi sigma
          e^fractextr0^22 sigma ^2 sqrtfractextr0^2sigma
          ^2 texterfleft(fracsqrtfractextr0^2sigma
          ^2sqrt2right)+sqrtpi textr0 e^fractextr0^22
          sigma ^2+sqrt2 sigma right)$$




          Integrate[ρ[x, y], x, -∞, ∞, y, -∞, ∞, Assumptions -> assumptions]



          1








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 47 mins ago









          ben18785

          1,398622




          1,398622










          answered 1 hour ago









          Henrik Schumacher

          39.5k254118




          39.5k254118











          • @ben18785 Good point. Thanks for the edit!
            – Henrik Schumacher
            46 mins ago










          • No problem. Thanks for your work here. My differential geometry isn't what it once (or ever) was!
            – ben18785
            38 mins ago










          • You're welcome!
            – Henrik Schumacher
            36 mins ago
















          • @ben18785 Good point. Thanks for the edit!
            – Henrik Schumacher
            46 mins ago










          • No problem. Thanks for your work here. My differential geometry isn't what it once (or ever) was!
            – ben18785
            38 mins ago










          • You're welcome!
            – Henrik Schumacher
            36 mins ago















          @ben18785 Good point. Thanks for the edit!
          – Henrik Schumacher
          46 mins ago




          @ben18785 Good point. Thanks for the edit!
          – Henrik Schumacher
          46 mins ago












          No problem. Thanks for your work here. My differential geometry isn't what it once (or ever) was!
          – ben18785
          38 mins ago




          No problem. Thanks for your work here. My differential geometry isn't what it once (or ever) was!
          – ben18785
          38 mins ago












          You're welcome!
          – Henrik Schumacher
          36 mins ago




          You're welcome!
          – Henrik Schumacher
          36 mins ago

















           

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