Calculate mean and variance of a cone-shaped distribution in N-dimensions

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I would like to calculate the mean and variance of a cone-shaped distribution,



$f(x) propto exp(-|x|)$,



where $xinmathbbR^d$, where $d$ can be any positive integer.



In two-dimensions, this distribution looks like,
enter image description here



a cone! I can calculate the normalising constant for this distribution using,



Integrate[Exp[-Norm[x,y]], x, -Infinity, Infinity, y, -Infinity, Infinity]


which is $2pi$. I can then calculate the mean normed distance using,



Integrate[Norm[x,y]Exp[-Norm[x,y]], x, -Infinity, Infinity,
 y, -Infinity, Infinity]


which is 2. Its second moment,



Integrate[Norm[x,y]^2Exp[-Norm[x,y]], x, -Infinity, Infinity,
 y, -Infinity, Infinity]


then allows me to calculate the variance ($var(x) = E(x^2) - E(x)^2$) = 2.



Calculating the normalising constants is easy enough in higher dimensions, but I run into trouble with finding the mean and variance.



Any ideas?



I'm guessing that something can maybe be done using polar coordinates in higher dimensions but this isn't something I know much about!






numerical-integration probability-or-statistics geometry







share|improve this question

























    up vote
    2
    down vote

    favorite












    I would like to calculate the mean and variance of a cone-shaped distribution,



    $f(x) propto exp(-|x|)$,



    where $xinmathbbR^d$, where $d$ can be any positive integer.



    In two-dimensions, this distribution looks like,
    enter image description here



    a cone! I can calculate the normalising constant for this distribution using,



    Integrate[Exp[-Norm[x,y]], x, -Infinity, Infinity, y, -Infinity, Infinity]


    which is $2pi$. I can then calculate the mean normed distance using,



    Integrate[Norm[x,y]Exp[-Norm[x,y]], x, -Infinity, Infinity,
     y, -Infinity, Infinity]


    which is 2. Its second moment,



    Integrate[Norm[x,y]^2Exp[-Norm[x,y]], x, -Infinity, Infinity,
     y, -Infinity, Infinity]


    then allows me to calculate the variance ($var(x) = E(x^2) - E(x)^2$) = 2.



    Calculating the normalising constants is easy enough in higher dimensions, but I run into trouble with finding the mean and variance.



    Any ideas?



    I'm guessing that something can maybe be done using polar coordinates in higher dimensions but this isn't something I know much about!






    numerical-integration probability-or-statistics geometry







    share|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I would like to calculate the mean and variance of a cone-shaped distribution,



      $f(x) propto exp(-|x|)$,



      where $xinmathbbR^d$, where $d$ can be any positive integer.



      In two-dimensions, this distribution looks like,
      enter image description here



      a cone! I can calculate the normalising constant for this distribution using,



      Integrate[Exp[-Norm[x,y]], x, -Infinity, Infinity, y, -Infinity, Infinity]


      which is $2pi$. I can then calculate the mean normed distance using,



      Integrate[Norm[x,y]Exp[-Norm[x,y]], x, -Infinity, Infinity,
       y, -Infinity, Infinity]


      which is 2. Its second moment,



      Integrate[Norm[x,y]^2Exp[-Norm[x,y]], x, -Infinity, Infinity,
       y, -Infinity, Infinity]


      then allows me to calculate the variance ($var(x) = E(x^2) - E(x)^2$) = 2.



      Calculating the normalising constants is easy enough in higher dimensions, but I run into trouble with finding the mean and variance.



      Any ideas?



      I'm guessing that something can maybe be done using polar coordinates in higher dimensions but this isn't something I know much about!






      numerical-integration probability-or-statistics geometry







      share|improve this question













      I would like to calculate the mean and variance of a cone-shaped distribution,



      $f(x) propto exp(-|x|)$,



      where $xinmathbbR^d$, where $d$ can be any positive integer.



      In two-dimensions, this distribution looks like,
      enter image description here



      a cone! I can calculate the normalising constant for this distribution using,



      Integrate[Exp[-Norm[x,y]], x, -Infinity, Infinity, y, -Infinity, Infinity]


      which is $2pi$. I can then calculate the mean normed distance using,



      Integrate[Norm[x,y]Exp[-Norm[x,y]], x, -Infinity, Infinity,
       y, -Infinity, Infinity]


      which is 2. Its second moment,



      Integrate[Norm[x,y]^2Exp[-Norm[x,y]], x, -Infinity, Infinity,
       y, -Infinity, Infinity]


      then allows me to calculate the variance ($var(x) = E(x^2) - E(x)^2$) = 2.



      Calculating the normalising constants is easy enough in higher dimensions, but I run into trouble with finding the mean and variance.



      Any ideas?



      I'm guessing that something can maybe be done using polar coordinates in higher dimensions but this isn't something I know much about!






      numerical-integration probability-or-statistics geometry




      numerical-integration probability-or-statistics geometry






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 46 mins ago









      ben18785

      1,351622




      1,351622




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          The expected value of the random variable should always be 0 due to symmetry.



          For the normalization, we need to determine $omega$ such that



          $$omega , int_mathbbR^n mathrme^- , mathrmdx = 1.$$



          The expected distance to the origin is



          $$omega , int_mathbbR^n |x|^1 , mathrme^- , mathrmdx.$$



          For the variance, we have to compute



          $$omega , int_mathbbR^n |x|^2 , mathrme^- , mathrmdx.$$



          All these integrals are radially symmetric.



          By introducing polar coorinates, we obtain
          $$int_mathbbR^n |x|^alpha , mathrme^- , mathrmdx =int_S^n-1int_0^infty mathrme^-r , r^alpha+n-1 , mathrmd r , mathrmdS = omega_n int_0^infty mathrme^-r , r^alpha+n-1 , mathrmd r,$$



          where $omega_n = frac(2 pi )^n/2Gamma left(fracn2right)$ is the surface area of the unit sphere.



          Such integrals can be computed symbolically by Mathematica:



          v[α_, n_] = (2 π)^(n/2)/Gamma[n/2] Integrate[r^α Exp[-r] r^(n - 1), r, 0, ∞, 
           Assumptions -> α + n > 0]



          $$ frac(2 pi )^n/2 Gamma (n+alpha )Gamma
          left(fracn2right)$$




          So, the variance should equal



          var[n_] = FullSimplify[
           v[2, n]/v[0, n],
           n ∈ Integers && n > 0]



          $n (1 + n)$




          var /@ Range[10]



          2, 6, 12, 20, 30, 42, 56, 72, 90, 110




          Edit



          The expectation of the norm can be obtained by



          normexpection[n_] = FullSimplify[
           v[1, n]/v[0, n], 
           n ∈ Integers && n > 0
           ]



          n







          share|improve this answer






















          • Thanks! Just a small thing, the expectation shouldn't be zero because I am calculating the expectation of the normed x, not x itself.
            – ben18785
            8 mins ago










          • You're welcome.
            – Henrik Schumacher
            3 mins ago

















          up vote
          2
          down vote













          ClearAll[aa, nc, dist]
          aa[n_Integer] := Array[x, n];
          nc[n_Integer] := 2^n*Pi^((n - 1)/2)*Gamma[(1 + n)/2]
          dist[n_Integer] := ProbabilityDistribution[Exp[-Norm[aa[n]]]/nc[n], 
           ## & @@ Thread[aa[n], -∞, ∞]]

          Expectation[Norm[aa[#]], Distributed[aa[#], dist[#]]] & /@ Range[2, 5]



          2, 3, 4, 5




          Expectation[Norm[aa[#]]^2, Distributed[aa[#], dist[#]]] & /@ Range[2, 5]



          6, 12, 20, 30




          Also:



          td[n_Integer] := TransformedDistribution[Norm[aa[n]], Distributed[aa[n], dist[n]]]

          Mean[td[#]] & /@ Range[2, 5]



          2, 3, 4, 5




          Moment[td[#], 2] & /@ Range[2, 5]



          6, 12, 20, 30




          Variance[td[#]] & /@ Range[2, 5]



          2, 3, 4, 5







          share|improve this answer






















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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote



            accepted










            The expected value of the random variable should always be 0 due to symmetry.



            For the normalization, we need to determine $omega$ such that



            $$omega , int_mathbbR^n mathrme^- , mathrmdx = 1.$$



            The expected distance to the origin is



            $$omega , int_mathbbR^n |x|^1 , mathrme^- , mathrmdx.$$



            For the variance, we have to compute



            $$omega , int_mathbbR^n |x|^2 , mathrme^- , mathrmdx.$$



            All these integrals are radially symmetric.



            By introducing polar coorinates, we obtain
            $$int_mathbbR^n |x|^alpha , mathrme^- , mathrmdx =int_S^n-1int_0^infty mathrme^-r , r^alpha+n-1 , mathrmd r , mathrmdS = omega_n int_0^infty mathrme^-r , r^alpha+n-1 , mathrmd r,$$



            where $omega_n = frac(2 pi )^n/2Gamma left(fracn2right)$ is the surface area of the unit sphere.



            Such integrals can be computed symbolically by Mathematica:



            v[α_, n_] = (2 π)^(n/2)/Gamma[n/2] Integrate[r^α Exp[-r] r^(n - 1), r, 0, ∞, 
             Assumptions -> α + n > 0]



            $$ frac(2 pi )^n/2 Gamma (n+alpha )Gamma
            left(fracn2right)$$




            So, the variance should equal



            var[n_] = FullSimplify[
             v[2, n]/v[0, n],
             n ∈ Integers && n > 0]



            $n (1 + n)$




            var /@ Range[10]



            2, 6, 12, 20, 30, 42, 56, 72, 90, 110




            Edit



            The expectation of the norm can be obtained by



            normexpection[n_] = FullSimplify[
             v[1, n]/v[0, n], 
             n ∈ Integers && n > 0
             ]



            n







            share|improve this answer






















            • Thanks! Just a small thing, the expectation shouldn't be zero because I am calculating the expectation of the normed x, not x itself.
              – ben18785
              8 mins ago










            • You're welcome.
              – Henrik Schumacher
              3 mins ago














            up vote
            1
            down vote



            accepted










            The expected value of the random variable should always be 0 due to symmetry.



            For the normalization, we need to determine $omega$ such that



            $$omega , int_mathbbR^n mathrme^- , mathrmdx = 1.$$



            The expected distance to the origin is



            $$omega , int_mathbbR^n |x|^1 , mathrme^- , mathrmdx.$$



            For the variance, we have to compute



            $$omega , int_mathbbR^n |x|^2 , mathrme^- , mathrmdx.$$



            All these integrals are radially symmetric.



            By introducing polar coorinates, we obtain
            $$int_mathbbR^n |x|^alpha , mathrme^- , mathrmdx =int_S^n-1int_0^infty mathrme^-r , r^alpha+n-1 , mathrmd r , mathrmdS = omega_n int_0^infty mathrme^-r , r^alpha+n-1 , mathrmd r,$$



            where $omega_n = frac(2 pi )^n/2Gamma left(fracn2right)$ is the surface area of the unit sphere.



            Such integrals can be computed symbolically by Mathematica:



            v[α_, n_] = (2 π)^(n/2)/Gamma[n/2] Integrate[r^α Exp[-r] r^(n - 1), r, 0, ∞, 
             Assumptions -> α + n > 0]



            $$ frac(2 pi )^n/2 Gamma (n+alpha )Gamma
            left(fracn2right)$$




            So, the variance should equal



            var[n_] = FullSimplify[
             v[2, n]/v[0, n],
             n ∈ Integers && n > 0]



            $n (1 + n)$




            var /@ Range[10]



            2, 6, 12, 20, 30, 42, 56, 72, 90, 110




            Edit



            The expectation of the norm can be obtained by



            normexpection[n_] = FullSimplify[
             v[1, n]/v[0, n], 
             n ∈ Integers && n > 0
             ]



            n







            share|improve this answer






















            • Thanks! Just a small thing, the expectation shouldn't be zero because I am calculating the expectation of the normed x, not x itself.
              – ben18785
              8 mins ago










            • You're welcome.
              – Henrik Schumacher
              3 mins ago












            up vote
            1
            down vote



            accepted







            up vote
            1
            down vote



            accepted






            The expected value of the random variable should always be 0 due to symmetry.



            For the normalization, we need to determine $omega$ such that



            $$omega , int_mathbbR^n mathrme^- , mathrmdx = 1.$$



            The expected distance to the origin is



            $$omega , int_mathbbR^n |x|^1 , mathrme^- , mathrmdx.$$



            For the variance, we have to compute



            $$omega , int_mathbbR^n |x|^2 , mathrme^- , mathrmdx.$$



            All these integrals are radially symmetric.



            By introducing polar coorinates, we obtain
            $$int_mathbbR^n |x|^alpha , mathrme^- , mathrmdx =int_S^n-1int_0^infty mathrme^-r , r^alpha+n-1 , mathrmd r , mathrmdS = omega_n int_0^infty mathrme^-r , r^alpha+n-1 , mathrmd r,$$



            where $omega_n = frac(2 pi )^n/2Gamma left(fracn2right)$ is the surface area of the unit sphere.



            Such integrals can be computed symbolically by Mathematica:



            v[α_, n_] = (2 π)^(n/2)/Gamma[n/2] Integrate[r^α Exp[-r] r^(n - 1), r, 0, ∞, 
             Assumptions -> α + n > 0]



            $$ frac(2 pi )^n/2 Gamma (n+alpha )Gamma
            left(fracn2right)$$




            So, the variance should equal



            var[n_] = FullSimplify[
             v[2, n]/v[0, n],
             n ∈ Integers && n > 0]



            $n (1 + n)$




            var /@ Range[10]



            2, 6, 12, 20, 30, 42, 56, 72, 90, 110




            Edit



            The expectation of the norm can be obtained by



            normexpection[n_] = FullSimplify[
             v[1, n]/v[0, n], 
             n ∈ Integers && n > 0
             ]



            n







            share|improve this answer














            The expected value of the random variable should always be 0 due to symmetry.



            For the normalization, we need to determine $omega$ such that



            $$omega , int_mathbbR^n mathrme^- , mathrmdx = 1.$$



            The expected distance to the origin is



            $$omega , int_mathbbR^n |x|^1 , mathrme^- , mathrmdx.$$



            For the variance, we have to compute



            $$omega , int_mathbbR^n |x|^2 , mathrme^- , mathrmdx.$$



            All these integrals are radially symmetric.



            By introducing polar coorinates, we obtain
            $$int_mathbbR^n |x|^alpha , mathrme^- , mathrmdx =int_S^n-1int_0^infty mathrme^-r , r^alpha+n-1 , mathrmd r , mathrmdS = omega_n int_0^infty mathrme^-r , r^alpha+n-1 , mathrmd r,$$



            where $omega_n = frac(2 pi )^n/2Gamma left(fracn2right)$ is the surface area of the unit sphere.



            Such integrals can be computed symbolically by Mathematica:



            v[α_, n_] = (2 π)^(n/2)/Gamma[n/2] Integrate[r^α Exp[-r] r^(n - 1), r, 0, ∞, 
             Assumptions -> α + n > 0]



            $$ frac(2 pi )^n/2 Gamma (n+alpha )Gamma
            left(fracn2right)$$




            So, the variance should equal



            var[n_] = FullSimplify[
             v[2, n]/v[0, n],
             n ∈ Integers && n > 0]



            $n (1 + n)$




            var /@ Range[10]



            2, 6, 12, 20, 30, 42, 56, 72, 90, 110




            Edit



            The expectation of the norm can be obtained by



            normexpection[n_] = FullSimplify[
             v[1, n]/v[0, n], 
             n ∈ Integers && n > 0
             ]



            n








            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 1 min ago

























            answered 16 mins ago









            Henrik Schumacher

            39k254115




            39k254115











            • Thanks! Just a small thing, the expectation shouldn't be zero because I am calculating the expectation of the normed x, not x itself.
              – ben18785
              8 mins ago










            • You're welcome.
              – Henrik Schumacher
              3 mins ago
















            • Thanks! Just a small thing, the expectation shouldn't be zero because I am calculating the expectation of the normed x, not x itself.
              – ben18785
              8 mins ago










            • You're welcome.
              – Henrik Schumacher
              3 mins ago















            Thanks! Just a small thing, the expectation shouldn't be zero because I am calculating the expectation of the normed x, not x itself.
            – ben18785
            8 mins ago




            Thanks! Just a small thing, the expectation shouldn't be zero because I am calculating the expectation of the normed x, not x itself.
            – ben18785
            8 mins ago












            You're welcome.
            – Henrik Schumacher
            3 mins ago




            You're welcome.
            – Henrik Schumacher
            3 mins ago










            up vote
            2
            down vote













            ClearAll[aa, nc, dist]
            aa[n_Integer] := Array[x, n];
            nc[n_Integer] := 2^n*Pi^((n - 1)/2)*Gamma[(1 + n)/2]
            dist[n_Integer] := ProbabilityDistribution[Exp[-Norm[aa[n]]]/nc[n], 
             ## & @@ Thread[aa[n], -∞, ∞]]

            Expectation[Norm[aa[#]], Distributed[aa[#], dist[#]]] & /@ Range[2, 5]



            2, 3, 4, 5




            Expectation[Norm[aa[#]]^2, Distributed[aa[#], dist[#]]] & /@ Range[2, 5]



            6, 12, 20, 30




            Also:



            td[n_Integer] := TransformedDistribution[Norm[aa[n]], Distributed[aa[n], dist[n]]]

            Mean[td[#]] & /@ Range[2, 5]



            2, 3, 4, 5




            Moment[td[#], 2] & /@ Range[2, 5]



            6, 12, 20, 30




            Variance[td[#]] & /@ Range[2, 5]



            2, 3, 4, 5







            share|improve this answer


























              up vote
              2
              down vote













              ClearAll[aa, nc, dist]
              aa[n_Integer] := Array[x, n];
              nc[n_Integer] := 2^n*Pi^((n - 1)/2)*Gamma[(1 + n)/2]
              dist[n_Integer] := ProbabilityDistribution[Exp[-Norm[aa[n]]]/nc[n], 
               ## & @@ Thread[aa[n], -∞, ∞]]

              Expectation[Norm[aa[#]], Distributed[aa[#], dist[#]]] & /@ Range[2, 5]



              2, 3, 4, 5




              Expectation[Norm[aa[#]]^2, Distributed[aa[#], dist[#]]] & /@ Range[2, 5]



              6, 12, 20, 30




              Also:



              td[n_Integer] := TransformedDistribution[Norm[aa[n]], Distributed[aa[n], dist[n]]]

              Mean[td[#]] & /@ Range[2, 5]



              2, 3, 4, 5




              Moment[td[#], 2] & /@ Range[2, 5]



              6, 12, 20, 30




              Variance[td[#]] & /@ Range[2, 5]



              2, 3, 4, 5







              share|improve this answer
























                up vote
                2
                down vote










                up vote
                2
                down vote









                ClearAll[aa, nc, dist]
                aa[n_Integer] := Array[x, n];
                nc[n_Integer] := 2^n*Pi^((n - 1)/2)*Gamma[(1 + n)/2]
                dist[n_Integer] := ProbabilityDistribution[Exp[-Norm[aa[n]]]/nc[n], 
                 ## & @@ Thread[aa[n], -∞, ∞]]

                Expectation[Norm[aa[#]], Distributed[aa[#], dist[#]]] & /@ Range[2, 5]



                2, 3, 4, 5




                Expectation[Norm[aa[#]]^2, Distributed[aa[#], dist[#]]] & /@ Range[2, 5]



                6, 12, 20, 30




                Also:



                td[n_Integer] := TransformedDistribution[Norm[aa[n]], Distributed[aa[n], dist[n]]]

                Mean[td[#]] & /@ Range[2, 5]



                2, 3, 4, 5




                Moment[td[#], 2] & /@ Range[2, 5]



                6, 12, 20, 30




                Variance[td[#]] & /@ Range[2, 5]



                2, 3, 4, 5







                share|improve this answer














                ClearAll[aa, nc, dist]
                aa[n_Integer] := Array[x, n];
                nc[n_Integer] := 2^n*Pi^((n - 1)/2)*Gamma[(1 + n)/2]
                dist[n_Integer] := ProbabilityDistribution[Exp[-Norm[aa[n]]]/nc[n], 
                 ## & @@ Thread[aa[n], -∞, ∞]]

                Expectation[Norm[aa[#]], Distributed[aa[#], dist[#]]] & /@ Range[2, 5]



                2, 3, 4, 5




                Expectation[Norm[aa[#]]^2, Distributed[aa[#], dist[#]]] & /@ Range[2, 5]



                6, 12, 20, 30




                Also:



                td[n_Integer] := TransformedDistribution[Norm[aa[n]], Distributed[aa[n], dist[n]]]

                Mean[td[#]] & /@ Range[2, 5]



                2, 3, 4, 5




                Moment[td[#], 2] & /@ Range[2, 5]



                6, 12, 20, 30




                Variance[td[#]] & /@ Range[2, 5]



                2, 3, 4, 5








                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 7 mins ago

























                answered 18 mins ago









                kglr

                161k8185385




                161k8185385



























                     

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                    One-line joke

                    A one-liner is a joke that is delivered in a single line. A good one-liner is said to be pithy - concise and meaningful. [1] Comedians and actors use this comedic method as part of their act, e.g. Jimmy Carr, Tommy Cooper, Rodney Dangerfield, Norm Macdonald, Ken Dodd, Stewart Francis, Zach Galifianakis, Mitch Hedberg, Anthony Jeselnik, Milton Jones, Shappi Khorsandi, Jay London, Mark Linn-Baker, Demetri Martin, Groucho Marx, Dan Mintz, Emo Philips, Tim Vine, Steven Wright and Henny Youngman and Arnold Schwarzenegger. [ citation needed ] Many fictional characters are also known to deliver one-liners, including James Bond, who usually includes pithy and laconic quips after disposing of a villain. [ citation needed ] Examples "A baby seal walks into a club." "A dyslexic man walks into a bra." (Anonymous) "There are three types of people, those who can count and those who can't." "Venison's dear, isn't it?" (Jimmy Carr) "An escala...
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