Expectation of the max operator Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Expectation of max of independent (unknown distribution) random variables.Marginal Distribution of Uniform Vector on SphereDiscrete probability distribution where (max - min) $not=$ averageExpectation of Gaussian random varible composed with max functionProbability of a gaussian lying in half-spaces?Complex Gaussian random vector, vec operator and expectationWhat is the pdf of $N(0,sigma^2)+maxN(0,sigma^2),…,N(0,sigma^2)$?What is the asymptotic distribution of the right singular vectors of a matrix with multivariate gaussian sampled rows?Expectation of linear transformation with noiseAbout a class of expectation calculation

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Expectation of the max operator



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Expectation of max of independent (unknown distribution) random variables.Marginal Distribution of Uniform Vector on SphereDiscrete probability distribution where (max - min) $not=$ averageExpectation of Gaussian random varible composed with max functionProbability of a gaussian lying in half-spaces?Complex Gaussian random vector, vec operator and expectationWhat is the pdf of $N(0,sigma^2)+maxN(0,sigma^2),…,N(0,sigma^2)$?What is the asymptotic distribution of the right singular vectors of a matrix with multivariate gaussian sampled rows?Expectation of linear transformation with noiseAbout a class of expectation calculation










0












$begingroup$


For some $n-$dimensional distribution $cal D$ and a vector $a in mathbbR^n$ can we exactly compute,



$$mathbbE_x sim cal D [ max 0, a^top x ] $$



?



  • At least for the Gaussian distribution on $mathbbR^n$ is this known?

  • At least for say the uniform distribution on $S^n-1$?









share|cite|improve this question









$endgroup$











  • $begingroup$
    It should be easy for any isotropic distribution, since without loss of generality you can assume $mathbfa = ahatmathbfx_n$
    $endgroup$
    – eyeballfrog
    Apr 1 at 14:40











  • $begingroup$
    Yes. And then what would the value of the integral be?
    $endgroup$
    – gradstudent
    Apr 1 at 14:44















0












$begingroup$


For some $n-$dimensional distribution $cal D$ and a vector $a in mathbbR^n$ can we exactly compute,



$$mathbbE_x sim cal D [ max 0, a^top x ] $$



?



  • At least for the Gaussian distribution on $mathbbR^n$ is this known?

  • At least for say the uniform distribution on $S^n-1$?









share|cite|improve this question









$endgroup$











  • $begingroup$
    It should be easy for any isotropic distribution, since without loss of generality you can assume $mathbfa = ahatmathbfx_n$
    $endgroup$
    – eyeballfrog
    Apr 1 at 14:40











  • $begingroup$
    Yes. And then what would the value of the integral be?
    $endgroup$
    – gradstudent
    Apr 1 at 14:44













0












0








0





$begingroup$


For some $n-$dimensional distribution $cal D$ and a vector $a in mathbbR^n$ can we exactly compute,



$$mathbbE_x sim cal D [ max 0, a^top x ] $$



?



  • At least for the Gaussian distribution on $mathbbR^n$ is this known?

  • At least for say the uniform distribution on $S^n-1$?









share|cite|improve this question









$endgroup$




For some $n-$dimensional distribution $cal D$ and a vector $a in mathbbR^n$ can we exactly compute,



$$mathbbE_x sim cal D [ max 0, a^top x ] $$



?



  • At least for the Gaussian distribution on $mathbbR^n$ is this known?

  • At least for say the uniform distribution on $S^n-1$?






probability probability-distributions expected-value






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 1 at 14:34









gradstudentgradstudent

19217




19217











  • $begingroup$
    It should be easy for any isotropic distribution, since without loss of generality you can assume $mathbfa = ahatmathbfx_n$
    $endgroup$
    – eyeballfrog
    Apr 1 at 14:40











  • $begingroup$
    Yes. And then what would the value of the integral be?
    $endgroup$
    – gradstudent
    Apr 1 at 14:44
















  • $begingroup$
    It should be easy for any isotropic distribution, since without loss of generality you can assume $mathbfa = ahatmathbfx_n$
    $endgroup$
    – eyeballfrog
    Apr 1 at 14:40











  • $begingroup$
    Yes. And then what would the value of the integral be?
    $endgroup$
    – gradstudent
    Apr 1 at 14:44















$begingroup$
It should be easy for any isotropic distribution, since without loss of generality you can assume $mathbfa = ahatmathbfx_n$
$endgroup$
– eyeballfrog
Apr 1 at 14:40





$begingroup$
It should be easy for any isotropic distribution, since without loss of generality you can assume $mathbfa = ahatmathbfx_n$
$endgroup$
– eyeballfrog
Apr 1 at 14:40













$begingroup$
Yes. And then what would the value of the integral be?
$endgroup$
– gradstudent
Apr 1 at 14:44




$begingroup$
Yes. And then what would the value of the integral be?
$endgroup$
– gradstudent
Apr 1 at 14:44










2 Answers
2






active

oldest

votes


















0












$begingroup$

If the distribution $f$ is isotropic (that is, $f = f(r)$), this can be calculated exactly whenever $E(|mathbfx|)$ can using $n$-dimensional spherical coordinates. Let $r$ represent the radial coordinate, $theta$ represent the polar angle, and $Omega$ represent the remaining spherical angles, which form the surface of a unit sphere in $n-1$ dimensions. Without loss of generality, assume $mathbfa$ is parallel to the polar axis. Then $mathbfacdot mathbfx = arcostheta$ and
beginmultline
E(maxmathbfacdotmathbfx,0) = int (mathbfacdotmathbfx)f d^nmathbfx \= int_0^inftyint_0^pileft[int_0^2piright]^n-2maxarcostheta,0 f(r)r^n-1sin^n-2theta ,d^n-2Omega ,dtheta, dr.
endmultline

where $r^n-1sin^n-2theta, d^n-2Omega,dtheta,dr$ is the n-dimensional spherical volume element. Since the factors all depend on one variable, we can separate this integral. Since $costheta < 0$ for $theta > pi/2$, we have
beginmultline
...= aint_0^infty rf(r)r^n-1drint_0^pi/2costhetasin^n-2theta dthetaleft[int_0^2piright]^n-2d^n-2Omega \= frac2api^(n-1)/2(n-1)Gamma((n-1)/2)int_0^infty r^n f(r)dr = fracapi^(n-1)/2Gamma((n+1)/2)int_0^infty r^nf(r)dr
endmultline

where we have used $int_0^pi/2sin^n-2thetacostheta,dtheta = 1/(n-1)$ and the fact that a sphere in $n$ dimensions has surface area $2pi^n/2/Gamma(n/2)$.



Lastly, from
$$
E(|mathbfx|) = int |mathbfx|f d^n mathbfx = 2fracpi^n/2Gamma(n/2)int_0^infty r^nf(r)dr
$$

(again using the $n$-dimensional surface area formula), we have
$$
E(maxmathbfacdotmathbfx,0) =fracapi^(n-1)/2Gamma((n+1)/2)int_0^infty r^nf(r)dr = fraca2sqrtpifracGamma(n/2)Gamma((n+1)/2)E(|mathbfx|)
$$

which is the general form for an isotropic distribution $f$.






share|cite|improve this answer











$endgroup$




















    0












    $begingroup$

    $ mathbbE_x sim cal D [ max 0, a^tophspace-0.2em x ] $ can also be calculated exactly whenever $ cal D=mathcalN_n(mu,,Sigma) $ is multivariate normal. In this case $ a^tophspace-0.2emx simmathcalN_1(a^tophspace-0.2emmu,,a^topSigma a) $, so putting $ alpha=a^tophspace-0.2emmu $ and $ sigma=sqrta^topSigma a $, we get
    begineqnarray mathbbE_x sim cal D [ max 0, a^tophspace-0.2em x ]&=&frac1sqrt2pisigmaint_-infty^inftymax 0, ye^-fracleft(y-alpharight)^22sigma^2dy\
    &=& frac1sqrt2pisigmaint_0^infty ye^-fracleft(y-alpharight)^22sigma^2dy\
    &=& frac1sqrt2piint_-fracalphasigma^inftyleft(alpha+sigma zright)e^-fracz^22dz\
    &=& alphaleft(1-mathcalN_1(0,,1)left(-fracalphasigmaright)right)+fracsigmasqrt2pi e^-fracalpha^22sigma^2
    endeqnarray






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      If the distribution $f$ is isotropic (that is, $f = f(r)$), this can be calculated exactly whenever $E(|mathbfx|)$ can using $n$-dimensional spherical coordinates. Let $r$ represent the radial coordinate, $theta$ represent the polar angle, and $Omega$ represent the remaining spherical angles, which form the surface of a unit sphere in $n-1$ dimensions. Without loss of generality, assume $mathbfa$ is parallel to the polar axis. Then $mathbfacdot mathbfx = arcostheta$ and
      beginmultline
      E(maxmathbfacdotmathbfx,0) = int (mathbfacdotmathbfx)f d^nmathbfx \= int_0^inftyint_0^pileft[int_0^2piright]^n-2maxarcostheta,0 f(r)r^n-1sin^n-2theta ,d^n-2Omega ,dtheta, dr.
      endmultline

      where $r^n-1sin^n-2theta, d^n-2Omega,dtheta,dr$ is the n-dimensional spherical volume element. Since the factors all depend on one variable, we can separate this integral. Since $costheta < 0$ for $theta > pi/2$, we have
      beginmultline
      ...= aint_0^infty rf(r)r^n-1drint_0^pi/2costhetasin^n-2theta dthetaleft[int_0^2piright]^n-2d^n-2Omega \= frac2api^(n-1)/2(n-1)Gamma((n-1)/2)int_0^infty r^n f(r)dr = fracapi^(n-1)/2Gamma((n+1)/2)int_0^infty r^nf(r)dr
      endmultline

      where we have used $int_0^pi/2sin^n-2thetacostheta,dtheta = 1/(n-1)$ and the fact that a sphere in $n$ dimensions has surface area $2pi^n/2/Gamma(n/2)$.



      Lastly, from
      $$
      E(|mathbfx|) = int |mathbfx|f d^n mathbfx = 2fracpi^n/2Gamma(n/2)int_0^infty r^nf(r)dr
      $$

      (again using the $n$-dimensional surface area formula), we have
      $$
      E(maxmathbfacdotmathbfx,0) =fracapi^(n-1)/2Gamma((n+1)/2)int_0^infty r^nf(r)dr = fraca2sqrtpifracGamma(n/2)Gamma((n+1)/2)E(|mathbfx|)
      $$

      which is the general form for an isotropic distribution $f$.






      share|cite|improve this answer











      $endgroup$

















        0












        $begingroup$

        If the distribution $f$ is isotropic (that is, $f = f(r)$), this can be calculated exactly whenever $E(|mathbfx|)$ can using $n$-dimensional spherical coordinates. Let $r$ represent the radial coordinate, $theta$ represent the polar angle, and $Omega$ represent the remaining spherical angles, which form the surface of a unit sphere in $n-1$ dimensions. Without loss of generality, assume $mathbfa$ is parallel to the polar axis. Then $mathbfacdot mathbfx = arcostheta$ and
        beginmultline
        E(maxmathbfacdotmathbfx,0) = int (mathbfacdotmathbfx)f d^nmathbfx \= int_0^inftyint_0^pileft[int_0^2piright]^n-2maxarcostheta,0 f(r)r^n-1sin^n-2theta ,d^n-2Omega ,dtheta, dr.
        endmultline

        where $r^n-1sin^n-2theta, d^n-2Omega,dtheta,dr$ is the n-dimensional spherical volume element. Since the factors all depend on one variable, we can separate this integral. Since $costheta < 0$ for $theta > pi/2$, we have
        beginmultline
        ...= aint_0^infty rf(r)r^n-1drint_0^pi/2costhetasin^n-2theta dthetaleft[int_0^2piright]^n-2d^n-2Omega \= frac2api^(n-1)/2(n-1)Gamma((n-1)/2)int_0^infty r^n f(r)dr = fracapi^(n-1)/2Gamma((n+1)/2)int_0^infty r^nf(r)dr
        endmultline

        where we have used $int_0^pi/2sin^n-2thetacostheta,dtheta = 1/(n-1)$ and the fact that a sphere in $n$ dimensions has surface area $2pi^n/2/Gamma(n/2)$.



        Lastly, from
        $$
        E(|mathbfx|) = int |mathbfx|f d^n mathbfx = 2fracpi^n/2Gamma(n/2)int_0^infty r^nf(r)dr
        $$

        (again using the $n$-dimensional surface area formula), we have
        $$
        E(maxmathbfacdotmathbfx,0) =fracapi^(n-1)/2Gamma((n+1)/2)int_0^infty r^nf(r)dr = fraca2sqrtpifracGamma(n/2)Gamma((n+1)/2)E(|mathbfx|)
        $$

        which is the general form for an isotropic distribution $f$.






        share|cite|improve this answer











        $endgroup$















          0












          0








          0





          $begingroup$

          If the distribution $f$ is isotropic (that is, $f = f(r)$), this can be calculated exactly whenever $E(|mathbfx|)$ can using $n$-dimensional spherical coordinates. Let $r$ represent the radial coordinate, $theta$ represent the polar angle, and $Omega$ represent the remaining spherical angles, which form the surface of a unit sphere in $n-1$ dimensions. Without loss of generality, assume $mathbfa$ is parallel to the polar axis. Then $mathbfacdot mathbfx = arcostheta$ and
          beginmultline
          E(maxmathbfacdotmathbfx,0) = int (mathbfacdotmathbfx)f d^nmathbfx \= int_0^inftyint_0^pileft[int_0^2piright]^n-2maxarcostheta,0 f(r)r^n-1sin^n-2theta ,d^n-2Omega ,dtheta, dr.
          endmultline

          where $r^n-1sin^n-2theta, d^n-2Omega,dtheta,dr$ is the n-dimensional spherical volume element. Since the factors all depend on one variable, we can separate this integral. Since $costheta < 0$ for $theta > pi/2$, we have
          beginmultline
          ...= aint_0^infty rf(r)r^n-1drint_0^pi/2costhetasin^n-2theta dthetaleft[int_0^2piright]^n-2d^n-2Omega \= frac2api^(n-1)/2(n-1)Gamma((n-1)/2)int_0^infty r^n f(r)dr = fracapi^(n-1)/2Gamma((n+1)/2)int_0^infty r^nf(r)dr
          endmultline

          where we have used $int_0^pi/2sin^n-2thetacostheta,dtheta = 1/(n-1)$ and the fact that a sphere in $n$ dimensions has surface area $2pi^n/2/Gamma(n/2)$.



          Lastly, from
          $$
          E(|mathbfx|) = int |mathbfx|f d^n mathbfx = 2fracpi^n/2Gamma(n/2)int_0^infty r^nf(r)dr
          $$

          (again using the $n$-dimensional surface area formula), we have
          $$
          E(maxmathbfacdotmathbfx,0) =fracapi^(n-1)/2Gamma((n+1)/2)int_0^infty r^nf(r)dr = fraca2sqrtpifracGamma(n/2)Gamma((n+1)/2)E(|mathbfx|)
          $$

          which is the general form for an isotropic distribution $f$.






          share|cite|improve this answer











          $endgroup$



          If the distribution $f$ is isotropic (that is, $f = f(r)$), this can be calculated exactly whenever $E(|mathbfx|)$ can using $n$-dimensional spherical coordinates. Let $r$ represent the radial coordinate, $theta$ represent the polar angle, and $Omega$ represent the remaining spherical angles, which form the surface of a unit sphere in $n-1$ dimensions. Without loss of generality, assume $mathbfa$ is parallel to the polar axis. Then $mathbfacdot mathbfx = arcostheta$ and
          beginmultline
          E(maxmathbfacdotmathbfx,0) = int (mathbfacdotmathbfx)f d^nmathbfx \= int_0^inftyint_0^pileft[int_0^2piright]^n-2maxarcostheta,0 f(r)r^n-1sin^n-2theta ,d^n-2Omega ,dtheta, dr.
          endmultline

          where $r^n-1sin^n-2theta, d^n-2Omega,dtheta,dr$ is the n-dimensional spherical volume element. Since the factors all depend on one variable, we can separate this integral. Since $costheta < 0$ for $theta > pi/2$, we have
          beginmultline
          ...= aint_0^infty rf(r)r^n-1drint_0^pi/2costhetasin^n-2theta dthetaleft[int_0^2piright]^n-2d^n-2Omega \= frac2api^(n-1)/2(n-1)Gamma((n-1)/2)int_0^infty r^n f(r)dr = fracapi^(n-1)/2Gamma((n+1)/2)int_0^infty r^nf(r)dr
          endmultline

          where we have used $int_0^pi/2sin^n-2thetacostheta,dtheta = 1/(n-1)$ and the fact that a sphere in $n$ dimensions has surface area $2pi^n/2/Gamma(n/2)$.



          Lastly, from
          $$
          E(|mathbfx|) = int |mathbfx|f d^n mathbfx = 2fracpi^n/2Gamma(n/2)int_0^infty r^nf(r)dr
          $$

          (again using the $n$-dimensional surface area formula), we have
          $$
          E(maxmathbfacdotmathbfx,0) =fracapi^(n-1)/2Gamma((n+1)/2)int_0^infty r^nf(r)dr = fraca2sqrtpifracGamma(n/2)Gamma((n+1)/2)E(|mathbfx|)
          $$

          which is the general form for an isotropic distribution $f$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Apr 1 at 16:16

























          answered Apr 1 at 16:10









          eyeballfrogeyeballfrog

          7,232633




          7,232633





















              0












              $begingroup$

              $ mathbbE_x sim cal D [ max 0, a^tophspace-0.2em x ] $ can also be calculated exactly whenever $ cal D=mathcalN_n(mu,,Sigma) $ is multivariate normal. In this case $ a^tophspace-0.2emx simmathcalN_1(a^tophspace-0.2emmu,,a^topSigma a) $, so putting $ alpha=a^tophspace-0.2emmu $ and $ sigma=sqrta^topSigma a $, we get
              begineqnarray mathbbE_x sim cal D [ max 0, a^tophspace-0.2em x ]&=&frac1sqrt2pisigmaint_-infty^inftymax 0, ye^-fracleft(y-alpharight)^22sigma^2dy\
              &=& frac1sqrt2pisigmaint_0^infty ye^-fracleft(y-alpharight)^22sigma^2dy\
              &=& frac1sqrt2piint_-fracalphasigma^inftyleft(alpha+sigma zright)e^-fracz^22dz\
              &=& alphaleft(1-mathcalN_1(0,,1)left(-fracalphasigmaright)right)+fracsigmasqrt2pi e^-fracalpha^22sigma^2
              endeqnarray






              share|cite|improve this answer









              $endgroup$

















                0












                $begingroup$

                $ mathbbE_x sim cal D [ max 0, a^tophspace-0.2em x ] $ can also be calculated exactly whenever $ cal D=mathcalN_n(mu,,Sigma) $ is multivariate normal. In this case $ a^tophspace-0.2emx simmathcalN_1(a^tophspace-0.2emmu,,a^topSigma a) $, so putting $ alpha=a^tophspace-0.2emmu $ and $ sigma=sqrta^topSigma a $, we get
                begineqnarray mathbbE_x sim cal D [ max 0, a^tophspace-0.2em x ]&=&frac1sqrt2pisigmaint_-infty^inftymax 0, ye^-fracleft(y-alpharight)^22sigma^2dy\
                &=& frac1sqrt2pisigmaint_0^infty ye^-fracleft(y-alpharight)^22sigma^2dy\
                &=& frac1sqrt2piint_-fracalphasigma^inftyleft(alpha+sigma zright)e^-fracz^22dz\
                &=& alphaleft(1-mathcalN_1(0,,1)left(-fracalphasigmaright)right)+fracsigmasqrt2pi e^-fracalpha^22sigma^2
                endeqnarray






                share|cite|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  $ mathbbE_x sim cal D [ max 0, a^tophspace-0.2em x ] $ can also be calculated exactly whenever $ cal D=mathcalN_n(mu,,Sigma) $ is multivariate normal. In this case $ a^tophspace-0.2emx simmathcalN_1(a^tophspace-0.2emmu,,a^topSigma a) $, so putting $ alpha=a^tophspace-0.2emmu $ and $ sigma=sqrta^topSigma a $, we get
                  begineqnarray mathbbE_x sim cal D [ max 0, a^tophspace-0.2em x ]&=&frac1sqrt2pisigmaint_-infty^inftymax 0, ye^-fracleft(y-alpharight)^22sigma^2dy\
                  &=& frac1sqrt2pisigmaint_0^infty ye^-fracleft(y-alpharight)^22sigma^2dy\
                  &=& frac1sqrt2piint_-fracalphasigma^inftyleft(alpha+sigma zright)e^-fracz^22dz\
                  &=& alphaleft(1-mathcalN_1(0,,1)left(-fracalphasigmaright)right)+fracsigmasqrt2pi e^-fracalpha^22sigma^2
                  endeqnarray






                  share|cite|improve this answer









                  $endgroup$



                  $ mathbbE_x sim cal D [ max 0, a^tophspace-0.2em x ] $ can also be calculated exactly whenever $ cal D=mathcalN_n(mu,,Sigma) $ is multivariate normal. In this case $ a^tophspace-0.2emx simmathcalN_1(a^tophspace-0.2emmu,,a^topSigma a) $, so putting $ alpha=a^tophspace-0.2emmu $ and $ sigma=sqrta^topSigma a $, we get
                  begineqnarray mathbbE_x sim cal D [ max 0, a^tophspace-0.2em x ]&=&frac1sqrt2pisigmaint_-infty^inftymax 0, ye^-fracleft(y-alpharight)^22sigma^2dy\
                  &=& frac1sqrt2pisigmaint_0^infty ye^-fracleft(y-alpharight)^22sigma^2dy\
                  &=& frac1sqrt2piint_-fracalphasigma^inftyleft(alpha+sigma zright)e^-fracz^22dz\
                  &=& alphaleft(1-mathcalN_1(0,,1)left(-fracalphasigmaright)right)+fracsigmasqrt2pi e^-fracalpha^22sigma^2
                  endeqnarray







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Apr 2 at 9:53









                  lonza leggieralonza leggiera

                  1,480128




                  1,480128



























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