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
$begingroup$
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 $S^n-1$?
probability probability-distributions expected-value
$endgroup$
add a comment |
$begingroup$
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 $S^n-1$?
probability probability-distributions expected-value
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $S^n-1$?
probability probability-distributions expected-value
$endgroup$
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
probability probability-distributions expected-value
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $f$.
$endgroup$
add a comment |
$begingroup$
$ 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
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 $f$.
$endgroup$
add a comment |
$begingroup$
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 $f$.
$endgroup$
add a comment |
$begingroup$
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 $f$.
$endgroup$
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 $f$.
edited Apr 1 at 16:16
answered Apr 1 at 16:10
eyeballfrogeyeballfrog
7,232633
7,232633
add a comment |
add a comment |
$begingroup$
$ 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
$endgroup$
add a comment |
$begingroup$
$ 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
$endgroup$
add a comment |
$begingroup$
$ 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
$endgroup$
$ 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 $, we get
begineqnarray mathbbE_x sim cal D [ max 0, a^tophspace-0.2em x ]&=&frac1sqrt2pisigmaint_-infty^inftymax 0, ye^-fracleft(y-alpharight)^22sigma^2dy\
&=& frac1sqrt2pisigmaint_0^infty ye^-fracleft(y-alpharight)^22sigma^2dy\
&=& frac1sqrt2piint_-fracalphasigma^inftyleft(alpha+sigma zright)e^-fracz^22dz\
&=& alphaleft(1-mathcalN_1(0,,1)left(-fracalphasigmaright)right)+fracsigmasqrt2pi e^-fracalpha^22sigma^2
endeqnarray
answered Apr 2 at 9:53
lonza leggieralonza leggiera
1,480128
1,480128
add a comment |
add a comment |
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$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