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Expected value of inner product of uniformly distributed random unit vectors
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I am trying to understand the solution to the following problem:
Let $mathbfx, mathbfy$ be two independent, uniformly distributed random unit vectors. Show that $mathbbElanglemathbfx, mathbfy rangle^2 = Theta(frac1n)$
The solution goes as follows:
Let $mathbfx$ and $mathbfy$ be two random unit vectors in $mathbbR^n$. By rotational invariance, all random variables of the form $langle mathbfx, yrangle^2$ have the same distribution, where $y$ is a unit vector $mathbbElangle mathbfx, e_1 rangle^2$ (i.e. possible realizations of $mathbfy)$. If follows that $mathbbElanglemathbfx, mathbfy rangle^2 = mathbbElanglemathbfx, e_1 rangle^2$ where $e_1$ is the first coordinate vector[*].
Knowing that we can represent $mathbfx$ in terms of a standard Gaussian vector, we write $mathbfx = fracu$ where $mathbfu sim N(0, mathrmId)$. Multiplying a Gaussian $ N(mu, Sigma)$ with a matrix $A$ gives another Gaussian $N(Amu, ASigma A^top)$. Therefore $langlemathbfx, e_1 rangle^2 sim N(0, 1)$. As a result, we want to compute the expectation of the random variable: $mathbfX = fracmathbfu_1^2mathbfu_1^2+mathbfu_2^2+dots+mathbfu_n^2$ with $mathbfu_i stackreliidsim N(0,1)$.
The random variables $mathbfX_i=fracmathbfu_i^2mathbfu_1^2+mathbfu_2^2+dots+mathbfu_n^2$ for $i in [n]$ have the same distribution and therefore the same expectation. We have that $sum_i mathbfX_i = fracmathbfu_1^2+mathbfu_2^2+dots+mathbfu_n^2mathbfu_1^2+mathbfu_2^2+dots+mathbfu_n^2 = 1$. By linearity of expectation we conclude that $mathbbEmathbfX_1 = frac1n$.
The part that I don't understand is the [*] one. How we can restrict $mathbfy$ to be just some coordinate vector? Can anyone give a more intuitive explanation that maybe will involve some calculations using the rotational invariance?
probability expectation
asked Aug 6 at 17:00
dimkou
535
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up vote
3
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favorite
I am trying to understand the solution to the following problem:
Let $mathbfx, mathbfy$ be two independent, uniformly distributed random unit vectors. Show that $mathbbElanglemathbfx, mathbfy rangle^2 = Theta(frac1n)$
The solution goes as follows:
Let $mathbfx$ and $mathbfy$ be two random unit vectors in $mathbbR^n$. By rotational invariance, all random variables of the form $langle mathbfx, yrangle^2$ have the same distribution, where $y$ is a unit vector $mathbbElangle mathbfx, e_1 rangle^2$ (i.e. possible realizations of $mathbfy)$. If follows that $mathbbElanglemathbfx, mathbfy rangle^2 = mathbbElanglemathbfx, e_1 rangle^2$ where $e_1$ is the first coordinate vector[*].
Knowing that we can represent $mathbfx$ in terms of a standard Gaussian vector, we write $mathbfx = fracu$ where $mathbfu sim N(0, mathrmId)$. Multiplying a Gaussian $ N(mu, Sigma)$ with a matrix $A$ gives another Gaussian $N(Amu, ASigma A^top)$. Therefore $langlemathbfx, e_1 rangle^2 sim N(0, 1)$. As a result, we want to compute the expectation of the random variable: $mathbfX = fracmathbfu_1^2mathbfu_1^2+mathbfu_2^2+dots+mathbfu_n^2$ with $mathbfu_i stackreliidsim N(0,1)$.
The random variables $mathbfX_i=fracmathbfu_i^2mathbfu_1^2+mathbfu_2^2+dots+mathbfu_n^2$ for $i in [n]$ have the same distribution and therefore the same expectation. We have that $sum_i mathbfX_i = fracmathbfu_1^2+mathbfu_2^2+dots+mathbfu_n^2mathbfu_1^2+mathbfu_2^2+dots+mathbfu_n^2 = 1$. By linearity of expectation we conclude that $mathbbEmathbfX_1 = frac1n$.
The part that I don't understand is the [*] one. How we can restrict $mathbfy$ to be just some coordinate vector? Can anyone give a more intuitive explanation that maybe will involve some calculations using the rotational invariance?
probability expectation
asked Aug 6 at 17:00
dimkou
535
1
Let $R$ be the rotation which brings $y$ to $e_1$. Then $langle x,yrangle = langle Rx, Ryrangle = langle Rx, e_1rangle$, but $Rx$ and $x$ are identically distributed.
– Jack D'Aurizio♦
Aug 6 at 17:07
Thanks! Can you elaborate a bit on why $Rx$ and $x$ have the same distribution? Is it because the uniform distribution doesn't change for affine tranformations?
– dimkou
Aug 6 at 17:12
1
A uniform distribution over a (hyper-)sphere stays uniform if such sphere is rotated.
– Jack D'Aurizio♦
Aug 6 at 17:14
$langle x,yrangle = |x|cdot|y|cdotcostheta$, where $theta$ is the angle between $x$ and $y$. $theta$ is invariant under isometric transformations (isometric $to$ conformal), and so they are $|x|$ and $|y|$.
– dimkou
Aug 6 at 17:16
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I am trying to understand the solution to the following problem:
Let $mathbfx, mathbfy$ be two independent, uniformly distributed random unit vectors. Show that $mathbbElanglemathbfx, mathbfy rangle^2 = Theta(frac1n)$
The solution goes as follows:
Let $mathbfx$ and $mathbfy$ be two random unit vectors in $mathbbR^n$. By rotational invariance, all random variables of the form $langle mathbfx, yrangle^2$ have the same distribution, where $y$ is a unit vector $mathbbElangle mathbfx, e_1 rangle^2$ (i.e. possible realizations of $mathbfy)$. If follows that $mathbbElanglemathbfx, mathbfy rangle^2 = mathbbElanglemathbfx, e_1 rangle^2$ where $e_1$ is the first coordinate vector[*].
Knowing that we can represent $mathbfx$ in terms of a standard Gaussian vector, we write $mathbfx = fracu$ where $mathbfu sim N(0, mathrmId)$. Multiplying a Gaussian $ N(mu, Sigma)$ with a matrix $A$ gives another Gaussian $N(Amu, ASigma A^top)$. Therefore $langlemathbfx, e_1 rangle^2 sim N(0, 1)$. As a result, we want to compute the expectation of the random variable: $mathbfX = fracmathbfu_1^2mathbfu_1^2+mathbfu_2^2+dots+mathbfu_n^2$ with $mathbfu_i stackreliidsim N(0,1)$.
The random variables $mathbfX_i=fracmathbfu_i^2mathbfu_1^2+mathbfu_2^2+dots+mathbfu_n^2$ for $i in [n]$ have the same distribution and therefore the same expectation. We have that $sum_i mathbfX_i = fracmathbfu_1^2+mathbfu_2^2+dots+mathbfu_n^2mathbfu_1^2+mathbfu_2^2+dots+mathbfu_n^2 = 1$. By linearity of expectation we conclude that $mathbbEmathbfX_1 = frac1n$.
The part that I don't understand is the [*] one. How we can restrict $mathbfy$ to be just some coordinate vector? Can anyone give a more intuitive explanation that maybe will involve some calculations using the rotational invariance?
probability expectation
asked Aug 6 at 17:00
dimkou
535
I am trying to understand the solution to the following problem:
Let $mathbfx, mathbfy$ be two independent, uniformly distributed random unit vectors. Show that $mathbbElanglemathbfx, mathbfy rangle^2 = Theta(frac1n)$
The solution goes as follows:
Let $mathbfx$ and $mathbfy$ be two random unit vectors in $mathbbR^n$. By rotational invariance, all random variables of the form $langle mathbfx, yrangle^2$ have the same distribution, where $y$ is a unit vector $mathbbElangle mathbfx, e_1 rangle^2$ (i.e. possible realizations of $mathbfy)$. If follows that $mathbbElanglemathbfx, mathbfy rangle^2 = mathbbElanglemathbfx, e_1 rangle^2$ where $e_1$ is the first coordinate vector[*].
Knowing that we can represent $mathbfx$ in terms of a standard Gaussian vector, we write $mathbfx = fracu$ where $mathbfu sim N(0, mathrmId)$. Multiplying a Gaussian $ N(mu, Sigma)$ with a matrix $A$ gives another Gaussian $N(Amu, ASigma A^top)$. Therefore $langlemathbfx, e_1 rangle^2 sim N(0, 1)$. As a result, we want to compute the expectation of the random variable: $mathbfX = fracmathbfu_1^2mathbfu_1^2+mathbfu_2^2+dots+mathbfu_n^2$ with $mathbfu_i stackreliidsim N(0,1)$.
The random variables $mathbfX_i=fracmathbfu_i^2mathbfu_1^2+mathbfu_2^2+dots+mathbfu_n^2$ for $i in [n]$ have the same distribution and therefore the same expectation. We have that $sum_i mathbfX_i = fracmathbfu_1^2+mathbfu_2^2+dots+mathbfu_n^2mathbfu_1^2+mathbfu_2^2+dots+mathbfu_n^2 = 1$. By linearity of expectation we conclude that $mathbbEmathbfX_1 = frac1n$.
The part that I don't understand is the [*] one. How we can restrict $mathbfy$ to be just some coordinate vector? Can anyone give a more intuitive explanation that maybe will involve some calculations using the rotational invariance?
probability expectation
asked Aug 6 at 17:00
dimkou
535
asked Aug 6 at 17:00
dimkou
535
asked Aug 6 at 17:00
dimkou
535
asked Aug 6 at 17:00
dimkou
535
535
1
Let $R$ be the rotation which brings $y$ to $e_1$. Then $langle x,yrangle = langle Rx, Ryrangle = langle Rx, e_1rangle$, but $Rx$ and $x$ are identically distributed.
– Jack D'Aurizio♦
Aug 6 at 17:07
Thanks! Can you elaborate a bit on why $Rx$ and $x$ have the same distribution? Is it because the uniform distribution doesn't change for affine tranformations?
– dimkou
Aug 6 at 17:12
1
A uniform distribution over a (hyper-)sphere stays uniform if such sphere is rotated.
– Jack D'Aurizio♦
Aug 6 at 17:14
$langle x,yrangle = |x|cdot|y|cdotcostheta$, where $theta$ is the angle between $x$ and $y$. $theta$ is invariant under isometric transformations (isometric $to$ conformal), and so they are $|x|$ and $|y|$.
– dimkou
Aug 6 at 17:16
add a comment |Â
1
Let $R$ be the rotation which brings $y$ to $e_1$. Then $langle x,yrangle = langle Rx, Ryrangle = langle Rx, e_1rangle$, but $Rx$ and $x$ are identically distributed.
– Jack D'Aurizio♦
Aug 6 at 17:07
Thanks! Can you elaborate a bit on why $Rx$ and $x$ have the same distribution? Is it because the uniform distribution doesn't change for affine tranformations?
– dimkou
Aug 6 at 17:12
1
A uniform distribution over a (hyper-)sphere stays uniform if such sphere is rotated.
– Jack D'Aurizio♦
Aug 6 at 17:14
$langle x,yrangle = |x|cdot|y|cdotcostheta$, where $theta$ is the angle between $x$ and $y$. $theta$ is invariant under isometric transformations (isometric $to$ conformal), and so they are $|x|$ and $|y|$.
– dimkou
Aug 6 at 17:16
1
1
Let $R$ be the rotation which brings $y$ to $e_1$. Then $langle x,yrangle = langle Rx, Ryrangle = langle Rx, e_1rangle$, but $Rx$ and $x$ are identically distributed.
– Jack D'Aurizio♦
Aug 6 at 17:07
Let $R$ be the rotation which brings $y$ to $e_1$. Then $langle x,yrangle = langle Rx, Ryrangle = langle Rx, e_1rangle$, but $Rx$ and $x$ are identically distributed.
– Jack D'Aurizio♦
Aug 6 at 17:07
Thanks! Can you elaborate a bit on why $Rx$ and $x$ have the same distribution? Is it because the uniform distribution doesn't change for affine tranformations?
– dimkou
Aug 6 at 17:12
Thanks! Can you elaborate a bit on why $Rx$ and $x$ have the same distribution? Is it because the uniform distribution doesn't change for affine tranformations?
– dimkou
Aug 6 at 17:12
1
1
A uniform distribution over a (hyper-)sphere stays uniform if such sphere is rotated.
– Jack D'Aurizio♦
Aug 6 at 17:14
A uniform distribution over a (hyper-)sphere stays uniform if such sphere is rotated.
– Jack D'Aurizio♦
Aug 6 at 17:14
$langle x,yrangle = |x|cdot|y|cdotcostheta$, where $theta$ is the angle between $x$ and $y$. $theta$ is invariant under isometric transformations (isometric $to$ conformal), and so they are $|x|$ and $|y|$.
– dimkou
Aug 6 at 17:16
$langle x,yrangle = |x|cdot|y|cdotcostheta$, where $theta$ is the angle between $x$ and $y$. $theta$ is invariant under isometric transformations (isometric $to$ conformal), and so they are $|x|$ and $|y|$.
– dimkou
Aug 6 at 17:16
add a comment |Â
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1
Let $R$ be the rotation which brings $y$ to $e_1$. Then $langle x,yrangle = langle Rx, Ryrangle = langle Rx, e_1rangle$, but $Rx$ and $x$ are identically distributed.
– Jack D'Aurizio♦
Aug 6 at 17:07
Thanks! Can you elaborate a bit on why $Rx$ and $x$ have the same distribution? Is it because the uniform distribution doesn't change for affine tranformations?
– dimkou
Aug 6 at 17:12
1
A uniform distribution over a (hyper-)sphere stays uniform if such sphere is rotated.
– Jack D'Aurizio♦
Aug 6 at 17:14
$langle x,yrangle = |x|cdot|y|cdotcostheta$, where $theta$ is the angle between $x$ and $y$. $theta$ is invariant under isometric transformations (isometric $to$ conformal), and so they are $|x|$ and $|y|$.
– dimkou
Aug 6 at 17:16