Distribution in Unit disk
Let $X$ and $Y$ be i.i.d random variables that are uniformly distributed on the positive orthant of the unit sphere in $mathbb{R}^n$. What is the distribution of $X - Y$? In particular, does $X-Y$ yield a uniform distribution over the angles of the vectors obtained by subtracting two uniformly random vectors on the positive orthant of the unit sphere?
probability probability-distributions
asked Nov 26 at 17:51
user114743
1026
add a comment |
Let $X$ and $Y$ be i.i.d random variables that are uniformly distributed on the positive orthant of the unit sphere in $mathbb{R}^n$. What is the distribution of $X - Y$? In particular, does $X-Y$ yield a uniform distribution over the angles of the vectors obtained by subtracting two uniformly random vectors on the positive orthant of the unit sphere?
probability probability-distributions
asked Nov 26 at 17:51
user114743
1026
Would you get a uniform distribution of angles with $n=2$?
– Henry
Nov 26 at 18:48
add a comment |
Let $X$ and $Y$ be i.i.d random variables that are uniformly distributed on the positive orthant of the unit sphere in $mathbb{R}^n$. What is the distribution of $X - Y$? In particular, does $X-Y$ yield a uniform distribution over the angles of the vectors obtained by subtracting two uniformly random vectors on the positive orthant of the unit sphere?
probability probability-distributions
asked Nov 26 at 17:51
user114743
1026
Let $X$ and $Y$ be i.i.d random variables that are uniformly distributed on the positive orthant of the unit sphere in $mathbb{R}^n$. What is the distribution of $X - Y$? In particular, does $X-Y$ yield a uniform distribution over the angles of the vectors obtained by subtracting two uniformly random vectors on the positive orthant of the unit sphere?
probability probability-distributions
probability probability-distributions
asked Nov 26 at 17:51
user114743
1026
asked Nov 26 at 17:51
user114743
1026
edited Nov 26 at 18:00
asked Nov 26 at 17:51
user114743
1026
asked Nov 26 at 17:51
user114743
1026
asked Nov 26 at 17:51
user114743
1026
1026
Would you get a uniform distribution of angles with $n=2$?
– Henry
Nov 26 at 18:48
add a comment |
Would you get a uniform distribution of angles with $n=2$?
– Henry
Nov 26 at 18:48
Would you get a uniform distribution of angles with $n=2$?
– Henry
Nov 26 at 18:48
Would you get a uniform distribution of angles with $n=2$?
– Henry
Nov 26 at 18:48
add a comment |
1 Answer
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The distribution isn't spherically symmetric. The probability of $X-Y$ to end in all-positive or all-negative orthant is zero, since there is no such two points $x,y$ on an orthant that $x_i>y_i$ for all $i$.
If you have meant the interior of unit sphere in the positive orthant, then it's less obvious, but the answer is the same. you have non-zero probability density near $(1,-1,0,0ldots0)$ and zero probabaility density near $(1,1,0,ldots0)$
answered Nov 26 at 19:00
Vasily Mitch
1,32837
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
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oldest
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active
oldest
votes
The distribution isn't spherically symmetric. The probability of $X-Y$ to end in all-positive or all-negative orthant is zero, since there is no such two points $x,y$ on an orthant that $x_i>y_i$ for all $i$.
If you have meant the interior of unit sphere in the positive orthant, then it's less obvious, but the answer is the same. you have non-zero probability density near $(1,-1,0,0ldots0)$ and zero probabaility density near $(1,1,0,ldots0)$
answered Nov 26 at 19:00
Vasily Mitch
1,32837
add a comment |
The distribution isn't spherically symmetric. The probability of $X-Y$ to end in all-positive or all-negative orthant is zero, since there is no such two points $x,y$ on an orthant that $x_i>y_i$ for all $i$.
If you have meant the interior of unit sphere in the positive orthant, then it's less obvious, but the answer is the same. you have non-zero probability density near $(1,-1,0,0ldots0)$ and zero probabaility density near $(1,1,0,ldots0)$
answered Nov 26 at 19:00
Vasily Mitch
1,32837
add a comment |
The distribution isn't spherically symmetric. The probability of $X-Y$ to end in all-positive or all-negative orthant is zero, since there is no such two points $x,y$ on an orthant that $x_i>y_i$ for all $i$.
If you have meant the interior of unit sphere in the positive orthant, then it's less obvious, but the answer is the same. you have non-zero probability density near $(1,-1,0,0ldots0)$ and zero probabaility density near $(1,1,0,ldots0)$
answered Nov 26 at 19:00
Vasily Mitch
1,32837
The distribution isn't spherically symmetric. The probability of $X-Y$ to end in all-positive or all-negative orthant is zero, since there is no such two points $x,y$ on an orthant that $x_i>y_i$ for all $i$.
If you have meant the interior of unit sphere in the positive orthant, then it's less obvious, but the answer is the same. you have non-zero probability density near $(1,-1,0,0ldots0)$ and zero probabaility density near $(1,1,0,ldots0)$
answered Nov 26 at 19:00
Vasily Mitch
1,32837
answered Nov 26 at 19:00
Vasily Mitch
1,32837
answered Nov 26 at 19:00
Vasily Mitch
1,32837
answered Nov 26 at 19:00
Vasily Mitch
1,32837
1,32837
add a comment |
add a comment |
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Would you get a uniform distribution of angles with $n=2$?
– Henry
Nov 26 at 18:48