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Uniform distribution over the unit disk


What is the probability density function of pairwise distances of random points in a ball?Distribution of $max(X_i)midmin(X_i)$ when $X_i$ are i.i.d uniform random variablesRelations between Order Statistics of Uniform RVs and Exponential RVsuniform distribution over diskUniform distribution on unit diskConditional expectation of $Y_1$ given that $sup Y_i=z$, for $(Y_i)$ i.i.d. uniform on $[0,theta]$Positivity of pdf of sum of non-iid random variablesUniform distribution density functionWhat is the distribution of the scalar product of two uniform, iid vectors?For a random sample from the distribution $f(x)=e^-(x-theta) , x>theta$ , show that $2n[X_(1)-theta]simchi^2_2$Joint PDF of min and max of iid uniform distributions













1












$begingroup$


Suppose that $U_1$ and $U_2$ are independent, and identically and uniformly distributed over the unit disk, i.e., for $i = 1,2$, $U_i = (X_i, Y_i)$ and the joint density is
beginequation
f_(X_i,Y_i)(x,y) = frac1pi mathbb1_(x,y):x^2+y^2 leq 1(x,y).
endequation
I want to know how to compute the probability that the Euclidean distance between $U_1$ and $U_2$ is smaller than, say, $frac12$, that is, $mathbbP(sqrt(X_1 - X_2)^2 + (Y_1 - Y_2)^2 < frac12)$? I am trying to see if the polar coordinates would help. E.g., let $U_i = (R_i cos Theta_i, R_isin Theta_i)$ and then relate the distance to the $R_i$. But would $R_i$ be uniform on $[0,1]$? Anyone has an idea? Thanks very much.






probability probability-distributions uniform-distribution







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Yes, you must suppose $U_i sim U[0,1]$ and $Theta_isim U[0,2pi)$
    $endgroup$
    – sinbadh
    Jan 4 '16 at 17:57










  • $begingroup$
    You may find this recent, related question interesting about a similar problem in 3-dimensions. math.stackexchange.com/q/1599950/74357
    $endgroup$
    – CommonerG
    Jan 4 '16 at 19:32















1












$begingroup$


Suppose that $U_1$ and $U_2$ are independent, and identically and uniformly distributed over the unit disk, i.e., for $i = 1,2$, $U_i = (X_i, Y_i)$ and the joint density is
beginequation
f_(X_i,Y_i)(x,y) = frac1pi mathbb1_(x,y):x^2+y^2 leq 1(x,y).
endequation
I want to know how to compute the probability that the Euclidean distance between $U_1$ and $U_2$ is smaller than, say, $frac12$, that is, $mathbbP(sqrt(X_1 - X_2)^2 + (Y_1 - Y_2)^2 < frac12)$? I am trying to see if the polar coordinates would help. E.g., let $U_i = (R_i cos Theta_i, R_isin Theta_i)$ and then relate the distance to the $R_i$. But would $R_i$ be uniform on $[0,1]$? Anyone has an idea? Thanks very much.






probability probability-distributions uniform-distribution







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Yes, you must suppose $U_i sim U[0,1]$ and $Theta_isim U[0,2pi)$
    $endgroup$
    – sinbadh
    Jan 4 '16 at 17:57










  • $begingroup$
    You may find this recent, related question interesting about a similar problem in 3-dimensions. math.stackexchange.com/q/1599950/74357
    $endgroup$
    – CommonerG
    Jan 4 '16 at 19:32













1












1








1


1



$begingroup$


Suppose that $U_1$ and $U_2$ are independent, and identically and uniformly distributed over the unit disk, i.e., for $i = 1,2$, $U_i = (X_i, Y_i)$ and the joint density is
beginequation
f_(X_i,Y_i)(x,y) = frac1pi mathbb1_(x,y):x^2+y^2 leq 1(x,y).
endequation
I want to know how to compute the probability that the Euclidean distance between $U_1$ and $U_2$ is smaller than, say, $frac12$, that is, $mathbbP(sqrt(X_1 - X_2)^2 + (Y_1 - Y_2)^2 < frac12)$? I am trying to see if the polar coordinates would help. E.g., let $U_i = (R_i cos Theta_i, R_isin Theta_i)$ and then relate the distance to the $R_i$. But would $R_i$ be uniform on $[0,1]$? Anyone has an idea? Thanks very much.






probability probability-distributions uniform-distribution







share|cite|improve this question











$endgroup$




Suppose that $U_1$ and $U_2$ are independent, and identically and uniformly distributed over the unit disk, i.e., for $i = 1,2$, $U_i = (X_i, Y_i)$ and the joint density is
beginequation
f_(X_i,Y_i)(x,y) = frac1pi mathbb1_(x,y):x^2+y^2 leq 1(x,y).
endequation
I want to know how to compute the probability that the Euclidean distance between $U_1$ and $U_2$ is smaller than, say, $frac12$, that is, $mathbbP(sqrt(X_1 - X_2)^2 + (Y_1 - Y_2)^2 < frac12)$? I am trying to see if the polar coordinates would help. E.g., let $U_i = (R_i cos Theta_i, R_isin Theta_i)$ and then relate the distance to the $R_i$. But would $R_i$ be uniform on $[0,1]$? Anyone has an idea? Thanks very much.






probability probability-distributions uniform-distribution




probability probability-distributions uniform-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 '16 at 17:44







Richie

















asked Jan 4 '16 at 17:28









RichieRichie

394210




394210







  • 1




    $begingroup$
    Yes, you must suppose $U_i sim U[0,1]$ and $Theta_isim U[0,2pi)$
    $endgroup$
    – sinbadh
    Jan 4 '16 at 17:57










  • $begingroup$
    You may find this recent, related question interesting about a similar problem in 3-dimensions. math.stackexchange.com/q/1599950/74357
    $endgroup$
    – CommonerG
    Jan 4 '16 at 19:32












  • 1




    $begingroup$
    Yes, you must suppose $U_i sim U[0,1]$ and $Theta_isim U[0,2pi)$
    $endgroup$
    – sinbadh
    Jan 4 '16 at 17:57










  • $begingroup$
    You may find this recent, related question interesting about a similar problem in 3-dimensions. math.stackexchange.com/q/1599950/74357
    $endgroup$
    – CommonerG
    Jan 4 '16 at 19:32







1




1




$begingroup$
Yes, you must suppose $U_i sim U[0,1]$ and $Theta_isim U[0,2pi)$
$endgroup$
– sinbadh
Jan 4 '16 at 17:57




$begingroup$
Yes, you must suppose $U_i sim U[0,1]$ and $Theta_isim U[0,2pi)$
$endgroup$
– sinbadh
Jan 4 '16 at 17:57












$begingroup$
You may find this recent, related question interesting about a similar problem in 3-dimensions. math.stackexchange.com/q/1599950/74357
$endgroup$
– CommonerG
Jan 4 '16 at 19:32




$begingroup$
You may find this recent, related question interesting about a similar problem in 3-dimensions. math.stackexchange.com/q/1599950/74357
$endgroup$
– CommonerG
Jan 4 '16 at 19:32










1 Answer
1






active

oldest

votes


















0












$begingroup$

We are computing the probability that two random points in the unit disk (here random means with respect to a uniform probability distribution) lie within a distance $leqfrac12$. We may assume without loss of generality that the first point lies on the segment joining the origin with the point $(1,0)$, with a probability density function supported on $[0,1]$ and given by $f(z)=2z$. If we assume to know the position of the first point, the second point lies within a distance $leqfrac12$ iff it belongs to the original disk and to a smaller disk. If $zin[0,1]$ and $g(z)$ is the Lebesgue measure of the set:
$$ E(z) = left(x,y): x^2+y^2leq 1, (x-z)^2+y^2leqfrac14right, $$
the wanted probability is just:
$$ P=frac1piint_0^1 f(z)g(z),dz = frac2piint_0^1 zcdot g(z),dz.$$
So the problem boils down to computing $g(z)$, that has a rather ugly closed form in terms of the $arcsin$ function. Numerically,
$$ Papprox 19.728%.$$






share|cite|improve this answer











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    1 Answer
    1






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    active

    oldest

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    0












    $begingroup$

    We are computing the probability that two random points in the unit disk (here random means with respect to a uniform probability distribution) lie within a distance $leqfrac12$. We may assume without loss of generality that the first point lies on the segment joining the origin with the point $(1,0)$, with a probability density function supported on $[0,1]$ and given by $f(z)=2z$. If we assume to know the position of the first point, the second point lies within a distance $leqfrac12$ iff it belongs to the original disk and to a smaller disk. If $zin[0,1]$ and $g(z)$ is the Lebesgue measure of the set:
    $$ E(z) = left(x,y): x^2+y^2leq 1, (x-z)^2+y^2leqfrac14right, $$
    the wanted probability is just:
    $$ P=frac1piint_0^1 f(z)g(z),dz = frac2piint_0^1 zcdot g(z),dz.$$
    So the problem boils down to computing $g(z)$, that has a rather ugly closed form in terms of the $arcsin$ function. Numerically,
    $$ Papprox 19.728%.$$






    share|cite|improve this answer











    $endgroup$

















      0












      $begingroup$

      We are computing the probability that two random points in the unit disk (here random means with respect to a uniform probability distribution) lie within a distance $leqfrac12$. We may assume without loss of generality that the first point lies on the segment joining the origin with the point $(1,0)$, with a probability density function supported on $[0,1]$ and given by $f(z)=2z$. If we assume to know the position of the first point, the second point lies within a distance $leqfrac12$ iff it belongs to the original disk and to a smaller disk. If $zin[0,1]$ and $g(z)$ is the Lebesgue measure of the set:
      $$ E(z) = left(x,y): x^2+y^2leq 1, (x-z)^2+y^2leqfrac14right, $$
      the wanted probability is just:
      $$ P=frac1piint_0^1 f(z)g(z),dz = frac2piint_0^1 zcdot g(z),dz.$$
      So the problem boils down to computing $g(z)$, that has a rather ugly closed form in terms of the $arcsin$ function. Numerically,
      $$ Papprox 19.728%.$$






      share|cite|improve this answer











      $endgroup$















        0












        0








        0





        $begingroup$

        We are computing the probability that two random points in the unit disk (here random means with respect to a uniform probability distribution) lie within a distance $leqfrac12$. We may assume without loss of generality that the first point lies on the segment joining the origin with the point $(1,0)$, with a probability density function supported on $[0,1]$ and given by $f(z)=2z$. If we assume to know the position of the first point, the second point lies within a distance $leqfrac12$ iff it belongs to the original disk and to a smaller disk. If $zin[0,1]$ and $g(z)$ is the Lebesgue measure of the set:
        $$ E(z) = left(x,y): x^2+y^2leq 1, (x-z)^2+y^2leqfrac14right, $$
        the wanted probability is just:
        $$ P=frac1piint_0^1 f(z)g(z),dz = frac2piint_0^1 zcdot g(z),dz.$$
        So the problem boils down to computing $g(z)$, that has a rather ugly closed form in terms of the $arcsin$ function. Numerically,
        $$ Papprox 19.728%.$$






        share|cite|improve this answer











        $endgroup$



        We are computing the probability that two random points in the unit disk (here random means with respect to a uniform probability distribution) lie within a distance $leqfrac12$. We may assume without loss of generality that the first point lies on the segment joining the origin with the point $(1,0)$, with a probability density function supported on $[0,1]$ and given by $f(z)=2z$. If we assume to know the position of the first point, the second point lies within a distance $leqfrac12$ iff it belongs to the original disk and to a smaller disk. If $zin[0,1]$ and $g(z)$ is the Lebesgue measure of the set:
        $$ E(z) = left(x,y): x^2+y^2leq 1, (x-z)^2+y^2leqfrac14right, $$
        the wanted probability is just:
        $$ P=frac1piint_0^1 f(z)g(z),dz = frac2piint_0^1 zcdot g(z),dz.$$
        So the problem boils down to computing $g(z)$, that has a rather ugly closed form in terms of the $arcsin$ function. Numerically,
        $$ Papprox 19.728%.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 4 '16 at 18:48

























        answered Jan 4 '16 at 18:33









        Jack D'AurizioJack D'Aurizio

        292k33284669




        292k33284669



























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