Given a probability density, how can I sample from the induced distribution?Let f be an integratable function such that ∫ f ( x ) d x = 1. If we want to take random samples from this, using whatever programming language one pleases, we should compute F ( t ) = ∫ − ∞ t f ( x ) d x, invert this function and feed it numbers drawn from a uniform distribution on [0,1].However I now want to sample coordinates ( u , v ) ∈ R 2 ∖ { 0 } such that the probability of (u,v) lying in a set E is given by ∫ E ( u 2 + v 2 ) − 3 2 d u d vI don't see how I can now try to find F ( s , t ) = ∫ − ∞ s ∫ − ∞ t ( u 2 + v 2 ) − 3 2 d u d vas I get into troubles close to the zero. What other way is there to obtain a sample following this distribution?A note for the context: If we consider all lines in the Euclidean plane with Cartesian coordinates, not passing through the origin, they can be represented via u x + v x = 1. If we impose the condition that the probability densitiy should be invariant under Euclidean transformations, then we arrive at the above distribution.

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Given a probability density, how can I sample from the induced distribution?Let f be an integratable function such that   ∫  f  (  x  )  d  x  =  1. If we want to take random samples from this, using whatever programming language one pleases, we should compute   F  (  t  )  =      ∫          −      ∞        t    f  (  x  )  d  x, invert this function and feed it numbers drawn from a uniform distribution on [0,1].However I now want to sample coordinates   (  u  ,  v  )  ∈            R        2    ∖  {  0  } such that the probability of (u,v) lying in a set E is given by      ∫    E    (      u    2    +      v    2        )          −              3        2              d  u  d  vI don&#039;t see how I can now try to find  F  (  s  ,  t  )  =      ∫          −      ∞        s        ∫          −      ∞        t    (      u    2    +      v    2        )          −              3        2              d  u  d  vas I get into troubles close to the zero. What other way is there to obtain a sample following this distribution?A note for the context: If we consider all lines in the Euclidean plane with Cartesian coordinates, not passing through the origin, they can be represented via   u  x  +  v  x  =  1. If we impose the condition that the probability densitiy should be invariant under Euclidean transformations, then we arrive at the above distribution.

Michelle Chavez · Accepted Answer

Step 1Your probability density on u,v is not a valid probability density.A probability density should have       ∫          −      ∞              ∞            ∫          −      ∞              ∞        f  (  u  ,  v  )      d    u      d    v  =  1Step 2If you convert to polar coordinates you will see that      ∫          −      ∞              ∞            ∫          −      ∞              ∞        (      u    2    +      v    2        )          −      3              /            2            d    u      d    v  =      ∫    0          ∞        2  π      r          −      2            d    r  =  ∞Removing the point {0} from             R        2   won&#039;t help either - you have to remove a neighborhood of zero in order to make your density normalisable.

pigskiniv · Accepted Answer

Step 1The book you cite describes the Haar measure on the space of lines in the plane, with a view to making it the intensity measure for a Poisson process. In the pages before the bit you cite it mentions the fact that an invariant measure might have infinite mass, and that if you want a probability distribution come from it you have to condition by restricting yourself to sets of finite measure. It then goes on to describe the polar coordinate representation that makes this process practical, described in Angela&#039;s answer.Step 2An intuition behind the non-finiteness of the integral is: the infinitely overwhelming majority of lines in the plane are very far from the origin. Such a line is described by an equation   u  x  +  v  y  =  1 for tiny u and v.

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