Question

I would like to get a different solution to a problem I have solved “symbolically” and through a little simulation. Now, I would like to know how could I have got the integration directly using Mathematica.

Please consider a target represented by a disk with r = 1, centered at (0,0).I want to do a simulation of my probability to hit this target throwing darts.

Now, I have no biases throwing them, that is on average I shall hit the center mu = 0 but my variance is 1.

Considering the coordinate of my dart as it hit the target (or the wall 🙂 ) I have the following distributions, 2 Gaussians:

XDistribution : 1/Sqrt[2 \[Pi]\[Sigma]^2] E^(-x^2/(2 \[Sigma]^2))

YDistribution : 1/Sqrt[2 \[Pi]\[Sigma]^2] E^(-y^2/(2 \[Sigma]^2))

With those 2 distribution centered at 0 with equal variance =1 , my joint distribution becomes a bivariate Gaussian such as :

1/(2 \[Pi]\[Sigma]^2) E^(-((x^2 + y^2)/(2 \[Sigma]^2)))

So I need to know my probability to hit the target or the probability of x^2 + y^2 to be inferior to 1.

An integration after a transformation in a polar coordinate system gave me first my solution : .39 . Simulation confirmed it using :

Total@ParallelTable[
   If[
      EuclideanDistance[{
                         RandomVariate[NormalDistribution[0, Sqrt[1]]], 
                         RandomVariate[NormalDistribution[0, Sqrt[1]]]
                        }, {0, 0}] < 1, 1,0], {1000000}]/1000000

I feel there were more elegant way to solve this problem using the integration capacities of Mathematica, but never got to map ether work.

Answer 1

There are actually several ways you can do this.

The simplest would be to use NIntegrate as:

JointDistrbution = 1/(2 \[Pi] \[Sigma]^2) E^(-((x^2 + y^2)/(2 \[Sigma]^2)));
NIntegrate[JointDistrbution /. \[Sigma] -> 1, {y, -1, 1}, 
    {x, -Sqrt[1 - y^2], Sqrt[1 - y^2]}] // Timing

Out[1]= {0.009625, 0.393469}

This is another way to do it empirically (similar to your example above), but a lot slower than using NIntegrate:

(EuclideanDistance[#, {0, 0}] <= 1 & /@ # // Boole // Total)/
     Length@# &@RandomVariate[NormalDistribution[0, 1], {10^6, 2}] // 
  N // Timing

Out[2]= {5.03216, 0.39281}