Solve this probability problem symbolically

Question

Consider a unit square, Pick two points P and Q uniformly at random inside the square, What is the probability that |PQ|>1?

I tried solve this problem

Integrate[Boole[(x1-x2)^2+(y1-y2)^2>1],{x1,0,1},{y1,0,1},{x2,0,1},{y2,0,1}]

above code with NIntegrate given 0.025074 , but I want a symbolically result as except is $\frac{19}{6}-\pi$ .

Answer 1

This is not quick (includes J.M. comment):

pdf = UniformDistribution[2];
td = TransformedDistribution[(x - y)^2, {x, y} \[Distributed] pdf];
zd = TransformedDistribution[
   a + b, {a \[Distributed] td, b \[Distributed] td}];

then

ans = 1 - FullSimplify[CDF[zd,1]]

yields the desired result.

Answer 2

This is an interesting problem, because the difficulty is not the concept, but rather how to compute it (efficiently). Given points $(X_i,Y_i)$ distributed Uniformly on the unit square, we are interested in $$P\big[\sqrt{(X_2-X_1)^2 + (Y_2-Y_1)^2} \; > \; 1\big] $$

Let $X = X_2 - X_1$ denote the difference of two standard Uniform random variables, which is well-known to be a symmetric Triangular distribution on (-1,1). Similarly, let $Y = Y_2-Y_1$. By independence, the joint pdf of $(X,Y)$, say $f(x,y)$ is then:

       f = (1-Abs[x]) (1-Abs[y]);       domain[f] = {{x,-1,1}, {y,-1,1}};

We seek:

_{(source: tri.org.au)}

All done. This takes just 2 seconds to evaluate, starting from a fresh kernel, where I am using the Prob function from the mathStatica package for Mathematica (how I roll, being one of the authors), but one can equally use in-built Mathematica functions to the same effect:

dist = ProbabilityDistribution[(1 - Abs[x]) (1 - Abs[y]), {x, -1, 1}, {y, -1, 1}]; 
Probability[Sqrt[x^2 + y^2] > 1, Distributed[{x, y}, dist]]

Answer 3

What's with all the heavy lifting and machinations?

d = ProductDistribution[{TriangularDistribution[{-1, 1}], 2}];
Probability[a^2 + b^2 > 1, {a, b} \[Distributed] d]

$\frac{19}{6}-\pi$

Finishes in a few seconds on a netbook...

Answer 4

This is a "please don't reinvent the wheel" type of answer. From Michael Trott, here

p[l_] := Piecewise[{
   {2 l (l^2 - 4 l + π), 0 <= l <= 1}, 
   {2 l (4 Sqrt[l^2 - 1] - (l^2 + 2 - π) - 4 ArcTan[Sqrt[l^2 - 1]]), 1 < l <= Sqrt[2]}}]

With[{n = 1}, Integrate[ p[l], {l, 1, Sqrt@2}]] // Simplify

(* 19/6 - π *)

Answer 5

The problem can also be solved in a generalized form. What is the probability that the distance between the two points is greater than t ?

The PDF for the distance of two random variables equally distributed between 0 and 1 is

f[u_] := 1 - u Sign[u]

Hence the probability in question is

p[t_] = Integrate[f[u] f[v] Boole[u^2 + v^2 > t^2], {v, -1, 1}, {u, -1, 1}, 
   Assumptions -> t > 0] // Simplify

(* Because Latex has unfavorable line breaks I write down the Mathematica expressions for the two regions of t *)

Simplify[p[t], 0 < t < 1]

(* Out[230]= 1 - π t^2 + (8 t^3)/3 - t^4/2 *)

Simplify[p[t], 1 < t < Sqrt[2]]

(* Out[231]= 1/12 (8 + 24 t^2 - 3 π t^2 + 6 t^4 - 
   16 Sqrt[-1 + t^2] - 32 t^2 Sqrt[-1 + t^2] - 18 t^2 ArcCsc[t] + 
   30 t^2 ArcTan[Sqrt[-1 + t^2]]) *)

(* Here's Latex *)

$$ \begin{array}{cc} -\frac{t^4}{2}+\frac{8 t^3}{3}-\pi t^2+1 & 0<t\leq 1 \\ \frac{1}{12} \left(6 t^4-32 \sqrt{t^2-1} t^2-3 \pi t^2+24 t^2-16\sqrt{t^2-1}+30 t^2 \tan^{-1}\left(\sqrt{t^2-1}\right)-18 t^2 \csc^{-1}(t)+8\right)& 1<t<\sqrt{2} \\ \end{array} $$

We recover

p[1]

(* Out[222]= 19/6 - π *)

The graph of p[t] is

Plot[p[t], {t, 0, Sqrt[2]}, AxesLabel -> {"t", "p(t)"}, 
 PlotLabel -> 
  "Probability that the distance between two points\nrandomly chosen in the \
unit square is greater than t"]