Probability in a square pyramide?

Suppose the pyramid has a square base defined by [-1, 1] x [-1, 1], and the top vertex is the point (0,0,1). Perhaps the only analytically interesting thing I can say about this problem is how to generate a point (X, Y, H) uniformly distributed over the pyramid: The random variable 1-H has a density that grows quadratically over [0,1], and hence its cumulative distribution grows cubically. So:

-Generate F uniformly over [0,1].

-Define H = 1-(F)^(1/3).

-Generate X independently and uniformly over the interval [(H-1), (1-H)].

-Generate Y independently and uniformly over the interval [(H-1), (1-H)].

You now have (X,Y,H) uniformly distributed over the pyramid. =)

In my simulation of 10 million iterations, using (rand()*1.0/RAND_MAX) in C to generate a random variable uniform over [0, 1], I get:

E[H] on a point generated over the pyramid = 0.249978 (exact is 1/4)

E[H^2] = 0.099979 (exact is 1/10)

Fraction of time squares 1 and 2 intersect: 0.349832

Fraction of time squares 1 and 3 intersect: 0.349881

Fraction of time squares 2 and 3 intersect: 0.349986

Fraction of time [Square 1 and 2 intersect] AND [Square 1 and 3 intersect] AND [Square 2 and 3 intersect] = 0.109451

The incorrect numbers I reported earlier were because my program was using the C command "abs()" for absolute value, when I just learned I should use "fabs()" as the previous one forces the output to an integer! Perhaps someone else could also simulate to verify...?

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An exact answer for the probability of square 1 intersecting square 2 can be computed as follows: Let H_1 and H_2 be the heights of each square, being independent and identically distributed with density f_H(h) = 3(1-h)^2 for h in [0, 1].

Let (X_1, Y_1) and (X_2, Y_2) be the centers of each square, where (X_1, Y_1, X_2, Y_2) are conditionally independent given (H_1, H_2). Then:

Pr[intersect] =

int_{h_1} int_{h_2} Pr[|X_1-X_2| <= h_1 + h_2]^2 9(1-h_1)^2(1-h_2)^2 dh_1 dh_2

Define Q(h_1, h_2) = Pr[|X_1-X_2|<= h_1 + h_2]. I believe the answer is:

Q(h_1, h_2) =

{1 , if h_1 + h_2 >=1

{1 - (1 - h_1-h_2)^2/((1-h_1)(1-h_2)) , if 0 <= h_1 + h_2 < 1.

Doing the double integral gives Pr[intersect] = 7/20 = 0.35, a good match for my simulation of 0.349832.

http://www.wolframalpha.com/input/?i=int+%28int+9%...

http://www.wolframalpha.com/input/?i=int+%28int+%2...

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I believe the exact answer to the question of pairwise non-empty intersections is:

value =

int_{h_1} int_{h_2} int_{h_3} 27(1-h_1)^2(1-h_2)^2(1-h_3)^2G(h_1, h_2, h_3)^2 dh_1 dh_2 dh_3

where:

G(h_1, h_2, h_2) =

Pr{|X_1-X_2|<=h_1+h_2 AND |X_1-X_3|<=h_1+h_3 AND |X_2-X_3|<=h_2+h_3}

The value G(h_1, h_2, h_3) is basically the probability that 3 random intervals have a non-empty intersection. I think G(h_1, h_2, h_3) can be computed, but it seems very complicated.

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@Gianlino: Yes, 23/210 would be a good guess.

Another idea: Note that G is defined over the cube [0,1]^3, and:

G(h_1, h_2, h_3) = 1 if h_1 + h_2 + h_3 >= 2.

Computation of the G function in the remaining case might be possible if one imagines the cube [0,1]^3, cuts it in half to consider h_1 + h_2 + h_3 < 2, and observes some nice properties of the geometry. Note that G(h_1, h_2, h_3) should be invariant to permutations in the arguments, similar to the 2-d case for Q(h_1, h_2).