Try playing with expressions using the clip operation: <x> is 0 for x<0, otherwise x. The distribution of the sum of two uniform samples from [0,1] is a trianglar wedge of height 1, width 2. /\ / \ _____/ \_____ which has the formula <x> - 2<x-1> + <x-2>. If you imagine a diagonal line sweeping across a square from the LL corner to the UR corner, representing this same distribution, the first <x> term is the situation while the line is growing uniformly in length; the -2<x-1> term is the deduction when the sweep line crosses the two corners, and subtracts the two squares adjacent to the starting square; and the + <x-2> term adds back the doubly deducted zone after the sweep line exits the original square. Rephrased: We've constructucted the square of interest by adding and subtracting four offset, but overlapping, copies of the first quadrant. The intersection of a diagonal line with the first quadrant is just <x>, and the <x-1> and <x-2> terms come from intersecting with the offset copies. The same trick works for intersecting a diagonal plane with a cube: The area of intersection is <x>^2 - 3<x-1>^2 + 3<x-2>^2 - <x-3>^2, scaled appropriately. The formula can be integrated to get volumes. It also works for doing rectangles & boxes, but you get expressions like <x> - <x-a> - <x-b> + <x-a-b> for an AxB rectangle. There's also a discrete version, which can tackle puzzles like "what's the chance of rolling 5 dice and getting a total of 14?", using the single die generating function x(1-x^6)/(1-x) raised to the 5th power. Rich ---- Quoting Erich Friedman <efriedma@stetson.edu>:
greetings.
does anyone know the n-dimensional volume V(k,n) that satisfies:
0 <= x_i <= 1 (for 1 <= i <= n) and \sum_{i=1}^n x_i <= k (where k is some integer 0 <= k <= n)?
this is just the portion of the unit hypercube that is underneath the hyperplane x_1+...+x_n <= k.
or, if you're a probabilist like me, V(k,n) is the chance that the sum of k iid uniform [0,1] random variables is larger than the sum of (n-k) other iid uniform [0,1] random variables.
surely these are important enough to have been determined before, though i could not find these in the OEIS.
easy multiple integrals and symmetry show:
V(0,n) = 0 V(n,n) = 1 V(1,n) = 1/n! V(n-1,n) = 1 - 1/n! V(n/2,n) = 1/2 V(n-k,n) = 1 - V(k,n)
is there a general formula for V(k,n)? i couldn't even find a good way to calculate V(2,5)....
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