The expected number of random values from the unit- interval [0,1] needed to obtain a sum exceeding 1

Graph of simulation results

Theory behind the simulation Let's try to see why Euler's Number e (= ) turns up as the average number of random values from the unit-interval [0,1] that need to be added in order to get a sum that exceeds 1 As background, let's recall that we showed in our previous lesson that e is equal to this sum: e = ∑ k=0,1,2,... 1/k!

Summing the reciprocal-factorials

Volume of an n-box Suppose you want to measure the “volume” in an n- dimensional rectangular box having sides s 1, s 2,..., s n For n>3 we can't visualize this object, but we can represent it as a product of n closed intervals B n = [0,s 1 ] x [0,s 2 ] x... x [0,s n ]

Special cases: n = 1, 2, 3 For n = 1 the measure is s 1 (i.e., “length”) For n = 2 the measure is s 1 s 2 (i.e., “area”) For n = 3 the measure is s 1 s 2 s 3 (i.e., “volume”) Do these facts suggest a pattern?

Integral calculus when n=2 The measure of [0,s 1 ] x [0,s 2 ] can be computed from the measure of [0,s 1 ] using calculus: m( B 2 ) = ∫ 0≤x≤s2 m( B 1 ) dx = s 1 s 2

Integral calculus when n=3 Similarly, the measure of B 3 can be computed from the measure of B 2 using calculus m( B 3 ) = ∫ 0≤x≤s3 m( B 2 ) dx = s 1 s 2 s 3

Mathematical Induction Continuing in this way we can obtain m( B n+1 ) from m( B n ) using integral calculus m( B n+1 ) = ∫ 0≤x≤sn+1 m( B n ) dx = s 1 s 2 s 3... s n s n+1

An n-dimensional simplex Let's consider the subset S n of the box B n consisting of those n-tuples (x 1, x 2,..., x n ) satisfying the inequality (1/s 1 )x 1 + (1/s 2 )x (1/s n )x n ≤ 1

When n = 2 it's a triangle s1s1 s2s2 (1/s 1 )x 1 + (1/s 2 )x 2 ≤ 1 0 “area”: m( S 2 ) = s 1 s 2 / 2

When n = 3 it's a tetrahedron s1s1 s2s2 s3s3 0 (1/s 1 )x 1 + (1/s 2 )x 2 + (1/s 3 )x 3 ≤ 1 “volume”: m( S 3 ) = s 1 s 2 s 3 / 6

“volume” as a sum of “areas” s1s1 s2s2 s3s3 0 (1/s 1 )x 1 + (1/s 2 )x 2 + (1/s 3 )x 3 ≤ 1 “volume”: m( S 3 ) = ∫ 0≤t≤s3 ½ (t/s)(t/s) dt = ½ [ t 3 / 3 s 1 s 2 ] 0≤t≤s3 = s 1 s 2 s 3 / 3! dt t/s 1 t/s 2 t

When n > 3 it's a hyper-simplex We can't truly visualize objects in n-dimensional space when n is greater than 3 But we can our imagination and memory to get a sense of some simple 4-dimensional objects through a “time- lapse” thought experiment, and we can use analogies from lower dimensions

“volume” by analogy m( S n ) = ∫ 0≤t≤sn (s 1 s 2...s n-1 /(n-1)!) dt = (s 1 s 2... s n-1 / (n-1)! ) s n /n = s 1 s 2... s n / n!

measures in unit-hypercubes If p(k) be the probability that the sum of k randomly selected values from [0,1] have a sum which does NOT exceed 1, then p(k) = 1/k! This fact is based on the formula for the “volume” of a k-dimensional rectangular simplex whose apex-edges all have 1-unit lengths (k = 1, 2, 3,... )

Probability distribution The probability f(n) that the sum of n randomly chosen values from [0,1] does exceed 1, but that the sum of the initial n-1 of these values does NOT exceed 1, will be given by f(n) = ( 1 – p(n) ) - ( 1 – p(n-1) ) Thus f(n) is the probability that the experiment requires selecting exactly n values from [0,1]

Expected number of selections μ The mean μ for this probability distribution f will be given by the (infinite) sum μ = Σ k=2,3,4,... k f(k) We will be able to recognize this sum once we have simplified our formula for f(k) f(k) = (1 – p(k)) – (1 – p(k-1))

Simplification for (k > 1) f(k) = ( 1 – p(k) ) - (1 - p(k-1)) = p(k-1) – p(k) = 1/(k-1)! - 1/k! = (k - 1) / k! = 1 / k(k-2)! Therefore μ = Σ k=2,3,4,... k f(k) = Σ k=2,3,4,... 1/(k-2)! = 1/0! + 1/1! + 1/2! +1/3! +... = e

Simulation agrees with theory μ =