Ana Sayfa
Matematikçiler
Makaleler
Matematik Seçkileri
Fraktallar
Paradokslar
=> Zeno's Paradoxes
=> Allais Paradox
=> Arnauld's Paradox
=> Banach-Tarski Paradox
=> Barber Paradox
=> Berry Paradox
=> Bottle Imp Paradox
=> Buchowski Paradox
=> Cantor's Paradox
=> Catalogue Paradox
=> Coin Paradox
=> Complex Number Paradox
=> Crocodile's Dilemma
=> Destructive Dilemma
=> Diagonal Paradox
=> Dilemma
=> Elevator Paradox
=> Epimenides Paradox
=> Eubulides Paradox
=> Grelling's Paradox
=> Hempel's Paradox
=> Liar's Paradox
=> Line Point Picking
=> Missing Dollar Paradox
=> Newcomb's Paradox
=> Parrondo's Paradox
=> Potato Paradox
=> Richard's Paradox
=> Russell's Antinomy
=> Skolem Paradox
=> Smarandache Paradox
=> Socrates' Paradox
=> Sorites Paradox
=> Strange Loop
=> Thompson Lamp Paradox
=> Unexpected Hanging Paradox
=> Fallacy
=> Plaindrome
Sayılar Teorisi
Ziyaretçi defteri
 

Line Point Picking

Consider a line segment of length 1, and pick a point x at random between [0,1]. This point x divides the line into line segments of length x and 1-x. If a set of points are thus picked at random, the resulting distribution of lengths has a uniform distribution on [0,1]. Similarly, separating the two pieces after each break, the larger piece has uniform distribution on [1/2,1] (with mean 3/4), and the smaller piece has uniform distribution on [0,1/2] (with mean 1/4).

The probability that the line segments resulting from cutting at two points picked at random on a unit line segment determine a triangle is given by 1/4.

LinePointPickingSmall

The probability and distribution functions for the ratio of small to large pieces are given by

P(x) = 2/((1+x)^2)
(1)
D(x) = (2x)/(1+x)
(2)

for x in [0,1]. The raw moments are therefore

 mu_n^'=n[psi_0(1/2(n+1))-psi_0(1/2n)]-1,
(3)

where psi_0(x) is the digamma function. The first few are therefore

mu_1^' = 2ln2-1
(4)
mu_2^' = -4ln2+3
(5)
mu_3^' = 6ln2-4
(6)
mu_4^' = -8ln2+(17)/3
(7)

(Sloane's A115388 and A115389). The central moments are therefore

 mu_n=((-1)^(n-1)(n-1)(2ln2)^n(1-n)_(n-1)(-n)_(n-1))/(Gamma(n)Gamma(n+1))
            +sum_(k=0)^(n-2)((ln2)^k(2^n-2^(k+1))(1-n)_k(-n)_k)/((n-1)k!(2-n)_k),
(8)

where (x)_n is a Pochhammer symbol. The first few are therefore

mu_2 = 2-4ln^22
(9)
mu_3 = 3-12ln2+16ln^32
(10)
mu_4 = (14)/3-24ln2+48ln^22-48ln^42.
(11)

This therefore gives mean, variance, skewness, and kurtosis excess of

mu = 2ln2-1
(12)
sigma^2 = 2-4ln^22
(13)
beta = (3-12ln2+16ln^32)/((2-4ln^22)^(3/2))
(14)
gamma = ((25)/6-6ln2)/((1-2ln^22)^2)-6.
(15)

The mean can be computed directly from

mu = int_0^1(min(x,1-x))/(max(x,1-x))dx
(16)
= 2int_0^(1/2)x/(1-x)dx
(17)
= 2ln2-1.
(18)
LinePointPickingLarge

The probability and distribution functions for the ratio of large to small pieces are given by

P(x) = 2/((1+x)^2)
(19)
D(x) = (x-1)/(x+1)
(20)

for x in [1,infty). Paradoxical though it may be, this distribution has infinite mean and other moments. The reason for this is that a theoretical bone can be cut extremely close to one end, thus giving huge ratio of largest to smallest pieces, whereas there is some limit for a real physical bone. Taking epsilon to be the smallest possible piece in which is bone cen be cut, the mean is then given by

 x^_=2ln(1/(2epsilon))+2epsilon-1.
(21)
 

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