PDFs of chord lengths and distances between random points

This webpage is typed up from undated notes (probably 1980s).  Although I haven’t yet found a reference to the main result in the literature, it is unlikely to be original.  MB, January 2014

Fig. 1 Notation for distances defined by two random points

Let K be a bounded plane convex region (fig. 1).  Let points U1 and U2 have independent uniform distributions inside K.  Let U1U2 produced meet the boundary of K at V1 and V2, where (to avoid ambiguity) V1V2 has azimuth in the range [0,π)

Let U1U2=u with PDF ƒ(u); let V1V2=v with PDF g(v); let V1U1=x and V1U2=y

1.  Given v and x, to find the PDF of y

i.e. to find the probability that y<V1U1<y+δy, where δy is small. 

The probability that V1V2=v and V1U1=x exactly is 0; so assume v<V1V2<v+δv and x<V1U1<x+δx, where δv and δx are small compared with δy

Fig. 2 Possible positions of second point

Given U1, the permissible set of azimuths of U1U2 will be a (possibly empty) union of small intervals like ε in fig. 2.  (The diagram is drawn for x<y; the change for x>y is straightforward.) The shaded area shows the posssible positions of U2.  Hence

prob(y1<V1U2<y+δy)= |yx|εδy = |yxy .
½εx2+½ε(vx)2 x2vx+½v2

Thus for 0yv the PDF of y for a given v and x is

|yx| .

2.  Given v, to find the PDFs of x and y

Clearly x and y have identical PDFs for a given v; call this PDF φv.  From section 1 we get

φv(y)= v |yxv(x)dx .
x=0 x2vx+½v2

Put φv(x)=ψ(x)(x2vx+½v2) and rewrite the last equation as

(y2vy+½v2)ψ(y)= y (yx)ψ(x)dx v (yx)ψ(x)dx.
x=0 x=y

Differentiate twice with respect to y:

(y2vy+½v2)ψ′(y)+(2yv)ψ(y)= y ψ(x)dx v ψ(x)dx,
x=0 x=y


d [(y2vy+½v2)2ψ′(y)]=0,

i.e., ψ′(y)=A/(y2vy+½v2)2, where A is some constant.  But ψ(y) is symmetrical in the range 0yv; hence ψ′(v/2)=0; hence A=0; hence ψ(y) is a constant.  Since 0vφv(x)dx=1 we conclude

φv(x)= 3 (x2vx+½v2).

3.  Relation between the PDFs of U1U2 and V1V2

First we find the PDF of u for a given v.  For vuv/2 the PDF is

(∫ vu + v ) uφv(x)dx = 6 u(vu).
0 u x2vx+½v2 v3

For 0uv/2 the PDF is

(∫ u +2 vu + v ) uφv(x)dx = 6 u(vu) also.
0 u vu x2vx+½v2 v3

Hence if u (=U1U2) and v (=V1V2) have respective PDFs ƒ and g then

ƒ(u)= 6 u(vu)g(v)dv.
u v3

Differentiate three times:

ƒ′(u)= 6 u(v2u)g(v)dv,
u v3
ƒ″(u)= 12 u(v2u)g(v)dv+ 6 g(u),
u v3 u2
ƒ‴(u)= 12 g(u) 12 g(u)+ 6 g′(u).
u3 u3 u2

On renaming the variable the desired relation can be expressed as:


Integration by parts gives


which can be expressed as:

g(t)= t3 d2 ( ƒ(t) ) .
6 dt2 t

4.  Relation between the moments of U1U2 and V1V2

Let k be a non-negative integer.  If we temporarily define h(t)=ƒ(t)/t, then the last equation gives

tkg(t)dt= 1 tk+3h″(t)dt
0 6 0
= 1 [ tk+3h′(t) ] 1 (k+3)tk+2h′(t)dt
6 0 6 0
=0 k+3 tk+2h′(t)dt
6 0
= k+3 [ tk+2h(t) ] + k+3 (k+2)tk+1h(t)dt
6 0 6 0
= (k+2)(k+3) tkƒ(t)dt.
6 0

Hence the main result of this webpage:

mean of (V1V2)k= (k+2)(k+3) × mean of (U1U2)k.

This result (which has been verified for a rectangle by simulations) is useful in the theory of ley statistics.  For similar results see the discussion in Luis A. Santaló, Integral Geometry and Geometric Probability, 1976, pp. 46–49.