Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 40 (1999), No. 2, 291-301.

Systems of Stochastically Independent and Normally Distributed Random Points

in the Euclidean Space E_{3}

Elena Bosetto

Dipartimento di Matematica, Università di Torino Via C. Alberto 10, I-10123 Torino, Italy

Abstract. We consider sets of three or four stochastically independent and normally distributed random points in E_{3}. We look at these points as the vertices of a triangle T_{3} or of a tetrahedron T_{4} and we study the random variables A, area of T_{3}, and V, volume of T_{4}. We also determine the probability that T_{3} is acute-angled. MSC 1991: 60D05, 52A22 Keywords: Geometric probability, stochastic geometry, random sets, random convex sets and integral geometry.

Random sets appear very early in probability theory, in the context of geometrical probability. The random segments (in Buffon’s terms “needles”) in Buffon’s experiment are one of the first examples of how the idea of randomness can be applied to elements that are geometric objects: a geometric element, of fixed shape and dimensions, can have a “random position” in the plane or in the space. The notion of random sets appears also in the famous book of Kolmogorov [1], where the mean area of a bounded random set is calculated by means of the Fubini theorem. Unlike the previously quoted Buffon’s problem, here the idea of randomness is associated to the shape of the studied object. In Kolmogorov’s own words: “We shall give an application of this theorem (i.e. the Fubini theorem) to one example in geometric probability. Let G be a measurable region of the plane whose shape depends on chance; in other words, let us assign to every elementary event ξ of a field of probability a definite measurable region G. We shall denote by J the area of the region G, and by p (x, y) the probability that the point (x, y) belongs to the region G. Then E (J) = ∫∫ p (x, y) dxdy.”

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 1999 Heldermann Verlag

292 E. Bosetto: Systems of stochastically independent and …

In this paper we shall be concerned with this latter notion of randomness. More precisely, we consider systems of three and four points, that are supposed to be stochastically independent and normally distributed in the Euclidean space E_{3}, and we look at these points as the vertices of a triangle T_{3} and of a tetrahedron T_{4}. Obviously the shapes of T_{3} and T_{4} depend on the random positions of their vertices, and the area of T_{3} and the volume of T_{4} are the most natural random variables that we can study in order to get some information about these “regions depending on chance”.

As important application fields we just mention geometrical statistics and image analysis, materials sciences, fuzzy data analysis and particle statistics.

1. On a random triangle with normally and independently distributed vertices

Let {P_{1}, P_{2}, P_{3}} be independent and identically distributed random points in the Euclidean space E_{3}, with common distribution

f (x, y, z) = \frac{1}{(2π)^{\frac{3}{2}}} e−\frac{x^{2}+y^{2}+z^{2}}{2}  (1)

where (x, y, z) ∈ ℝ^{3} are the Cartesian coordinates in E_{3}. We denote by T_{3} the triangle of vertices P_{i} ≡ (x_{i}, y_{i}, z_{i}), i = 1, 2, 3, and by A the area of T_{3}.

If we define the set Δ_{T_{3}} as

Δ_{T_{3}} = {P_{3} ∈ E_{3} : A = area (T_{3}) < α}

then the distribution function of A can be written:

F (α) = P {A < α}
= \frac{1}{(2π)^{\frac{9}{2}}} ∫_{ℝ^{3}×ℝ^{3}} e−^{\frac{∑_{i=1}^{2}(x_{i}^{2}+y_{i}^{2}+z_{i}^{2})}{2}} dP_{1}dP_{2} ∫_{Δ_{T_{3}}} e−\frac{x_{3}^{2}+y_{3}^{2}+z_{3}^{2}}{2} dP_{3}

where dP_{i} = dx_{i} ∧ dy_{i} ∧ dz_{i}, i = 1, 2, 3. From now on we choose α > 0, as for α ≤ 0 we obviously have F (α) ≡ 0.

In order to compute F (α) we consider the following change of variables:

\begin{cases}x_{i} = p cos φ sin ϑ + q_{i} cos φ cos ϑ + r_{i} sin φ, & ϑ ∈ [0, π] , φ ∈ [0, 2π] \\ y_{i} = p sin φ sin ϑ + q_{i} sin φ cos ϑ − r_{i} cos φ, & q_{i}, r_{i} ∈ ℝ, i = 1, 2, 3 \\ z_{i} = p cos ϑ − q_{i} sin ϑ, & p ∈ ℝ^{+} \end{cases}  (2)

with

\begin{aligned}|J| & = |\frac{D (x_{1}, y_{1}, z_{1}, x_{2}, y_{2}, z_{2}, x_{3}, y_{3}, z_{3})}{D (p, q_{1}, q_{2}, q_{3}, r_{1}, r_{2}, r_{3}, ϑ, φ)}| \\ & = | ∑_{i≠j=1}^{2} (−1)^{i} (q_{i} − q_{3}) (r_{j} − r_{3})| sin ϑ \end{aligned}  (3)

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so that

F (α) = \frac{1}{(2π)^{\frac{9}{2}}} ∫_{0}^{π} sin ϑdϑ ∫_{0}^{2π}dφ ∫_{0}^{+∞}e−\frac{3}{2}p^{2}dp

· ∫_{Δ′_{T_{3}}} | ∑_{i≠j=1}^{2} (−1)^{i} (q_{i} − q_{3}) (r_{j} − r_{3})| e−^{\frac{∑_{i=1}^{3}(q_{i}^{2}+r_{i}^{2})}{2}} dq⃗dr⃗

= \frac{1}{\sqrt{3} (2π)^{3}} ∫_{Δ′_{T_{3}}} | ∑_{i≠j=1}^{2} (−1)^{i} (q_{i} − q_{3}) (r_{j} − r_{3})| e−^{\frac{∑_{i=1}^{3}(q_{i}^{2}+r_{i}^{2})}{2}} dq⃗dr⃗

where dq⃗ = dq_{1}dq_{2}dq_{3}, dr⃗ = dr_{1}dr_{2}dr_{3} and

Δ′_{T_{3}} = {q_{i}, r_{i} ∈ ℝ, i = 1, 2, 3 : |(q_{2} − q_{3}) (r_{1} − r_{3}) − (q_{1} − q_{3}) (r_{2} − r_{3})| < 2α}. Next we let

q_{1} = \frac{u_{2} + v_{2}}{2} + q_{3} q_{2} = \frac{u_{1} + v_{1}}{2} + q_{3} q_{3} = q_{3}
r_{1} = \frac{u_{1} − v_{1}}{2} + r_{3} r_{2} = \frac{u_{2} − v_{2}}{2} + r_{3} r_{3} = r_{3}

which yields

F (α) = \frac{1}{2^{7}\sqrt{3}π^{3}} ∫_{Δ′′_{T_{3}}} |u_{1}^{2} − u_{2}^{2} − v_{1}^{2} + v_{2}^{2}| e−\frac{1}{4} ∑_{i=1}^{2} (u_{i}^{2} + v_{i}^{2})

· (∫_{−∞}^{+∞}e−^{\frac{3q_{3}^{2}+q_{3} ∑_{i=1}^{2}(u_{i}+v_{i})}{2}} dq_{3} ∫_{−∞}^{+∞}e−^{\frac{3r_{3}^{2}+r_{3} ∑_{i=1}^{2}(u_{i}−v_{i})}{2}} dr_{3}) du_{1}du_{2}dv_{1}dv_{2}

with

Δ′′_{T_{3}} = {(u_{1}, u_{2}, v_{1}, v_{2}) ∈ ℝ^{4} : |u_{1}^{2} − u_{2}^{2} − v_{1}^{2} + v_{2}^{2}| < 8α}.

Since

−\frac{3}{2}q_{3}^{2} − \frac{q_{3}}{2} ∑_{i=1}^{2} (u_{i} + v_{i}) = −\frac{3}{2} [q_{3} + \frac{1}{6} ∑_{i=1}^{2} (u_{i} + v_{i})]^{2} + \frac{1}{24} [∑_{i=1}^{2} (u_{i} + v_{i})]^{2}

integration with respect to dq_{3} gives

∫_{−∞}^{+∞}e−^{\frac{3q_{3}^{2}+q_{3} ∑_{i=1}^{2}(u_{i}+v_{i})}{2}} dq_{3} = \sqrt{\frac{2π}{3}} e^{\frac{(u_{1}+u_{2}+v_{1}+v_{2})^{2}}{24}} .

294 E. Bosetto: Systems of stochastically independent and …

Likewise, integrating dr_{3} we find

∫_{−∞}^{+∞}e−^{\frac{3r_{3}^{2}+r_{3} ∑_{i=1}^{2}(u_{i}−v_{i})}{2}} dr_{3} = \sqrt{\frac{2π}{3}} e^{\frac{(u_{1}+u_{2}−v_{1}−v_{2})^{2}}{24}}

and therefore

F (α) = \frac{1}{3\sqrt{3} (8π)^{2}} · ∫_{Δ′′_{T_{3}}} |u_{1}^{2} − u_{2}^{2} − v_{1}^{2} + v_{2}^{2}| e−^{\frac{∑_{i=1}^{2}(u_{i}^{2}+v_{i}^{2})−u_{1}u_{2}−v_{1}v_{2}}{6}} du_{1}du_{2}dv_{1}dv_{2}.

Finally we let

\{ \begin{matrix}u_{1} = ρ_{1} cos ϑ_{1} + ρ_{2} cos ϑ_{2} \\ u_{2} = ρ_{1} cos ϑ_{1} − ρ_{2} cos ϑ_{2} \\ v_{1} = ρ_{1} sin ϑ_{1} + ρ_{2} sin ϑ_{2} \\ v_{2} = ρ_{1} sin ϑ_{1} − ρ_{2} sin ϑ_{2} \end{matrix} ϑ_{i} ∈ [0, 2π] , ρ_{i} > 0, i = 1, 2

and we find

F (α) = \frac{1}{3\sqrt{3} (2π)^{2}} ∫_{Δ′′′_{T_{3}}} ρ_{1}^{2}ρ_{2}^{2} |cos (ϑ_{1} + ϑ_{2})| e−\frac{ρ_{1}^{2}+3ρ_{2}^{2}}{6} dϑ_{1}dϑ_{2}dρ_{1}dρ_{2}

where now

Δ′′′_{T_{3}} = {ϑ_{i} ∈ [0, 2π] , ρ_{i} > 0, i = 1, 2 : ρ_{1}ρ_{2} |cos (ϑ_{1} + ϑ_{2})| < 2α} .

The previous expression of F (α) is equivalent to

F (α) = \frac{2}{3\sqrt{3}π} ∫_{\{ρ_{1}ρ_{2}<2α\}} ρ_{1}^{2}ρ_{2}^{2}e−\frac{ρ_{1}^{2}+3ρ_{2}^{2}}{6} dρ_{1}dρ_{2}

+ ∫_{\{ρ_{1}ρ_{2}>2α\}} \frac{ρ_{1}^{2}ρ_{2}^{2}e−\frac{ρ_{1}^{2}+3ρ_{2}^{2}}{6}}{3\sqrt{3}π^{2}} ∫_{\{ξ<ϑ_{1}+ϑ_{2}< \frac{π}{2}\}} cos (ϑ_{1} + ϑ_{2}) dϑ_{1}dϑ_{2}dρ_{1}dρ_{2}

− ∫_{\{ρ_{1}ρ_{2}>2α\}} \frac{ρ_{1}^{2}ρ_{2}^{2}e−\frac{ρ_{1}^{2}+3ρ_{2}^{2}}{6}}{3\sqrt{3}π^{2}} ∫_{\{\frac{π}{2} <ϑ_{1}+ϑ_{2}<π−ξ\}} cos (ϑ_{1} + ϑ_{2}) dϑ_{1}dϑ_{2}dρ_{1}dρ_{2}

where ξ = arccos (\frac{2α}{ρ_{1}ρ_{2}} ). Integration with respect to dϑ_{1} and dϑ_{2} yields

F (α) = \frac{1}{2} + \frac{1}{3\sqrt{3}π} [I_{1} (α) − I_{2} (α)]  (4)

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having denoted by I_{1} (α) and I_{2} (α) the integrals

I_{1} (α) = ∫_{\{ρ_{1}ρ_{2}<2α\}} ρ_{1}^{2}ρ_{2}^{2}e−\frac{ρ_{1}^{2}+3ρ_{2}^{2}}{6} dρ_{1}dρ_{2}
I_{2} (α) = ∫_{\{ρ_{1}ρ_{2}>2α\}} ρ_{1}ρ_{2}\sqrt{ρ_{1}^{2}ρ_{2}^{2} − 4α^{2}} · e−\frac{ρ_{1}^{2}+3ρ_{2}^{2}}{6} dρ_{1}dρ_{2}.

Let us consider first I_{1} (α). By letting

\{ \begin{matrix}ρ_{1} = \sqrt{6}r cos φ \\ ρ_{2} = \sqrt{2}r sin φ \end{matrix} r > 0, φ ∈ [0, \frac{π}{2} ]

we obtain

I_{1} (α) = 24\sqrt{3} ∫0^{\begin{matrix}^{√} & \overline{2α} \\ ^{4} & \sqrt{3} \end{matrix}} dr ∫0^{\frac{π}{2}} r^{5}e^{−r^{2}} sin^{2} φ cos^{2} φdφ

+24\sqrt{3} ∫ +∞_{\begin{matrix}^{√} & \overline{2α} \\ ^{4} & \sqrt{3} \end{matrix}} dr ∫_{0}^{\frac{1}{2} arcsin \frac{2α}{\sqrt{3}r^{2}}} r^{5}e^{−r^{2}} sin^{2} φ cos^{2} φdφ

+24\sqrt{3} ∫ +∞_{\begin{matrix}^{√} & \overline{2α} \\ ^{4} & \sqrt{3} \end{matrix}} dr ∫\frac{π}{2} − \frac{1}{2} arcsin \frac{2α}{\sqrt{3}r^{2}}^{\frac{π}{2}} r^{5}e^{−r^{2}} sin^{2} φ cos^{2} φdφ

= \frac{3\sqrt{3}}{2}π − (\sqrt{3}α^{2} + 3α + \frac{3\sqrt{3}}{2}) πe−\frac{2\sqrt{3}}{3}α

+3\sqrt{3} ∫ +∞_{\begin{matrix}^{√} & \overline{2α} \\ ^{4} & \sqrt{3} \end{matrix}} r^{5} arcsin (\frac{2α}{\sqrt{3}r^{2}} ) e^{−r^{2}}dr

−3α ∫ +∞_{\begin{matrix}^{√} & \overline{2α} \\ ^{4} & \sqrt{3} \end{matrix}} r^{3}\sqrt{1 − \frac{4α^{2}}{3r^{4}}} e^{−r^{2}}dr

and the final substitution \frac{\sqrt{3}}{2α}r^{2} = u gives

I_{1} (α) = \frac{3\sqrt{3}}{2}π − ∫_{1}^{+∞} \frac{2α^{3}u (u^{2} − 1) + 2\sqrt{3}α^{2}u^{2} + 6αu + 3\sqrt{3}}{u\sqrt{u^{2} − 1}}e−\frac{2α}{\sqrt{3}}udu.

Taking into account the following integral representation for the modified Bessel function of the second kind K_{n} (see, e.g., [4]):

K_{n} (x) = \frac{\sqrt{π}x^{n}}{2^{n}Γ (n + \frac{1}{2})} ∫_{1}^{+∞} (u^{2} − 1)^{n− \frac{1}{2}} e^{−xu}du  (5)

we can write

I_{1} (α) = \frac{3\sqrt{3}}{2}π − 3\sqrt{3}α^{2}K_{1} (\frac{2α}{\sqrt{3}}) − 6αK_{0} (\frac{2α}{\sqrt{3}}) − 3\sqrt{3} ∫_{1}^{+∞} \frac{e−\frac{2α}{\sqrt{3}}u}{u\sqrt{u^{2} − 1}}du.

As far as I_{2} (α) is concerned we consider the change of variables:

\{ \begin{matrix}ρ_{1} = u \\ ρ_{2} = \frac{1}{u} \sqrt{v^{2} + 4α^{2}} \end{matrix}

and we get

I_{2} (α) = \frac{\sqrt{2π}}{2} ∫_{0}^{+∞} u^{2}e− (\frac{u^{2}}{6} + \frac{2α^{2}}{u^{2}} )du.

As ∫_{0}^{+∞} exp (−a^{2}x^{2} − b^{2}/x^{2}) dx = \frac{\sqrt{π}}{2a} exp (−2ab) for any a, b > 0, we find

I_{2} (α) = ∫_{0}^{α} \frac{d}{ds}I_{2} (s) ds + I_{2} (0)

= \frac{\sqrt{2π}}{2} ∫_{0}^{+∞}u^{2}e−\frac{u^{2}}{6} du − 2\sqrt{2π} ∫_{0}^{α} ds ∫_{0}^{+∞}se− (\frac{u^{2}}{6} + \frac{2s^{2}}{u^{2}} )du

= \frac{3\sqrt{3}π}{2} − 2π\sqrt{3} ∫_{0}^{α} se−\frac{2s}{\sqrt{3}} ds = \frac{3π}{2} (2α + \sqrt{3}) e−\frac{2}{\sqrt{3}}α.

Thus formula (4) becomes

F (α) = 1 − \frac{α^{2}}{π} K_{1} (\frac{2α}{\sqrt{3}}) − \frac{2α}{\sqrt{3}π} K_{0} (\frac{2α}{\sqrt{3}}) − \frac{2α+3}{2\sqrt{3}}e−\frac{2α}{\sqrt{3}} − \frac{1}{π} ∫_{1}^{+∞}\frac{e−\frac{2}{\sqrt{3}}α}{u\sqrt{u^{2} − 1}}du  (6)

and we can state the following:

Theorem 1. Let {P_{1}, P_{2}, P_{3}} be independent and identically distributed random points in the Euclidean space E_{3}, with normal distribution (1). The random variable A, area of the triangle T_{3} of vertices P_{i}, i = 1, 2, 3, has density function

f (α) = \frac{2α^{2}}{\sqrt{3}π} K_{0} (\frac{2α}{\sqrt{3}}) + \frac{α}{3π} K_{1} (\frac{2α}{\sqrt{3}}) + \frac{2}{3}αe−\frac{2α}{\sqrt{3}} ,  α > 0.  (7)

The mean value and the variance of A are:

E [A] = \frac{\sqrt{3}}{2} + \frac{7\sqrt{3}}{4π} Var [A] = \frac{69}{16} − \frac{21}{4π} − \frac{147}{16π^{2}} .

Proof. In order to get the density function of A it suffices to take the derivative of its distribution function F (α) given in (6) and to apply the following properties of the modified Bessel functions of second kind K_{0} and K_{1}:

\frac{d}{dα}K_{0} (α) = −K_{1} (α) \frac{d}{dα}K_{1} (α) = −K_{0} (α) − \frac{1}{α}K_{1} (α)

Figure 1: density function of the area A of T_{3}

as well as the integral representation (5). The values of E [A] and Var [A] are consequences of (7). □

A plot of the density function f (α) is shown in Figure 1.

Remark. If {P_{1}, P_{2}, P_{3}} are independent and identically distributed random points in the

Euclidean plane E_{2}, with normal distribution f (x, y) = \frac{1}{2π}e−\frac{x^{2}+y^{2}}{2} , then a result of Stoka (see [3], p. 540-541) gives

f (α) = 2\sqrt{\frac{2}{π}} α ∫_{0}^{+∞} \frac{1}{q^{2}} e− (\frac{q^{2}}{2} + \frac{α^{2}}{q^{2}} )dq,  E [A] = \frac{1}{\sqrt{2}},  Var [A] = \frac{3}{2}.

The shape of the triangle T_{3} “depends on chance”, so it is also interesting to determine whether “T_{3} is more likely to be acute-angled or obtuse-angled”. The answer is that it is slightly more probable to find an obtuse-angled triangle. More precisely, with the same hypothesis and notation of theorem 1, we have

Theorem 2. The probability that T_{3} is an acute-angled triangle is p = \frac{3\sqrt{3}}{4π^{2}} .

Proof. The triangle T_{3} is acute-angled if and only if

\begin{matrix}|P_{1}P_{2}|^{2} + |P_{1}P_{3}|^{2} > |P_{2}P_{3}|^{2} ,  |P_{1}P_{2}|^{2} + |P_{2}P_{3}|^{2} > |P_{1}P_{3}|^{2} , \\ |P_{1}P_{3}|^{2} + |P_{2}P_{3}|^{2} > |P_{1}P_{2}|^{2} , \end{matrix}  (8)

where |P_{i}P_{j}| denotes the euclidean distance between P_{i} and P_{j} (see, e.g., [2] p. 99).

If we use the change of variables (2) to express the cartesian coordinates (x_{i}, y_{i}, z_{i}) of P_{i}, i = 1, 2, 3, then condition (8) becomes

(q_{i} − q_{j}) (q_{i} − q_{k}) + (r_{i} − r_{j}) (r_{i} − r_{k}) > 0 i ≠ j ≠ k, i, j, k = 1, 2, 3  (9)

and the probability p is

p = P {|P_{i}P_{j}|^{2} + |P_{i}P_{k}|^{2} > |P_{j}P_{k}|^{2} i ≠ j ≠ k, i, j, k = 1, 2, 3}
= \frac{1}{(2π)^{\frac{9}{2}}} ∫_{0}^{π} sin ϑdϑ ∫_{0}^{2π}dφ ∫_{0}^{+∞}e−\frac{3p^{2}}{2} dp

· ∫_{D} | ∑_{i≠j=1}^{2} (−1)^{i} (q_{i} − q_{3}) (r_{j} − r_{3})| e−^{\frac{∑_{i=1}^{3}(q_{i}^{2}+r_{i}^{2})}{2}} dq⃗dr⃗

= \frac{\sqrt{3}}{24π^{3}} ∫_{D} | ∑_{i≠j=1}^{2} (−1)^{i} (q_{i} − q_{3}) (r_{j} − r_{3})| e−^{\frac{∑_{i=1}^{3}(q_{i}^{2}+r_{i}^{2})}{2}} dq⃗dr⃗

where D is the set of real numbers (q_{1}, q_{2}, q_{3}, r_{1}, r_{2}, r_{3}) that satisfies (9).

In order to compute the above integral we let

\{ \begin{matrix}q_{1} = ρ_{1} cos ϑ_{1} + ρ_{2} cos ϑ_{2} + q_{3} \\ q_{2} = ρ_{1} cos ϑ_{1} − ρ_{2} cos ϑ_{2} + q_{3} \\ q_{3} = q_{3} \\ r_{1} = ρ_{1} sin ϑ_{1} + ρ_{2} sin ϑ_{2} + r_{3} \\ r_{1} = ρ_{1} sin ϑ_{1} − ρ_{2} sin ϑ_{2} + r_{3} \\ r_{3} = r_{3} \end{matrix}  \begin{matrix}ϑ_{1}, ϑ_{2} ∈ [0, 2π] \\ ρ_{1}, ρ_{2} > 0 \end{matrix}

and we obtain

p = \frac{2\sqrt{3}}{9π^{2}} ∫ ; \{ \frac{ρ_{2}}{ρ_{1}} <1\} ρ_{1}^{2}ρ_{2}^{2}e−\frac{ρ_{1}^{2}}{3} − ρ_{2}^{2}dρ_{1}dρ_{2}  ∫ ; \{|cos(ϑ_{1}−ϑ_{2})|< \frac{ρ_{2}}{ρ_{1}} \} |sin (ϑ_{1} − ϑ_{2})| dϑ_{1}dϑ_{2}

= \frac{16\sqrt{3}}{9π} ∫ ; \{ \frac{ρ_{2}}{ρ_{1}} <1\} ρ_{1}ρ_{2}^{3}e−\frac{ρ_{1}^{2}}{3} − ρ_{2}^{2}dρ_{1}dρ_{2} = \frac{3\sqrt{3}}{4π}

and the proof is complete. □

2. On a random tetrahedron with normally independently distributed vertices

Let us add now a fourth point P_{4} to the set {P_{1}, P_{2}, P_{3}} which we considered in the previous section and let us suppose that {P_{i}}_{i=1,... ,4} are still independent and identically distributed.

We denote by T_{4} the tetrahedron of vertices P_{i}, i = 1, …, 4, and by V the volume of T_{4}. Once again our aim is to determine the density function, the mean value and the variance of V. In order to be able to exploit the results of Section 1, we define the set

Δ_{T_{4}} = {P_{4} ∈ E_{3} : V = volume (T_{4}) < v}

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thus for any v > 0

G (v) = P {V < v}
= \frac{1}{(2π)^{6}} ∫_{ℝ^{3}×ℝ^{3}×ℝ^{3}} e−\frac{1}{2} ∑_{i=1}^{3} (x_{i}^{2} + y_{i}^{2} + z_{i}^{2})dP_{1}dP_{2}dP_{3} ∫_{Δ_{T_{4}}} e−\frac{x_{4}^{2}+y_{4}^{2}+z_{4}^{2}}{2} dP_{4}.

Following (2), we introduce the change of variables: x_{i} = p cos φ sin ϑ + q_{i} cos φ cos ϑ + r_{i} sin φ y_{i} = p sin φ sin ϑ + q_{i} sin φ cos ϑ − r_{i} cos φ

z_{i} = p cos ϑ − q_{i} sin ϑ  ϑ ∈ [0, π] , φ ∈ [0, 2π]
x_{4} = (p + s) cos φ sin ϑ + q_{4} cos φ cos ϑ + r_{4} sin φ  p ∈ ℝ^{+}, s ∈ ℝ
y_{4} = (p + s) sin φ sin ϑ + q_{4} sin φ cos ϑ − r_{4} cos φ  q_{i}, r_{i} ∈ ℝ, i = 1, 2, 3, 4
z_{4} = (p + s) cos ϑ − q_{4} sin ϑ

with |J| again given by (3). We have

\begin{matrix}G (v) = \frac{1}{(2π)^{6}} & ∫_{0}^{π} sin ϑdϑ ∫_{0}^{2π} dφ ∫_{−∞}^{+∞} e−\frac{q_{4}^{2}}{2} dq_{4} ∫_{−∞}^{+∞} e−\frac{r_{4}^{2}}{2} dr_{4} \\ · & ∫_{Δ′_{T_{4}}} | ∑_{i≠j=1}^{2} (−1)^{i} (q_{i} − q_{3}) (r_{j} − r_{3})| e−\frac{4p^{2}+2ps+s^{2}}{2} − ∑_{i=1}^{3} \frac{q_{i}^{2}+r_{i}^{2}}{2} dpdsdq⃗dr⃗ \\ = \frac{1}{8π^{4}} & ∫_{Δ′_{T_{4}}} | ∑_{i≠j=1}^{2} (−1)^{i} (q_{i} − q_{3}) (r_{j} − r_{3})| e−2p^{2} − ps − \frac{s^{2}}{2} + ∑_{i=1}^{3} \frac{q_{i}^{2}+r_{i}^{2}}{2} dpdsdq⃗dr⃗ \end{matrix}

where

Δ′_{T_{4}} = {q_{i}, r_{i}, s ∈ ℝ, i = 1, 2, 3, p ∈ ℝ^{+} : | ∑_{i≠j=1}^{2} (−1)^{i} (q_{i} − q_{3}) (r_{j} − r_{3}) s| < 6v}.

Writing

e− (2p^{2} + ps + \frac{s^{2}}{2} ) = e−^{\frac{(2p+ \frac{s}{2})^{2}}{2}} e−\frac{3s^{2}}{8}

and integrating dp we get

G (v) = \frac{1}{(2π)^{\frac{7}{2}}} ∫_{Δ′′_{T_{4}}} | ∑_{i≠j=1}^{2} (−1)^{i} (q_{i} − q_{3}) (r_{j} − r_{3})| e−\frac{3s^{2}}{8} − ∑_{i=1}^{3} \frac{(q_{i}^{2}+r_{i}^{2})}{2} dsdq⃗dr⃗

having denoted by Δ′′_{T_{4}} the set

Δ′′_{T_{4}} = {q_{i}, r_{i} ∈ ℝ, i = 1, 2, 3, s > 0 : | ∑_{i≠j=1}^{2} (−1)^{i} (q_{i} − q_{3}) (r_{j} − r_{3})| s < 6v} .

If we compare the sets Δ′_{T_{3}} and Δ′′_{T_{4}} it is not difficult to realize that

G (v) = \sqrt{\frac{3}{2π}} ∫_{0}^{+∞} F (\frac{3v}{s} ) e−\frac{3s^{2}}{8} ds

where F (·), that is the distribution function of the area of the triangle T_{3} of vertices {P_{1}, P_{2}, P_{3}}, is given by (6).

Using the change of variable s = \frac{6v}{\sqrt{3}u} we find

G (v) = 1 − \frac{9v}{(2π)^{\frac{3}{2}}} ∫_{0}^{+∞} K_{1} (u) e−\frac{9v^{2}}{2u^{2}} du − \frac{6v}{π\sqrt{2π}} ∫_{0}^{+∞} \frac{1}{u}K_{0} (u) e−\frac{9v^{2}}{2u^{2}} du

− \frac{3v}{\sqrt{2π}} ∫_{0}^{+∞} \frac{1 + u}{u^{2}}e−u − \frac{9v^{2}}{2u^{2}} du − \frac{6v}{π\sqrt{2π}} ∫_{0}^{+∞} du ∫_{1}^{+∞} \frac{e−ut − \frac{9v^{2}}{2u^{2}}}{u^{2}t\sqrt{t^{2} − 1}} dt (10)

and we can now state

Theorem 3. Let {P_{1}, P_{2}, P_{3}, P_{4}} be independent and identically distributed random points in the Euclidean space E_{3}, with normal distribution (1). The random variable V, volume of the tetrahedron T_{4} of vertices P_{i}, i = 1, 2, 3, 4, has density function

g (v) = \frac{3}{\sqrt{2π}} ∫_{0}^{+∞} (\frac{3u}{2π}K_{0} (u) + \frac{1}{2π} K_{1} (u) + e^{−u}) e−\frac{9v^{2}}{2u^{2}} du v > 0

where K_{0} (·) and K_{1} (·) are the modified Bessel functions of second kind of order n = 0 and n = 1. The mean value and the variance of V are

E [V] = \frac{1}{3\sqrt{2π}} (2 + \frac{7}{π} ) Var [V] = \frac{3}{4} − \frac{2}{9π} − \frac{14}{9π^{2}} − \frac{49}{18π^{3}} .

Proof. The expression of the density function g (v) = \frac{d}{dv} G (v) is easily obtained from (10). For the computation of E [V] and Var [V] it is useful to remind the integral representation (5) of the functions K_{0} (u) and K_{1} (u). □

References

[1] Kolmogorov, A. N.: Grundbegriffe der Wahrscheinlichkeitsrechnung. Springer-Verlag, Berlin 1933.

[2] Stoka, M. I.: Probabilità e Geometria. Herbita Editrice, Palermo 1982.

[3] Stoka, M. I.: Alcuni problemi di probabilità geometriche non uniformi. Ann. Fac. Econ. e Comm. Univ. Bari, nuova serie, vol. XXVIII parte II, 1989.

[4] Watson, G. N.: A Treatise on the Theory of Bessel Functions. Cambridge University Press, 2nd ed., 1966.

Received July 2, 1998