Geometric Probabilities and Continuous Spaces

Classical probability assumes a finite number of equally probable outcomes. For continuous spaces (interval, circle, ℝⁿ), geometric probabilities are needed, defined via the Lebesgue measure.

Geometric Probabilities

Principle: In a continuous space, “equally possible” means a uniform distribution according to the measure (length, area, volume). P(A) = measure(A)/measure(Ω).

Bertrand’s Paradox (1889): “Random chord” of a circle — is it longer than the side of an inscribed equilateral triangle? Three reasonable definitions of “random chord” yield P = 1/2, P = 1/3, P = 1/4. Demonstrates that “equally possible” is ambiguous without precise definition.

Buffon’s Needle: A needle of length l is thrown onto a floor with parallel lines spaced d apart (l < d). P(crossing a line) = 2l/(πd). Allows estimation of π via experiment!

Continuous Probability Spaces

Borel σ-algebra: B(ℝ) — the minimal σ-algebra containing all open intervals of ℝ. Contains: closed sets, countable unions/intersections, everything “reasonable”.

Lebesgue Measure on ℝ: λ((a,b)) = b−a — length of the interval. Extends to B(ℝ). P((a,b)) = (b−a)/(β−α) for uniform U[α,β].

Uniform Distribution U[a,b]: P(X ∈ [c,d]) = (d−c)/(b−a). Density: f(x) = 1/(b−a) for x ∈ [a,b].

Exercise: (a) A point is chosen uniformly in the square [0,1]². P(x² + y² < 1) — P(inside the unit circle) = π/4. Use for Monte Carlo estimation of π. (b) Waiting time for a bus is uniform on [0, 30] min. Do you wait the average amount of time? P(wait < 10 min)?

Bertrand’s Paradox and the Measure Problem

Bertrand’s paradox demonstrates a fundamental issue: the concept of “random” has no unique meaning without precise specification of the probability space. Three methods for constructing a random chord give different answers:

Method 1 (random endpoints): Choose uniformly two points on the circle. Probability of a long chord = 1/3.

Method 2 (random midpoint): Choose uniformly a point inside the circle — this is the midpoint of the chord. Probability = 1/4.

Method 3 (random projection): Fix a direction, choose uniformly the distance from the center to the chord. Probability = 1/2.

All three methods are “reasonable,” but yield different answers (1/3, 1/4, 1/2). This shows that the phrase “random chord” is incomplete — it must be clarified with respect to which measure. The problem is resolved by modern axiomatic theory: the probability space (Ω, F, P) must be explicitly specified.

Lebesgue Measure and “Zero” Probabilities

In continuous spaces, the probability of a single point is zero: P(X = c) = 0 for any continuous random variable. This seems paradoxical — if the probability of each outcome is zero, how does anything actually happen? The answer is that countable additivity holds for countable sets of events, but continuous spaces are not correlated. Probability on an interval is “density,” not “point mass.”

From this it follows: zero probability ≠ impossibility. The event “random point exactly equals 0.7” has probability zero, but is not impossible — in each particular experiment, some value is realized anyway. This nuance is often missed when interpreting probabilities.

Monte Carlo Method and Geometric Probabilities

The Monte Carlo method uses geometric probabilities for numerical integration and modeling. Idea: if P(point falls in region A) = |A|/|Ω|, then |A| ≈ |Ω| × (fraction of points in simulation falling into A).

Estimating π: throw n points uniformly in the square [−1,1]². The number k falling inside the unit circle x²+y²≤1 gives π ≈ 4k/n. For n = 10⁶, error ≈ 0.001. The Monte Carlo method is used in physics (nuclear reactions), finance (option pricing), medicine (dosimetry), and machine learning (integral estimation in Bayesian models).

Convergence Rate: Error decreases as O(1/√n) — independently of dimension! Unlike deterministic methods (cubature), where error increases with dimension. Thus, Monte Carlo is preferable in high dimensions (d > 5).

Continuous Probability Spaces and σ-algebras

Constructing a continuous probability space requires care. For ℝ, we take the Borel σ-algebra B(ℝ) — minimal, containing all open sets. It is closed under countable unions and intersections, which ensures measurability of all “reasonable” subsets of ℝ.

But there exist non-measurable sets (Vitali’s construction): partition [0,1] into equivalence classes x ~ y ⟺ x − y ∈ ℚ. Choose one representative from each class — this is a non-measurable set V. P(V) cannot be defined correctly without breaking additivity. Existence of such sets depends on the axiom of choice — without it they may not exist.

Lebesgue Measure and Absolute Continuity

Lebesgue measure λ on ℝ — extension of interval length: λ([a,b]) = b−a. Countable sets have measure 0. The Cantor set (construction: remove middle thirds) has measure 0 and is uncountable — a paradoxical object.

A distribution X has a density f(x) if and only if P_X is absolutely continuous with respect to λ: for any A with λ(A) = 0, we have P(X∈A) = 0. Radon-Nikodym theorem: then there exists f = dP_X/dλ (Radon-Nikodym derivative). For mixed distributions (combination of discrete and continuous), the Lebesgue decomposition is used: P_X = P_ac + P_d.

Convergence of Random Variables: In probability: X_n →_P X ⟺ P(|X_n−X|>ε)→0. Almost surely: X_n →_ac X ⟺ P(X_n→X)=1. In L²: E[(X_n−X)²]→0. In distribution (weak): F_n(x)→F(x) at points of continuity of F. Hierarchy: almost surely ⟹ in probability ⟹ in distribution; L² ⟹ in probability.

Numerical Example: Calculating Probability via Density

Task: X ~ Uniform(0, 4). Find P(1 < X ≤ 3).

Step 1: Density of the uniform distribution: f(x) = 1/(4−0) = 1/4 on [0,4], and 0 outside this interval.

Step 2: P(1 < X ≤ 3) = ∫₁³ (1/4) dx = (1/4)·(3−1) = (1/4)·2 = 1/2.

Step 3: Via CDF: F(x) = x/4 on [0,4]. P = F(3)−F(1) = 3/4 − 1/4 = 2/4 = 0.5. ✓

Step 4: Connection with measure: Lebesgue measure λ([1,3]) = 2. Density f is the Radon-Nikodym derivative dP_X/dλ = 1/4. Sum: P(X∈[0,4]) = (1/4)·λ([0,4]) = (1/4)·4 = 1. ✓