This chapter closes with three examples of the use of probability in simple geometrical calcu- lations, namely Bertrand’s paradox, Buffon’s needle, and stick breaking.
The paradox of Joseph Louis Franc¸ois Bertrand. A chord of the unit circle is picked at random. What is the probability that an equilateral triangle with the chord as base fits inside the circle?
There are several ways of ‘choosing a chord at random’, and the ‘paradox’ lies in the mult- iplicity of answers arising from different interpretations. Here are three such interpretations, and a fourth is found in Exercise 5.70.
D A
C
Fig. 5.4 The triangle can be drawn inside the circle if and only if D ≥12.
I. Let D be the perpendicular distance between the centre of the circle and the chord, as illustrated in Figure 5.4. The triangle can be drawn inside the circle if and only if D ≥ 12.
Assumethat D is uniformly distributed on the interval (0, 1). Then the answer is P(D ≥ 1
2) = 1 2.
5.7 Geometrical probability 77
II. Assume the acute angle A between the chord and the tangent is uniform on the interval (0,12π ). Since D =12 when A =13π, the answer is
P(A ≤ 1 3π ) = 1 3π 1 2π = 2 3.
III. Assume the centre C of the chord is chosen uniformly in the interior of the circle. Then
D ≤ d if and only if C lies within a circle of radius d, so that P(D ≤ d) = πd2/π = d2. The answer is P(D ≥ 1 2) = 1 − P(D ≤ 1 2) = 3 4.
The needle of Georges Louis Leclerc, Comte de Buffon. This is more interesting. A plane is ruled by straight lines which are unit distance apart, as in Figure 5.5. What is the probability that a unit needle, dropped at random, intersects a line?5
We may position the plane so that the x -axis is along a line. Let (X, Y ) be the coordinates of the midpoint of the needle, and let 2 be its inclination to horizontal, as in the figure. It is reasonable to assume that
(a) Z := Y − ⌊Y ⌋ is uniformly distributed on [0, 1], (b) 2 is uniform on [0, π], and
(c) Z and 2 are independent. Thus, fZ ,2(z, θ ) = 1 π for z ∈ [0, 1], θ ∈ [0, π]. 2 Z
Fig. 5.5 The needle and the ruled plane in Buffon’s problem.
An intersection occurs if and only if (z, θ ) ∈ B, where
B =(z, θ ) : either z ≤ 12sin θ or 1 − z ≤ 12sin θ . Therefore, the probability of an intersection satisfies
5The original problem of Buffon was slightly different, namely the following. A coin is dropped on a tiled floor.
P(intersection) = Z Z B 1 π d z dθ = 1 π Z π θ =0 dθ Z 1 2sin θ 0 d z + Z 1 1−12sin θ d z ! = π1 Z π 0 dθ sin θ = 2 π.
This motivates a Monte Carlo experiment to estimate the value of π . Drop the needle n times, and let Inbe the number of throws that result in intersections. The natural estimate of π
isbπn:= (2n)/In. It may be shown thatE(bπn) → π as n → ∞, and furthermore the variance
ofbπnis of order n−1. Thus, there is a sense in which the accuracy of this experiment increases
as n → ∞. There are, however, better ways to estimate π than by Monte Carlo methods. Stick breaking. Here is an everyday problem of broken sticks. A stick of unit length is brok- en at the two places X , Y , each chosen uniformly at random along the stick. What is the probability that the three pieces can be used to make a triangle? We assume that X and Y are independent.
The lengths of the three substicks are
U = min{X, Y }, V = |Y − X|, W = 1 − U − V,
and we are asked for the probability that no substick is longer than the sum of the other two lengths. Since U + V + W = 1, this is equivalent to requiring that U, V, W ≤ 12. The region
of the (X, Y )-plane satisfying U, V , 1 − U − V ≤ 12 is shaded in Figure 5.6, and it has area 1
4. Therefore, the answer is 1 4.
1
1 X
Y
5.8 Problems 79
Exercise 5.70 What is the answer to Bertrand’s question if the chord is P Q, where P and Q are chosen uniformly and independently at random on the circumference of the circle?
Exercise 5.71 Suppose Buffon junior uses a needle of length ℓ (< 1). Show that the probability of an intersection is 2ℓ/π .
Exercise 5.72 One of Hugo’s longer noodles, of length ℓ, falls at random from his bowl onto Buffon’s ruled plane. Show that the mean number of intersections of the noodle with lines is 2ℓ/π .
5.8
Problems
1. The bilateral (or double) exponential distribution has density function
f (x) = 12ce−c|x| for x ∈ R,
where c (> 0) is a parameter of the distribution. Show that the mean and variance of this
distribution are 0 and 2c−2, respectively.
2. Let X be a random variable with the Poisson distribution, parameter λ. Show that, for w = 1, 2, 3, . . . ,
P(X ≥ w) = P(Y ≤ λ),
where Y is a random variable having the gamma distribution with parameters w and 1. 3. The random variable X has density function proportional to g(x), where g is a function satis-
fying
g(x) =
(
|x|−n if |x| ≥ 1,
0 otherwise,
and n (≥ 2) is an integer. Find and sketch the density function of X, and determine the values of n for which both the mean and variance of X exist.
4. If X has the normal distribution with mean 0 and variance 1, find the density function of
Y = |X|, and find the mean and variance of Y .
5. Let X be a random variable whose distribution function F is a continuous function. Show that the random variable Y , defined by Y = F(X), is uniformly distributed on the interval (0, 1). * 6. Let F be a distribution function, and let X be a random variable which is uniformly distributed
on the interval (0, 1). Let F−1be the inverse function of F, defined by
F−1(y) = inf{x : F(x) ≥ y}.
Show that the random variable Y = F−1(X )has distribution function F. This observation may
be used in practice to generate pseudorandom numbers drawn from any given distribution. 7. If X is a continuous random variable taking non-negative values only, show that
E(X) =
Z ∞
0 [1 − FX
(x)] dx,
* 8. Use the result of Problem 5.8.7 to show that E(g(X)) =
Z ∞
−∞
g(x) fX(x) d x
whenever X and g(X ) are continuous random variables and g : R → [0, ∞).
9. The random variable X′is said to be obtained from the random variable X by truncation at the
point a if
X′(ω) =
(
X (ω) if X (ω) ≤ a,
a if X (ω) > a.
Express the distribution function of X′in terms of the distribution function of X .
10. Let X have the exponential distribution with parameter 1. Find the density function of Y = (X − 2)/(X + 1).
11. William Tell is a very bad shot. In practice, he places a small green apple on top of a straight wall which stretches to infinity in both directions. He then takes up position at a distance of one perch from the apple, so that his line of sight to the target is perpendicular to the wall. He now selects an angle uniformly at random from his entire field of view and shoots his arrow in this direction. Assuming that his arrow hits the wall somewhere, what is the distribution function of the horizontal distance (measured in perches) between the apple and the point which the arrow strikes? There is no wind.
* 12. Buffon–Laplace needle. Let a, b > 0. The Cartesian plane is ruled with two sets of parallel lines of the form x = ma and y = nb for integers m and n. A needle of length ℓ (< min{a, b}) is dropped at random. Show that the probability it intersects some line is ℓ(2a +2b−ℓ)/(πab). * 13. A unit stick is broken at n random places, each uniform on [0, 1], and different breaks are chosen independently. Show that the resulting n + 1 substicks can form a closed polygon with
probability 1 − (n + 1)/2n.
14. The random variable X is uniformly distributed on the interval [0, 1]. Find the distribution and probability density function of Y , where
Y = 3X
1 − X.
Part B
6
Multivariate distributions and
independence
Summary. A random vector is studied via its joint distribution func- tion, and this leads to a discussion of the independence of random variables. The joint, marginal, and conditional density functions of continuous variables are defined, and their theory explored. Sums of independent variables are studied via the convolution formula, and transformations of random vectors via the Jacobian method. The basic properties of the bivariate normal distribution are described.
6.1
Random vectors and independence
Given two random variables X and Y , acting on a probability space (, F ,P), it is often useful to think of them acting together as a random vector (X, Y ) taking values inR2. If X and Y are discrete, we may study this random vector by using the joint mass function of X and
Y, but this method is not always available. In the general case of arbitrary random variables
X, Y , we study instead their joint distribution function, defined as follows.
Definition 6.1 The joint distribution function of the pair X , Y of random variables is
the mapping FX,Y : R2→ [0, 1] given by
FX,Y(x , y) = P(X ≤ x, Y ≤ y). (6.2)
Joint distribution functions have certain elementary properties which are exactly analogous to those of ordinary distribution functions. For example, it is easy to see that
lim
x ,y→−∞FX,Y(x , y) = 0, (6.3)
lim
x ,y→∞FX,Y(x , y) = 1, (6.4)
just as in (5.6) and (5.7). Similarly, FX,Y is non-increasing in each variable in that
FX,Y(x1,y1) ≤ FX,Y(x2,y2) if x1≤ x2and y1≤ y2. (6.5)
The joint distribution function FX,Y contains a great deal more information than the two or-
particular, the distribution functions of X and Y may be found from their joint distribution function in a routine way. It is intuitively attractive to write
FX(x ) = P(X ≤ x)
= P(X ≤ x, Y ≤ ∞) = FX,Y(x , ∞)
and similarly,
FY(y) = FX,Y(∞, y),
but the mathematically correct way of expressing this is
FX(x ) = lim
y→∞FX,Y(x , y), FY(y) = limx →∞FX,Y(x , y). (6.6)
These distribution functions are called the marginal distribution functions of the joint distri- bution function FX,Y.
The idea of ‘independence’ of random variables X and Y follows naturally from this dis- cussion.
Definition 6.7 We call X and Y independent if, for all x , y ∈ R, the events {X ≤ x}
and {Y ≤ y} are independent.
That is to say, X and Y are independent if and only if
P(X ≤ x, Y ≤ y) = P(X ≤ x)P(Y ≤ y) for x , y ∈ R,
which is to say that their joint distribution function factorizes as the product of the two marginal distribution functions:
FX,Y(x , y) = FX(x )FY(y) for x , y ∈ R. (6.8)
It is a straightforward exercise to show that this is a genuine extension of the notion of ind- ependent discrete random variables. Random variables which are not independent are called
dependent.
We study families of random variables in very much the same way. Briefly, if X1,X2, . . . ,
Xn are random variables on (, F ,P), theirjoint distribution functionis the function FX :
Rn
→ [0, 1] given by
FX(x) = P X1≤ x1, X2≤ x2, . . . , Xn ≤ xn
(6.9) for x = (x1,x2, . . . ,xn) ∈ Rn. The variables X1,X2, . . . ,Xnare called independent if
P X1≤ x1, . . . ,Xn≤ xn= P(X1≤ x1) · · · P(Xn≤ xn) for x ∈ Rn,
or equivalently if
FX(x) = FX1(x1) · · · FXn(xn) for x ∈ R