A Modern Introduction to Probability and Statistics, Understanding Why and How - Dekking, Kraaikamp, Lopuhaa, Meester (Современное введение в теорию вероятностей и статистику - Как? и Почему? )
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110 8 Computations with random variables
8.5 Solutions to the quick exercises
8.1 Clearly Z can take the values 1,...,150. The value 150 is special:
the plane is full if 150 or more people buy a ticket. Hence P(Z = 150) =
P(X ≥ 150) = 51/200. For the other values we have P(Z = i)=P(X = i)=
1/200, for i =1,...,149. Clearly, here g(x)=min{150,x}.
8.2 The probability density of Y =1/X is
f
Y
(y)=
1
y
2
1
π(1 + (
1
y
)
2
)
=
1
π(1 + y
2
)
.
We see that 1/X has the same distribution as X! (This distribution is called
the standard Cauchy distribution, it will be introduced in Chapter 11.)
8.3 First define Z =(X −4)/5, which has an N (0, 1) distribution. Then from
Table B.1
P(X ≤ 5) = P
Z ≤
5 − 4
5
=P(Z ≤ 0.20) = 1 − 0.4207 = 0.5793.
Similarly, using the symmetry of the normal distribution,
P(X ≥ 2) = P
Z ≥
2 − 4
5
=P(Z ≥−0.40) = P(Z ≤ 0.40) = 0.6554.
8.4 If g(x)=e
−x
,theng
(x)=e
−x
> 0; hence g is strictly convex. It follows
from Jensen’s inequality that
e
−E[X]
≤ E
e
−X
.
Moreover, if Var(X) > 0, then the inequality is strict.
8.5 The distribution function of the X
i
is given by F (x)=x on [0, 1]. There-
fore the distribution function F
Z
of the maximum Z is equal to F
Z
(a)=
(F (a))
n
= a
n
. Its probability density function is
f
Z
(z)=
d
dz
F
Z
(z)=nz
n−1
for 0 ≤ z ≤ 1.
8.6 The distribution function of the X
i
is given by F (x)=x on [0, 1]. There-
fore the distribution function F
V
of the minimum V is equal to F
V
(a)=
1 − (1 − a)
n
. Its probability density function is
f
V
(v)=
d
dv
F
V
(v)=n(1 − v)
n−1
for 0 ≤ v ≤ 1.
8.6 Exercises 111
8.6 Exercises
8.1 Often one is interested in the distribution of the deviation of a random
variable X from its mean µ =E[X]. Let X take the values 80, 90, 100, 110,
and 120, all with probability 0.2; then E [X]=µ = 100. Determine the dis-
tribution of Y = |X −µ|. That is, specify the values Y can take and give the
corresponding probabilities.
8.2 Suppose X has a uniform distribution over the points {1, 2, 3, 4, 5, 6}
and that g(x)=sin(
π
2
x).
a. Determine the distribution of Y = g(X)=sin(
π
2
X), that is, specify the
values Y can take and give the corresponding probabilities.
b. Let Z =cos(
π
2
X). Determine the distribution of Z.
c. Determine the distribution of W = Y
2
+ Z
2
. Warning: in this example
there is a very special dependency between Y and Z, and in general it is
much harder to determine the distribution of a random variable that is a
function of two other random variables. This is the subject of Chapter 11.
8.3 The continuous random variable U is uniformly distributed over [0, 1].
a. Determine the distribution function of V =2U + 7. What kind of distri-
bution does V have?
b. Determine the distribution function of V = rU + s for all real numbers
r>0ands. See Exercise 8.9 for what happens for negative r.
8.4 Transforming exponential distributions.
a. Let X have an Exp(
1
2
) distribution. Determine the distribution function
of
1
2
X. What kind of distribution does
1
2
X have?
b. Let X have an Exp (λ) distribution. Determine the distribution function
of λX. What kind of distribution does λX have?
8.5 Let X be a continuous random variable with probability density func-
tion
f
X
(x)=
3
4
x(2 − x)for0≤ x ≤ 2
0elsewhere.
a. Determine the distribution function F
X
.
b. Let Y =
√
X. Determine the distribution function F
Y
.
c. Determine the probability density of Y .
8.6 Let X be a continuous random variable with probability density f
X
that
takes only positive values and let Y =1/X.
112 8 Computations with random variables
a. Determine F
Y
(y)andshowthat
f
Y
(y)=
1
y
2
f
X
1
y
for y>0.
b. Let Z =1/Y .Usinga, determine the probability density f
Z
of Z,interms
of f
X
.
8.7 Let X have a Par(α) distribution. Determine the distribution function of
ln X. What kind of a distribution does ln X have?
8.8 Let X have an Exp(1) distribution, and let α and λ be positive numbers.
Determine the distribution function of the random variable
W =
X
1/α
λ
.
The distribution of the random variable W is called the Weibull distribution
with parameters α and λ.
8.9 Let X be a continuous random variable. Express the distribution function
and probability density of the random variable Y = −X in terms of those of X.
8.10 Let X be an N (3, 4) distributed random variable. Use the rule for
normal random variables under change of units and Table B.1 to determine
the probabilities P(X ≥ 3) and P(X ≤ 1).
8.11 Let X be a random variable, and let g be a twice differentiable function
with g
(x) ≤ 0 for all x. Such a function is called a concave function. Show
that for concave functions always
g(E[X]) ≥ E[g(X)] .
8.12 Let X be a random variable with the following probability mass func-
tion:
x 0 1 100 10 000
P(X = x)
1
4
1
4
1
4
1
4
a. Determine the distribution of Y =
√
X.
b. Which is larger E
√
X
or
E[X]?
Hint: use Exercise 8.11, or start by showing that the function g(x)=−
√
x
is convex.
c. Compute
E[X]andE
√
X
to check your answer (and to see that it
makes a big difference!).
8.13 Let W have a U (π,2π) distribution. What is larger: E [sin(W )] or
sin(E[W ])? Check your answer by computing these two numbers.
8.6 Exercises 113
8.14 In this exercise we take a look at Jensen’s inequality for the function
g(x)=x
3
(which is neither convex nor concave on (−∞, ∞)).
a. Can you find a (discrete) random variable X with Var(X) > 0 such that
E
X
3
=(E[X])
3
?
b. Under what kind of conditions on a random variable X will the inequality
E
X
3
> (E [X])
3
certainly hold?
8.15 Let X
1
,X
2
,...,X
n
be independent random variables, all with a U(0, 1)
distribution. Let Z =max{X
1
,...,X
n
} and V =min{X
1
,...,X
n
}.
a. Compute E[max{X
1
,X
2
}]andE[min{X
1
,X
2
}].
b. Compute E [Z]andE[V ] for general n.
c. Can you argue directly (using the symmetry of the uniform distribu-
tion (see Exercise 6.3) and not the result of the computation in b)that
1 − E[max{X
1
,...,X
n
}]=E[min{X
1
,...,X
n
}]?
8.16 In this exercise we derive a kind of Jensen inequality for the minimum.
a. Let a and b be real numbers. Show that
min{a, b} =
1
2
(a + b −|a − b|).
b. Let X and Y be independent random variables with the same distribution
and finite expectation. Deduce from a that
E[min{X, Y }]=E[X] −
1
2
E[|X − Y |] .
c. Show that
E[min{X, Y }] ≤ min{E[X] , E[Y ]}.
Remark: this is not so interesting, since min{E[X] , E[Y ]} =E[X]=E[Y ],
but we will see in the exercises of Chapter 11 that this inequality is also true
for X and Y, which do not have the same distribution.
8.17 Let X
1
,...,X
n
be n independent random variables with the same dis-
tribution function F .
a. Convince yourself that for any numbers x
1
,...,x
n
it is true that
min{x
1
,...,x
n
} = −max{−x
1
,...,−x
n
}.
b. Let Z =max{X
1
,X
2
,...,X
n
} and V =min{X
1
,X
2
,...,X
n
}.UseExer-
cise 8.9 and the observation in a to deduce the formula
114 8 Computations with random variables
F
V
(a)=1−(1 − F (a))
n
directly from the formula
F
Z
(a)=(F (a))
n
.
8.18 Let X
1
,X
2
,...,X
n
be independent random variables, all with an
Exp(λ) distribution. Let V =min{X
1
,...,X
n
}. Determine the distribution
function of V . What kind of distribution is this?
8.19 From the “north pole” N of a circle with diameter 1, a point Q on
the circle is mapped to a point t on the line by its projection from N,as
illustrated in Figure 8.2.
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Fig. 8.2. Mapping the circle to the line.
Suppose that the point Q is uniformly chosen on the circle. This is the same
as saying that the angle ϕ is uniformly chosen from the interval [−
π
2
,
π
2
](can
you see this?). Let X be this angle, so that X is uniformly distributed over
the interval [−
π
2
,
π
2
]. This means that P(X ≤ ϕ)=1/2+ϕ/π (cf. Quick
exercise 5.3). What will be the distribution of the projection of Q on the line?
Let us call this random variable Z. Then it is clear that the event {Z ≤ t} is
equal to the event {X ≤ ϕ},wheret and ϕ correspond to each other under
the projection. This means that tan(ϕ)=t, which is the same as saying that
arctan(t)=ϕ.
a. What part of the circle is mapped to the interval [1, ∞)?
b. Compute the distribution function of Z using the correspondence between
t and ϕ.
c. Compute the probability density function of Z.
The distribution of Z is called the Cauchy distribution (which will be discussed
in Chapter 11).
9
Joint distributions and independence
Random variables related to the same experiment often influence one another.
In order to capture this, we introduce the joint distribution of two or more
random variables. We also discuss the notion of independence for random
variables, which models the situation where random variables do not influence
each other. As with single random variables we treat these topics for discrete
and continuous random variables separately.
9.1 Joint distributions of discrete random variables
In a census one is usually interested in several variables, such as income, age,
and gender. In itself these variables are interesting, but when two (or more) are
studied simultaneously, detailed information is obtained on the society where
the census is performed. For instance, studying income, age, and gender jointly
might give insight to the emancipation of women.
Without mentioning it explicitly, we already encountered several examples of
joint distributions of discrete random variables. For example, in Chapter 4 we
defined two random variables S and M , the sum and the maximum of two
independent throws of a die.
Quick exercise 9.1 List the elements of the event {S =7,M =4} and
compute its probability.
In general, the joint distribution of two discrete random variables X and Y ,
defined on the same sample space Ω, is given by prescribing the probabilities
of all possible values of the pair (X, Y ).
116 9 Joint distributions and independence
Definition. The joint probability mass function p of two discrete
random variables X and Y is the function p : R
2
→ [0, 1], defined by
p(a, b)=P(X = a, Y = b)for−∞<a,b<∞.
To stress the dependence on (X, Y ), we sometimes write p
X,Y
instead of p.
If X and Y take on the values a
1
,a
2
,...,a
k
and b
1
,b
2
,...,b
, respectively,
the joint distribution of X and Y can simply be described by listing all the
possible values of p(a
i
,b
j
). For example, for the random variables S and M
from Chapter 4 we obtain Table 9.1.
Table 9.1. Joint probability mass function p(a, b)=P(S = a, M = b).
b
a 123456
2 1/36 00000
3 0 2/36 0000
4 0 1/36 2/36 0 0 0
5 0 0 2/36 2/36 0 0
6 0 0 1/36 2/36 2/36 0
7 0 0 0 2/36 2/36 2/36
8 0 0 0 1/36 2/36 2/36
9 00002/36 2/36
1000001/36 2/36
11000002/36
12000001/36
From this table we can retrieve the distribution of S and of M . For example,
because
{S =6} = {S =6,M =1}∪{S =6,M =2}∪···∪{S =6,M =6},
and because the six events
{S =6,M =1}, {S =6,M =2},...,{S =6,M =6}
are mutually exclusive, we find that
p
S
(6) = P(S =6)=P(S =6,M =1)+···+P(S =6,M =6)
= p(6, 1) + p(6, 2) + ···+ p(6, 6)
=0+0+
1
36
+
2
36
+
2
36
+0
=
5
36
.
9.1 Joint distributions of discrete random variables 117
Table 9.2. Joint distribution and marginal distributions of S and M.
b
a 12345 6 p
S
(a)
2 1/36 0000 0 1/36
3 0 2/36 0 0 0 0 2/36
4 0 1/36 2/36 0 0 0 3/36
5 0 0 2/36 2/36 0 0 4/36
6 0 0 1/36 2/36 2/36 0 5/36
7 0 0 0 2/36 2/36 2/36 6/36
8 0 0 0 1/36 2/36 2/36 5/36
9 00002/36 2/36 4/36
10 00001/36 2/36 3/36
11 000002/36 2/36
12 000001/36 1/36
p
M
(b) 1/36 3/36 5/36 7/36 9/36 11/36 1
Thus we see that the probabilities of S can be obtained by taking the sum
of the joint probabilities in the rows of Table 9.1. This yields the probability
distribution of S, i.e., all values of p
S
(a)fora =2,...,12. We speak of the
marginal distribution of S. In Table 9.2 we have added this distribution in the
right “margin” of the table. Similarly, summing over the columns of Table 9.1
yields the marginal distribution of M, in the bottom margin of Table 9.2.
The joint distribution of two random variables contains a lot more information
than the two marginal distributions. This can be illustrated by the fact that in
many cases the joint probability mass function of X and Y cannot be retrieved
from the marginal probability mass functions p
X
and p
Y
.Asimpleexample
is given in the following quick exercise.
Quick exercise 9.2 Let X and Y be two discrete random variables, with
joint probability mass function p, given by the following table, where ε is an
arbitrary number between −1/4and1/4.
b
a 01p
X
(a)
01/4 − ε 1/4+ε ...
11/4+ε 1/4 − ε ...
p
Y
(b) ... ... ...
Complete the table, and conclude that we cannot retrieve p from p
X
and p
Y
.
118 9 Joint distributions and independence
The joint distribution function
As in the case of a single random variable, the distribution function enables
us to treat pairs of discrete and pairs of continuous random variables in the
same way.
Definition. The joint distribution function F of two random vari-
ables X and Y is the function F : R
2
→ [0, 1] defined by
F (a, b)=P(X ≤ a, Y ≤ b)for−∞<a,b<∞.
Quick exercise 9.3 Compute F (5, 3) for the joint distribution function F
of the pair (S, M ).
The distribution functions F
X
and F
Y
can be obtained from the joint distri-
bution function of X and Y . As before, we speak of the marginal distribution
functions. The following rule holds.
From joint to marginal distribution function. Let F be
the joint distribution function of random variables X and Y .Then
the marginal distribution function of X is given for each a by
F
X
(a)=P(X ≤ a)=F (a, +∞) = lim
b→∞
F (a, b), (9.1)
and the marginal distribution function of Y is given for each b by
F
Y
(b)=P(Y ≤ b)=F (+∞,b) = lim
a→∞
F (a, b). (9.2)
9.2 Joint distributions of continuous random variables
We saw in Chapter 5 that the probability that a single continuous random
variable X lies in an interval [a, b], is equal to the area under the probability
density function f of X over the interval (see also Figure 5.1). For the joint
distribution of continuous random variables X and Y the situation is analo-
gous: the probability that the pair (X, Y ) falls in the rectangle [a
1
,b
1
]×[a
2
,b
2
]
is equal to the volume under the joint probability density function f(x, y)of
(X, Y ) over the rectangle. This is illustrated in Figure 9.1, where a chunk of
a joint probability density function f(x, y)isdisplayedforx between −0.5
and 1 and for y between −1.5 and 1. Its volume represents the probability
P(−0.5 ≤ X ≤ 1, −1.5 ≤ Y ≤ 1). As the volume under f on [−0.5, 1]×[−1.5, 1]
is equal to the integral of f over this rectangle, this motivates the following
definition.
9.2 Joint distributions of continuous random variables 119
-3
-2
-1
0
1
2
3
x
-3
-2
-1
0
1
2
3
y
0
0.05
0.1
0.15
f(x,y)
Fig. 9.1. Volume under a joint probability density function f on the rectangle
[−0.5, 1] × [−1.5, 1].
Definition. Random variables X and Y have a joint continuous
distribution if for some function f : R
2
→ R and for all numbers
a
1
,a
2
and b
1
,b
2
with a
1
≤ b
1
and a
2
≤ b
2
,
P(a
1
≤ X ≤ b
1
,a
2
≤ Y ≤ b
2
)=
b
1
a
1
b
2
a
2
f(x, y)dx dy.
The function f has to satisfy f (x, y) ≥ 0 for all x and y,and
∞
−∞
∞
−∞
f(x, y)dx dy =1.Wecallf the joint probability density
function of X and Y .
As in the one-dimensional case there is a simple relation between the joint
distribution function F and the joint probability density function f:
F (a, b)=
a
−∞
b
−∞
f(x, y)dx dy and f(x, y)=
∂
2
∂x∂y
F (x, y).
A joint probability density function of two random variables is also called
a bivariate probability density. An explicit example of such a density is the
function
f(x, y)=
30
π
e
−50x
2
−50y
2
+80xy
for −∞ <x<∞ and −∞ <y<∞; see Figure 9.2. This is an example of
a bivariate normal density (see Remark 11.2 for a full description of bivariate
normal distributions).
We illustrate a number of properties of joint continuous distributions by means
of the following simple example. Suppose that X and Y have joint probability