A Modern Introduction to Probability and Statistics, Understanding Why and How - Dekking, Kraaikamp, Lopuhaa, Meester (Современное введение в теорию вероятностей и статистику

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130 9 Joint distributions and independence

b −10 1

4 η −

− η 0

6 η +

− η

a. Which are the values η can attain?

b. Is there a value of η for which X and Y are independent?

9.6  Let X and Y be two independent Ber(

) random variables. Deﬁne

random variables U and V by:

U = X + Y and V = |X −Y |.

a. Determine the joint and marginal probability distributions of U and V .

b. Find out whether U and V are dependent or independent.

9.7 To investigate the relation between hair color and eye color, the hair color

and eye color of 5383 persons was recorded. The data are given in the following

table:

Hair color

Eye color Fair/red Medium Dark/black

Light 1168 825 305

Dark 573 1312 1200

Source: B. Everitt and G. Dunn. Applied multivariate data analysis. Second

edition Hodder Arnold, 2001; Table 4.12. Reproduced by permission of Hodder

&Stoughton.

Eye color is encoded by the values 1 (Light) and 2 (Dark), and hair color by

1 (Fair/red), 2 (Medium), and 3 (Dark/black). By dividing the numbers in

the table by 5383, the table is turned into a joint probability distribution for

random variables X (hair color) taking values 1 to 3 and Y (eye color) taking

values 1 and 2.

a. Determine the joint and marginal probability distributions of X and Y .

b. Find out whether X and Y are dependent or independent.

9.8  Let X and Y be independent random variables with probability distri-

butions given by

P(X =0)=P(X =1)=

and P(Y =0)=P(Y =2)=

9.7 Exercises 131

a. Compute the distribution of Z = X + Y .

b. Let

Y and

Z be independent random variables, where

Y has the same

distribution as Y ,and

Z the same distribution as Z. Compute the distri-

bution of

X =

Z −

Y .

9.9  Suppose that the joint distribution function of X and Y is given by

F (x, y)=1− e

−2x

− e

−y

−(2x+y)

if x>0,y>0,

and F (x, y)=0otherwise.

a. Determine the marginal distribution functions of X and Y .

b. Determine the joint probability density function of X and Y .

c. Determine the marginal probability density functions of X and Y .

d. Find out whether X and Y are independent.

9.10  Let X and Y be two continuous random variables with joint proba-

bility density function

f(x, y)=

xy(1 + y)for0≤ x ≤ 1and0≤ y ≤ 1,

and f (x, y)=0otherwise.

a. Find the probability P



≤ X ≤

≤ Y ≤



b. Determine the joint distribution function of X and Y for a and b between

0and1.

c. Useyouranswerfromb to ﬁnd F

(a)fora between 0 and 1.

d. Apply the rule on page 122 to ﬁnd the probability density function of X

from the joint probability density function f(x, y). Use the result to verify

your answer from c.

e. Find out whether X and Y are independent.

9.11  Let X and Y be two continuous random variables, with the same

joint probability density function as in Exercise 9.10. Find the probability

P(X<Y)thatX is smaller than Y .

9.12 The joint probability density function f of the pair (X, Y )isgivenby

f(x, y)=K(3x

+8xy)for0≤ x ≤ 1and0≤ y ≤ 2,

and f (x, y) = 0 for all other values of x and y.HereK is some positive

constant.

a. Find K.

b. Determine the probability P(2X ≤ Y ).

132 9 Joint distributions and independence

9.13  On a disc with origin (0, 0) and radius 1, a point (X,Y ) is selected by

throwing a dart that hits the disc in an arbitrary place. This is best described

by the joint probability density function f of X and Y ,givenby

f(x, y)=



c if x

+ y

≤ 1

0otherwise,

where c is some positive constant.

a. Determine c.

b. Let R =

√

+ Y

be the distance from (X, Y ) to the origin. Determine

the distribution function F

c. Determine the marginal density function f

. Without doing any calcula-

tions, what can you say about f

9.14 An arbitrary point (X, Y ) is drawn from the square [−1, 1] × [−1, 1].

This means that for any region G in the plane, the probability that (X, Y )is

in G,isgivenbytheareaofG ∩ divided by the area of ,where denotes

the square [−1, 1] ×[−1, 1]:

P((X,Y ) ∈ G)=

area of G ∩ 

area of 

a. Determine the joint probability density function of the pair (X, Y ).

b. Check that X and Y are two independent, U(−1, 1) distributed random

variables.

9.15  Let the pair (X, Y ) be drawn arbitrarily from the triangle ∆ with

vertices (0, 0), (0, 1), and (1, 1).

a. Use Figure 9.5 to show that the joint distribution function F of the pair

(X, Y )satisﬁes

F (a, b)=

⎧

⎪

⎨

⎪

⎩

0fora or b less than 0

a(2b − a)for(a, b) in the triangle ∆

for b between0and1anda larger than b

2a − a

for a between 0 and 1 and b larger than 1

1fora and b larger than 1.

b. Determine the joint probability density function f of the pair (X,Y ).

c. Show that f

(x)=2−2x for x between0and1andthatf

(y)=2y for

y between 0 and 1.

9.16 (Continuation of Exercise 9.15) An arbitrary point (U, V )isdrawnfrom

the unit square [0, 1]×[0, 1]. Let X and Y be deﬁned as in Exercise 9.15. Show

that min{U, V } has the same distribution as X and that max{U, V } has the

same distribution as Y .

9.7 Exercises 133

(0, 0)

(0, 1) (1, 1)

(a, b)

•

∆

←− Rectangle (−∞,a] × (−∞,b]

Fig. 9.5. Drawing (X, Y )from(−∞,a] ×(−∞,b] ∩ ∆.

9.17 Let U

and U

be two independent random variables, both uniformly

distributed over [0,a]. Let V =min{U

} and Z =max{U

}. Show

that the joint distribution function of V and Z is given by

F (s, t)=P(V ≤ s, Z ≤ t)=

− (t − s)

for 0 ≤ s ≤ t ≤ a.

Hint:notethatV ≤ s and Z ≤ t happens exactly when both U

≤ t and

≤ t, but not both s<U

≤ t and s<U

≤ t.

9.18 Suppose a vase contains balls numbered 1, 2,...,N.Wedrawn balls

without replacement from the vase. Each ball is selected with equal probability,

i.e., in the ﬁrst draw each ball has probability 1/N , in the second draw each

of the N − 1 remaining balls has probability 1/(N − 1), and so on. For i =

1, 2,...,n,letX

denote the number on the ball in the ith draw. We have

shown that the marginal probability mass function of X

is given by

(k)=

, for k =1, 2,...,N.

a. Show that

E[X

N +1

b. Compute the variance of X

.Youmayusetheidentity

1+4+9+···+ N

N(N + 1)(2N +1).

9.19  Let X and Y be two continuous random variables, with joint proba-

bility density function

f(x, y)=

−50x

−50y

+80xy

for −∞ <x<∞ and −∞ <y<∞; see also Figure 9.2.

134 9 Joint distributions and independence

a. Determine positive numbers a, b,andc such that

50x

− 80xy +50y

=(ay − bx)

+ cx

b. Setting µ =

x,andσ =

, show that

(

√

50y −

√

32x)



y − µ



and use this to show that



∞

−∞

−(

√

50y−

√

32x)

dy =

√

2π

c. Use the results from b to determine the probability density function f

of X. What kind of distribution does X have?

9.20 Suppose we throw a needle on a large sheet of paper, on which horizontal

lines are drawn, which are at needle-length apart (see also Exercise 21.16).

Choose one of the horizontal lines as x-axis, and let (X, Y ) be the center of the

needle. Furthermore, let Z be the distance of this center (X, Y ) to the nearest

horizontal line under (X, Y ), and let H be the angle between the needle and

the positive x-axis.

a. Assuming that the length of the needle is equal to 1, argue that Z has

a U (0, 1) distribution. Also argue that H has a U(0,π) distribution and

that Z and H are independent.

b. Show that the needle hits a horizontal line when

Z ≤

sin H or 1 − Z ≤

sin H.

c. Show that the probability that the needle will hit one of the horizontal

lines equals 2/π.

Covariance and correlation

In this chapter we see how the joint distribution of two or more random vari-

ables is used to compute the expectation of a combination of these random

variables. We discuss the expectation and variance of a sum of random vari-

ables and introduce the notions of covariance and correlation, which express

to some extent the way two random variables inﬂuence each other.

10.1 Expectation and joint distributions

China vases of various shapes are produced in the Delftware factories in the

old city of Delft. One particular simple cylindrical model has height H and

radius R centimeters. Due to all kinds of circumstances—the place of the vase

in the oven, the fact that the vases are handmade, etc.—H and R are not

constants but are random variables. The volume of a vase is equal to the

random variable V = πHR

, and one is interested in its expected value E[V ].

When f

denotes the probability density of V , then by deﬁnition

E[V ]=



∞

−∞

(v)dv.

However, to obtain E [V ], we do not necessarily need to determine f

from

the joint probability density f of H and R!SinceV is a function of H and R,

we can use a rule similar to the change-of-variable formula from Chapter 7:

E[V ]=E



πHR





∞

−∞



∞

−∞

πhr

f(h, r)dh dr.

Suppose that H has a U (25, 35) distribution and that R has a U(7.5, 12.5)

distribution. In the case that H and R are also independent, we have

136 10 Covariance and correlation

E[V ]=



∞

−∞



∞

−∞

πhr

(h)f

(r)dh dr =



12.5

7.5

πhr

dh dr



h dh



12.5

7.5

dr = 9621.127 cm

This illustrates the following general rule.

Two-dimensional change-of-variable formula. Let X and

Y be random variables, and let g : R

→ R be a function.

If X and Y are discrete random variables with values a

,... and

,..., respectively, then

E[g(X, Y )] =



g(a

)P(X = a

,Y = b

) .

If X and Y are continuous random variables with joint probability

density function f ,then

E[g(X, Y )] =



∞

−∞



∞

−∞

g(x, y)f (x, y)dx dy.

As an example, take g(x, y)=xy for discrete random variables X and Y with

the joint probability distribution given in Table 10.1. The expectation of XY

is computed as follows:

E[XY ]=(0·0) · 0+(1· 0) ·

+(2· 0) · 0

+(0·1) ·

+(1· 1) ·0+(2· 1) ·

+(0·2) · 0+(1· 2) ·

+(2· 2) · 0=1.

A natural question is whether this value can also be obtained from E[X]E[Y ].

We return to this question later in this chapter. First we address the expec-

tation of the sum of two random variables.

Table 10.1. Joint probabilities P(X = a, Y = b).

b 012

0 0 1/4 0

1 1/4 0 1/4

2 0 1/4 0

10.1 Expectation and joint distributions 137

Quick exercise 10.1 Compute E[X + Y ] for the random variables with the

joint distribution given in Table 10.1.

For discrete X and Y with values a

,... and b

,..., respectively, we

see that

E[X + Y ]=



+ b

)P(X = a

,Y = b

)



P(X = a

,Y = b



P(X = a

,Y = b

)







P(X = a

,Y = b

)









P(X = a

,Y = b

)





P(X = a



P(Y = b

)

=E[X]+E[Y ] .

A similar line of reasoning applies in case X and Y are continuous random

variables. The following general rule holds.

Linearity of expectations. For all numbers r, s,andt and

random variables X and Y , one has

E[rX + sY + t]=rE[X]+sE[Y ]+t.

Quick exercise 10.2 Determine the marginal distributions for the random

variables X and Y with the joint distribution given in Table 10.1, and use

them to compute E[X]enE[Y ]. Check that E[X]+E[Y ]isequaltoE[X + Y ],

which was computed in Quick exercise 10.1.

More generally, for random variables X

,...,X

and numbers s

,...,s

and t,

E[s

+ ···+ s

+ t]=s

E[X

]+···+ s

E[X

]+t.

This rule is a powerful instrument. For example, it provides an easy way to

compute the expectation of a random variable X with a Bin (n, p) distribution.

If we would use the deﬁnition of expectation, we have to compute

E[X]=



k=0

kP(X = k)=



k=0





(1 − p)

n−k

To determine this sum is not straightforward. However, there is a simple alter-

native. Recall the multiple-choice example from Section 4.3. We represented

138 10 Covariance and correlation

the number of correct answers out of 10 multiple-choice questions as a sum of

10 Bernoulli random variables. More generally, any random variable X with

a Bin(n, p) distribution can be represented as

X = R

+ R

+ ···+ R

where R

,...,R

are independent Ber(p) random variables, i.e.,



1 with probability p

0 with probability 1 − p.

Since E[R

]=0· (1 − p)+1· p = p, for every i =1, 2,...,n, the linearity-of-

expectations rule yields

E[X]=E[R

]+E[R

]+···+E[R

]=np.

Hence we conclude that the expectation of a Bin(n, p) distribution equals np.

Remark 10.1 (More than two random variables). In both the discrete

and continuous cases, the change-of-variable formula for n random variables

is a straightforward generalization of the change-of-variable formula for two

random variables. For instance, if X

,...,X

are continuous random

variables, with joint probability density function f,andg is a function from

to R,then

E[g(X

,...,X

)] =



∞

−∞

···



∞

−∞

g(x

,...,x

)f(x

,...,x

)dx

···dx

10.2 Covariance

In the previous section we have seen that for two random variables X and Y

always

E[X + Y ]=E[X]+E[Y ] .

Does such a simple relation also hold for the variance of the sum Var(X + Y )

or for expectation of the product E [XY ]? We will investigate this in the

current section.

For the variables X and Y from the example in Section 9.2 with joint proba-

bility density

f(x, y)=



y + xy



for 0 ≤ x ≤ 3and1≤ y ≤ 2,

one can show that

Var(X + Y )=

939

2000

and Var(X)+Var(Y )=

989

2500

791

10 000

4747

10 000

10.2 Covariance 139

(see Exercise 10.10). This shows, in contrast to the linearity-of-expectations

rule, that Var(X + Y ) is generally not equal to Var(X)+ Var(Y ). To deter-

mine Var(X + Y ), we exploit its deﬁnition:

Var(X + Y )=E



(X + Y − E[X + Y ])



Now X + Y − E[X + Y ]=(X − E[X]) + (Y − E[Y ]), so that

(X + Y − E[X + Y ])

=(X −E[X])

+(Y − E[Y ])

+2(X − E[X]) (Y − E[Y ]) .

Taking expectations on both sides, another application of the linearity-of-

expectations rule gives

Var(X + Y )=Var(X)+Var(Y )+2E[(X −E[X])(Y − E[Y ])] .

That is, the variance of the sum X + Y equals the sum of the variances of X

and Y , plus an extra term 2E [(X − E[X])(Y − E[Y ])]. To some extent this

term expresses the way X and Y inﬂuence each other.

Definition. Let X and Y be two random variables. The covariance

between X and Y is deﬁned by

Cov(X,Y )=E[(X − E[X])(Y − E[Y ])] .

Loosely speaking, if the covariance of X and Y is positive, then if X has a

realization larger than E[X], it is likely that Y will have a realization larger

than E [Y ], and the other way around. In this case we say that X and Y are

positively correlated. In case the covariance is negative, the opposite eﬀect oc-

curs; X and Y are negatively correlated.IncaseCov(X, Y )=0wesaythatX

and Y are uncorrelated . An easy consequence of the linearity-of-expectations

property (see Exercise 10.19) is the following rule.

An alternative expression for the covariance. Let X and

Y be two random variables, then

Cov(X,Y )=E[XY ] − E[X]E[Y ] .

For X and Y from the example in Section 9.2, we have E [X] = 109/50,

E[Y ] = 157/100, and E[XY] = 171/50 (see Exercise 10.10). Thus we see that

X and Y are negatively correlated:

Cov(X, Y )=

171

−

109

157

100

= −

5000

< 0.

Moreover, this also illustrates that, in contrast to the expectation of the sum,

for the expectation of the product, in general E[XY ]isnot equal to E [X]E[Y ].

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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