A Course in Mathematical Statistics phần 3 docx

90 4 Distribution Functions, Probability Densities, and Their Relationship4.1.3 Refer to Exercise 3.3.13, in Chapter 3, and determine the d.f.’s sponding to the p.d.f.’s given there.. 94

90 4 Distribution Functions, Probability Densities, and Their Relationship

4.1.3 Refer to Exercise 3.3.13, in Chapter 3, and determine the d.f.’s sponding to the p.d.f.’s given there

corre-4.1.4 Refer to Exercise 3.3.14, in Chapter 3, and determine the d.f.’s sponding to the p.d.f.’s given there

corre-4.1.5 Let X be an r.v with d.f F Determine the d.f of the following r.v.’s:

i) Show that the following function F is a d.f (Logistic distribution) and

derive the corresponding p.d.f., f.

F x e

x x

ii) Show that f(x)=αF(x)[1 − F(x)]

4.1.9 Refer to Exercise 3.3.17 in Chapter 3 and determine the d.f F sponding to the p.d.f f given there Write out the expressions of F and f for

corre-n = 2 and n = 3.

4.1.10 If X is an r.v distributed as N(3, 0.25), use Table 3 in Appendix III in

order to compute the following probabilities:

approxi-i) At least 150?

ii) At most 80?

iii) Between 95 and 125?

4.1.12 A certain manufacturing process produces light bulbs whose life

length (in hours) is an r.v X distributed as N(2,000, 2002

) A light bulb issupposed to be defective if its lifetime is less than 1,800 If 25 light bulbs are

4.1 The Cumulative Distribution Function 91

tested, what is the probability that at most 15 of them are defective? (Use therequired independence.)

4.1.13 A manufacturing process produces 1

2-inch ball bearings, which areassumed to be satisfactory if their diameter lies in the interval 0.5 ± 0.0006 anddefective otherwise A day’s production is examined, and it is found that thedistribution of the actual diameters of the ball bearings is approximatelynormal with mean μ = 0.5007 inch and σ = 0.0005 inch Compute the propor-tion of defective ball bearings

4.1.14 If X is an r.v distributed as N(μ, σ2

), find the value of c (in terms of

μ and σ) for which P(X < c) = 2 − 9P(X > c).

4.1.15 Refer to the Weibull p.d.f., f, given in Exercise 3.3.19 in Chapter 3 and

( )= ( )( ), and draw its graphforα = 1 and β = 1

2, 1, 2;

iii) For s and t > 0, calculate the probability P(X > s + t|X > t) where X is an r.v.

having the Weibull distribution;

iv) What do the quantities F(x), (x), H(x) and the probability in

part (iii) become in the special case of the Negative Exponentialdistribution?

4.2 The d.f of a Random Vector and Its Properties—Marginal and Conditional d.f.’s and p.d.f.’s

For the case of a two-dimensional r vector, a result analogous to Theorem

1 can be established So consider the case that k = 2 We then have X =

(X1, X2) ′ and the d.f F(or FX or F X1,X2 ) of X, or the joint distribution function

of X1, X2, is F(x1, x2) = P(X1 ≤ x1, X2 ≤ x2) Then the following theorem holds

iii) F is continuous from the right with respect to each of the coordinates x1, x2,

or both of them jointly

THEOREM 4

4.2 The d.f of a Random Vector and Its Properties 91

92 4 Distribution Functions, Probability Densities, and Their Relationship

iv) If both x1, x2,→ ∞, then F(x1, x2)→ 1, and if at least one of the x1, x2→

PROOF i) Obvious.

ii) V = P(x1< X1≤ x2, y1< X2≤ y2) and is hence, clearly, ≥ 0

iii) Same as in Theorem 3 (If x= (x1, x2)′, and zn = (x 1n , x 2n)′, then zn↓ x means

REMARK 3 The function F(x1,∞) = F1(x1) is the d.f of the random variable

REMARK 4 It should be pointed out here that results like those discussed

in parts (i)–(iv) in Remark 1 still hold true here (appropriately interpreted)

also remarks such as Remark 1(i)–(iv) following Theorem 3 In particular, the

4.1 The Cumulative Distribution Function 93

at continuity points of f, where F, or FX, or F X1, · · · , Xk , is the d.f of X, or the joint

distribution function of X1, , Xk As in the two-dimensional case,

F(∞ ⋅ ⋅ ⋅ ∞, , , x j,∞ ⋅ ⋅ ⋅ ∞, , )=F x j( )j

is the d.f of the random variable X j , and if m x j’s are replaced by ∞ (1 < m < k),

then the resulting function is the joint d.f of the random variables

correspond-ing to the remaincorrespond-ing (k − m) X j ’s All these d.f.’s are called marginal tion functions.

distribu-In Statement 2, we have seen that if X= (X1, , X k)′ is an r vector, then

Xj , j = 1, 2, , k are r.v.’s and vice versa Then the p.d.f of X, f(x) =

f(x1, , xk ), is also called the joint p.d.f of the r.v.’s X1, , X k

Consider first the case k = 2; that is, X = (X1, X2) ′, f(x) = f(x1, x2) and set

Then f1, f2 are p.d.f.’s In fact, f1(x1)≥ 0 and

x x x

2 1 1

94 4 Distribution Functions, Probability Densities, and Their Relationship

This is considered as a function of x2, x1 being an arbitrary, but fixed, value of

X1 ( f1(x1)> 0) Then f(·|x1 ) is a p.d.f In fact, f(x2|x1)≥ 0 and

f x x

f x x f x x f x f x x

and show that f(·|x2) is a p.d.f Furthermore, if X1, X2 are both discrete, the

f(x2|x1) has the following interpretation:

Hence P(X2 ∈ B|X1 = x1) = ∑x2∈B f(x2|x1) For this reason, we call f(·|x2) the

conditional p.d.f of X2, given that X1= x1 (provided f1(x1)> 0) For a similar

reason, we call f(·|x2) the conditional p.d.f of X1, given that X2 = x2 (provided

f2(x2)> 0) For the case that the p.d.f.’s f and f 2 are of the continuous type,

the conditional p.d.f f (x1|x2) may be given an interpretation similar to the one given above By assuming (without loss of generality) that h1, h2> 0, onehas

11

11

where F is the joint d.f of X1, X2 and F2 is the d.f of X2 By letting h1, h2→ 0

and assuming that (x1, x2) ′ and x2 are continuity points of f and f2, respectively, the last expression on the right-hand side above tends to f(x1, x2)/f2(x2) which was denoted by f(x1|x2) Thus for small h1, h2, h1 f(x1|x2) is approximately equal

to P(x1 < X1 ≤ x1 + h1|x2 < X2 ≤ x2 + h2), so that h1 f(x1|x2) is approximately the

conditional probability that X1 lies in a small neighborhood (of length h1) of x1, given that X2 lies in a small neighborhood of x2 A similar interpretation may

be given to f(x2|x1) We can also define the conditional d.f of X2, given X1 = x1,

by means of

4.1 The Cumulative Distribution Function 95

F x x

f x x

f x x dx

x x x

2 1

2 1

2 1 2

2 2 2

and similarly for F(x1|x2).

The concepts introduced thus far generalize in a straightforward way for

k > 2 Thus if X = (X1, , X k)′ with p.d.f f(x1, , x k), then we have called

f(x1, , xk ) the joint p.d.f of the r.v.’s X1, X2, , X k If we sum (integrate)

over t of the variables x1, , x k keeping the remaining s fixed (t + s = k), the resulting function is the joint p.d.f of the r.v.’s corresponding to the remaining

s variables; that is,

1 , ⋅ ⋅ ⋅ , ⋅ ⋅ ⋅

such p.d.f.’s which are also called marginal p.d.f.’s Also if x i1 , , x is are such

that f i1, , it (x i1 , , x is)> 0, then the function (of x j1 , , x jt) defined by

is a p.d.f called the joint conditional p.d.f of the r.v.’s X j1 , , X jt , given X i1=

x i1 , · · · , X js = x js , or just given X i1 , , X is Again there are 2k− 2 joint

condi-tional p.d.f.’s involving all k r.v.’s X1, , X k Conditional distribution

func-tions are defined in a way similar to the one for k= 2 Thus

Let the r.v.’s X1, , X k have the Multinomial distribution with parameters

n and p1, , pk Also, let s and t be integers such that 1 ≤ s, t < k and

s + t = k Then in the notation employed above, we have:

4.2 The d.f of a Random Vector and Its Properties 95

EXAMPLE 1

96 4 Distribution Functions, Probability Densities, and Their Relationship

that is, the r.v.’s X i1 , , X is and Y = n − (X i1 + · · · + X is) have the Multinomial

distribution with parameters n and p i1 , , p is , q.

p q

p q

j x

j x

that is, the (joint) conditional distribution of X j1, , X jt given X i1 , , X is is

Multinomial with parameters n − r and p j1 /q, , p jt /q.

i x i

x n r

i x i x j x j x

k

s

i s

i s

i s

i s j

t jt

1 1

1 1

i x i

t jt p q

1

,

as was to be seen

Let the r.v.’s X1 and X2 have the Bivariate Normal distribution, and recall that

their (joint) p.d.f is given by:

EXAMPLE 2

4.1 The Cumulative Distribution Function 97

2

1 1 1

2 2 2

2 2 2 2

μσ

μσ

We saw that the marginal p.d.f.’s f1, f2 are N(μ1,σ2

1), N(μ2,σ2

2), respectively;

that is, X1, X2 are also normally distributed Furthermore, in the process of

proving that f(x1, x2) is a p.d.f., we rewrote it as follows:

1 2

2

1 1 2

1 2

2 2

1 1 2

2 2

Exercises

4.2.1 Refer to Exercise 3.2.17 in Chapter 3 and:

i) Find the marginal p.d.f.’s of the r.v.’s X j , j= 1, · · · , 6;

ii) Calculate the probability that X1≥ 5

4.2.2 Refer to Exercise 3.2.18 in Chapter 3 and determine:

ii) The marginal p.d.f of each one of X1, X2, X3;

ii) The conditional p.d.f of X1, X2, given X3; X1, X3, given X2; X2, X3, given

X;

Exercises 97

98 4 Distribution Functions, Probability Densities, and Their Relationship

iii) The conditional p.d.f of X1, given X2, X3; X2, given X3, X1; X3, given X1,

i) Determine the constant c;

ii) Find the marginal p.d.f.’s of X and Y;

iii) Find the conditional p.d.f of X, given Y, and the conditional p.d.f of Y,

given X;

iv) Calculate the probability that X≤ 1

4.2.4 Let the r.v.’s X, Y be jointly distributed with p.d.f f given by f(x, y)=

e −x−y I(0,∞)×(0,∞)(x, y) Compute the following probabilities:

i) Determine the constant c;

ii) Find the marginal p.d.f of each one of the r.v.’s X j , j= 1, 2, 3;

iii) Find the conditional (joint) p.d.f of X1, X2, given X3, and the conditional

4.1 The Cumulative Distribution Function 99

i) Show that for each fixed y, f(·|y) is a p.d.f., the conditional p.d.f of an r.v.

X, given that another r.v Y equals y;

ii) If the marginal p.d.f of Y is Negative Exponential with parameter λ = 1,

what is the joint p.d.f of X, Y?

iii) Show that the marginal p.d.f of X is given by f(x)= (1

2)x+1I A (x), where

A= {0, 1, 2, }

4.2.7 Let Y be an r.v distributed as P(λ) and suppose that the conditional

distribution of the r.v X, given Y = n, is B(n, p) Determine the p.d.f of X and the conditional p.d.f of Y, given X = x.

4.2.8 Consider the function f defined as follows:

1 2

1 2 2 2

1 3 2 3

1 1 1 1 1 2

1

14

4.3 Quantiles and Modes of a Distribution

Let X be an r.v with d.f F and consider a number p such that 0 < p < 1 A pth quantile of the r.v X, or of its d.f F, is a number denoted by x pand having

the following property: P(X ≤ x p)≥ p and P(X ≥ x p)≥ 1 − p For p = 0.25 we get a quartile of X, or its d.f., and for p = 0.5 we get a median of X, or its

d.f For illustrative purposes, consider the following simple examples

Let X be an r.v distributed as U(0, 1) and let p= 0.10, 0.20, 0.30, 0.40, 0.50,

0.60, 0.70, 0.80 and 0.90 Determine the respective x0.10, x0.20, x0.30, x0.40, x0.50, x0.60,

x0.70, x0.80, and x0.90.Since for 0 ≤ x ≤ 1, F(x) = x, we get: x0.10 = 0.10, x0.20 = 0.20, x0.30= 0.30,

x0.40= 0.40, x0.50 = 0.50, x0.60 = 0.60, x0.70 = 0.70, x0.80 = 0.80, and x0.90= 0.90

Let X be an r.v distributed as N(0, 1) and let p= 0.10, 0.20, 0.30, 0.40, 0.50,

0.60, 0.70, 0.80 and 0.90 Determine the respective x0.10, x0.20, x0.30, x0.40, x0.50, x0.60,

100 4 Distribution Functions, Probability Densities, and Their Relationship

0 (c)

x

x p F(x)

Figure 4.3 Observe that the figures demonstrate that, as defined, xp need not be unique.

From the Normal Tables (Table 3 in Appendix III), by linear interpolation

to how the unit probability mass is distributed over the real line In Fig 4.3

various cases are demonstrated for determining graphically the pth quantile of

4.1 The Cumulative Distribution Function 101

Let X be B(n, p); that is,

Consider the number (n + 1)p and set m = [(n + 1)p], where [y] denotes the

largest integer which is ≤ y Then if (n + 1)p is not an integer, f(x) has a unique mode at x = m If (n + 1)p is an integer, then f(x) has two modes obtained for

( )

−

( )1 = − + ⋅

1

Hence f(x) > f(x − 1) (f is increasing) if and only if

Thus if (n + 1)p is not an integer, f(x) keeps increasing for x ≤ m and then decreases so the maximum occurs at x = m If (n + 1)p is an integer, then the maximum occurs at x = (n + 1)p, where f(x) = f(x − 1) (from above calcula-

tions) Thus

x=( )n+1p−1

is a second point which gives the maximum value 

Let X be P(λ); that is,

102 4 Distribution Functions, Probability Densities, and Their Relationship

λ

!

!

Hence f(x) > f(x − 1) if and only if λ > x Thus if λ is not an integer, f(x) keeps increasing for x≤ [λ] and then decreases Then the maximum of f(x) occurs

at x= [λ] If λ is an integer, then the maximum occurs at x = λ But in this case

f(x) = f(x − 1) which implies that x = λ − 1 is a second point which givesthe maximum value to the p.d.f 

Exercises

4.3.1 Determine the pth quantile x p for each one of the p.d.f.’s given inExercises 3.2.13–15, 3.3.13–16 (Exercise 3.2.14 for α = 1

4) in Chapter 3 if p=0.75, 0.50

4.3.2 Let X be an r.v with p.d.f f symmetric about a constant c (that is, f(c − x) = f(c + x) for all x ∈ ) Then show that c is a median of f.

4.3.3 Draw four graphs—two each for B(n, p) and P(λ)—which represent

the possible occurrences for modes of the distributions B(n, p) and P(λ)

4.3.4 Consider the same p.d.f.’s mentioned in Exercise 4.3.1 from the point

of view of a mode

4.4* Justification of Statements 1 and 2

In this section, a rigorous justification of Statements 1 and 2 made in Section4.1 will be presented For this purpose, some preliminary concepts and resultsare needed and will be also discussed

A set G in  is called open if for every x in G there exists an open interval containing x and contained in G Without loss of generality, such intervals may

be taken to be centered at x.

It follows from this definition that an open interval is an open set, theentire real line  is an open set, and so is the empty set (in a vacuous manner).Every open set in  is measurable

PROOF Let G be an open set in , and for each x ∈ G, consider an open interval centered at x and contained in G Clearly, the union over x, as x varies

in G, of such intervals is equal to G The same is true if we consider only those intervals corresponding to all rationals x in G These intervals are countably

many and each one of them is measurable; then so is their union 

LEMMA 1

DEFINITION 1

4.1 The Cumulative Distribution Function 103

A set G in m

, m ≥ 1, is called open if for every x in G there exists an open cube

inm containing x and contained in G; by the term open “cube” we mean the Cartesian product of m open intervals of equal length Without loss of gener- ality, such cubes may be taken to be centered at x.

Every open set in n

is measurable

PROOF It is analogous to that of Lemma 1 Indeed, let G be an open set in

m , and for each x ∈ G, consider an open cube centered at x and contained in

G The union over x, as x varies in G, of such cubes clearly is equal to G The same is true if we restrict ourselves to x’s in G whose m coordinates are

rationals Then the resulting cubes are countably many, and therefore theirunion is measurable, since so is each cube 

Recall that a function g: S⊆ →  is said to be continuous at x0 ∈ S if for

everyε > 0 there exists a δ = δ(ε, x0)> 0 such that |x − x0|<ε implies |g(x) − g(x0)|

<δ The function g is continuous in S if it is continuous for every x ∈ S.

It follows from the concept of continuity that ε → 0 implies δ → 0

Let g:  →  be continuous Then g is measurable.

PROOF By Theorem 5 in Chapter 1 it suffices to show that g−1(G) are urable sets for all open intevals G in  Set B = g−1(G) Thus if B = ∅, the

meas-assertion is valid, so let B ≠ ∅ and let x0 be an arbitrary point of B, so that g(x0)

∈ G Continuity of g at x0 implies that for every ε > 0 there exists δ = δ(ε, x0)

> 0 such that |x − x0|<ε implies |g(x) − g(x0)|<δ Equivalently, x ∈ (x0−ε, x0

+ε) implies g(x) ∈ (g(x0)−δ, g(x0)+δ) Since g(x0)∈ G and G is open, by

choosingε sufficiently small, we can make δ so small that (g(x0)−δ, g(x0)+δ)

is contained in G Thus, for such a choice of ε and δ, x ∈ (x0−ε, x0+ε) implies

that (g(x0)−δ, g(x0)+δ) ⊂ G But B(= g−1(G)) is the set of all x∈ for which

g(x) ∈ G As all x ∈ (x0−ε, x0+ε) have this property, it follows that (x0−ε, x0

+ε) ⊂ B Since x0 is arbitrary in B, it follows that B is open Then by Lemma

1, it is measurable 

The concept of continuity generalizes, of course, to Euclidean spaces ofhigher dimensions, and then a result analogous to the one in Lemma 3 alsoholds true

A function g : S⊆k→m

(k, m ≥ 1) is said to be continuous at x0∈k

if foreveryε > 0 there exists a δ = δ (ε, x0) > 0 such that ||x − x0||<ε implies ||g(x) − g(x0)||<δ The function g is continuous in S if it is continuous for every x ∈ S.

Here ||x|| stands for the usual norm in k

; i.e., for x = (x1, , x k)′, ||x|| =

x i i

( ) , and similarly for the other quantities

Once again, from the concept of continuity it follows that ε → 0 implies

104 4 Distribution Functions, Probability Densities, and Their Relationship

PROOF The proof is similar to that of Lemma 3 The details are presented

here for the sake of completeness Once again, it suffices to show that g−1(G) are measurable sets for all open cubes G in m

Set B = g−1(G) If B= ∅ the

assertion is true, and therefore suppose that B≠ ∅ and let x0 be an arbitrary

point of B Continuity of g at x0 implies that for every ε > 0 there exists a δ =

δ(ε, x0)> 0 such that ||x − x0||<ε implies ||g(x) − g(x0)||<δ; equivalently, x ∈

S(x0,ε) implies g(x) ∈ S(g(x0),δ), where S(c, r) stands for the open sphere with

center c and radius r Since g(x0)∈ G and G is open, we can choose ε so small

that the corresponding δ is sufficiently small to imply that g(x) ∈ S(g(x0),δ).Thus, for such a choice of ε and δ, x ∈ S(x0,ε) implies that g(x) ∈ S(g(x0),δ)

Since B( = g−1(G)) is the set of all x∈k

for which g(x) ∈ G, and x ∈ S(x0,ε)

implies that g(x)∈ S(g(x0),δ), it follows that S(x0,ε) ⊂ B At this point, observe

that it is clear that there is a cube containing x0 and contained in S(x0,ε); call

it C(x0,ε) Then C(x0,ε) ⊂ B, and therefore B is open By Lemma 2, it is also

measurable 

We may now proceed with the justification of Statement 1

Let X : (S, A) → (k

,B k ) be a random vector, and let g : (k

) and is a random vector (That

is, measurable functions of random vectors are random vectors.)

PROOF To prove that [g(X)]−1(B) ∈ A if B ∈ B m

To this theorem, we have the following

Let X be as above and g be continuous Then g(X) is a random vector (That

is, continuous functions of random vectors are random vectors.)

PROOF The continuity of g implies its measurability by Lemma 3, and

there-fore the theorem applies and gives the result 

For j = 1, , k, the jth projection function g j is defined by: g j:k

→ and

gj(x)= g j (x1, , x k)= x j

It so happens that projection functions are continuous; that is,

The coordinate functions g j , j = 1, , k, as defined above, are continuous.

PROOF For an arbitrary point x0 in K

of continuity of g j is satisfied here for δ = ε 

Now consider a k-dimensional function X defined on the sample space S.

Then X may be written as X= (X1, , X k)′, where X j , j = 1, , k are

real-valued functions The question then arises as to how X and X , j = 1, , k are

THEOREM 7

LEMMA 5

DEFINITION 5

COROLLARY

4.1 The Cumulative Distribution Function 105

related from a measurability point of view To this effect, we have the ing result

Then g j’s are continuous by Lemma 5 and therefore

measurable by Lemma 4 Then for each j = 1, , k, g j(X)= g j (X1, , X k)=

X j is measurable and hence an r.v

Next, assume that X j , j = 1, , k are r.v.’s To show that X is an r vector,

by special case 3 in Section 2 of Chapter 1, it suffices to show that X−1(B)∈ A

for each B = (−∞, x1]× · · · × (−∞, x k ], x1, , x k∈ Indeed,

where Q is the set of rationals in 

4.4.2 Use Exercise 4.4.1 in order to show that, if X and Y are r.v.’s, then so

is the function X + Y.

4.4.3

ii) If X is an r.v., then show that so is the function −X.

ii) Use part (i) and Exercise 4.4.2 to show that, if X and Y are r.v.’s, then so is

) in conjunction with part (i)

and Exercises 4.4.2 and 4.4.3(ii) to show that, if X and Y are r.v.’s, then so

is the function XY.

4.4.5

ii) If X is an r.v., then show that so is the function 1

ii) Use part (i) in conjunction with Exercise 4.4.4(ii) to show that, if X and Y

THEOREM 8

106 5 Moments of Random Variables—Some Moment and Probability Inequalities

106

5.1 Moments of Random Variables

In the definitions to be given shortly, the following remark will prove useful

REMARK 1 We say that the (infinite) series ∑xh(x), where x = (x1, , x k)′varies over a discrete set in k

, k ≥ 1, converges absolutely if ∑x|h(x)|< ∞ Also

we say that the integral ∫ ⋅ ⋅ ⋅∫−∞ ( ⋅ ⋅ ⋅ ) ⋅ ⋅ ⋅

be defined below exist.

Let X= (X1 , , Xk)′ be an r vector with p.d.f f and consider the (measurable) function g:k

→, so that g(X) = g(X1 , , X k) is an r.v Then we give the

iii) For n = 1, 2, , the nth moment of g(X) is denoted by E[g(X)] n

and isdefined by:

5.1 Moments of Random Variables 107

and call it the mathematical expectation or mean value or just mean of g(X) Another notation for E[g(X)] which is often used is μg(X), or μ[g(X)], or just

μ, if no confusion is possible

iii) For r > 0, the rth absolute moment of g(X) is denoted by E|g(X)| r

and isdefined by:

standard deviation (s.d.) of g(X) and is also denoted by σg(X), or just σ, if no

confusion is possible The variance of an r.v is referred to as the moment of inertia in Mechanics.

1 ⋅ ⋅ ⋅ ) is called the (n1, , n k )-joint moment of X1, , X k In

par-ticular, for n1 = · · · = n−1= n+1= · · · = n = 0, n = n, we get

108 5 Moments of Random Variables—Some Moment and Probability Inequalities

j n

j n

k

j n

j n

x

j n

n x n

which is the mathematical expectation or mean value or just mean of X This

quantity is also denoted by μX or μ(X) or just μ when no confusion is possible.

The quantity μX can be interpreted as follows: It follows from the

defini-tion that if X is a discrete uniform r.v., then μX is just the arithmetic average of

the possible outcomes of X Also, if one recalls from physics or elementary calculus the definition of center of gravity and its physical interpretation as the

point of balance of the distributed mass, the interpretation of μX as the mean

or expected value of the random variable is the natural one, provided the probability distribution of X is interpreted as the unit mass distribution.

REMARK 2 In Definition 1, suppose X is a continuous r.v Then E[g(X)] =

∫ ( ) There seems to be a discrepancy between these two definitions

More specifically, in the definition of E[g(X)], one would expect to use the p.d.f of g(X) rather than that of X Actually, the definition of E[g(X)], as given, is correct and its justification is roughly as follows: Consider E[g(x)] =

5.1 Moments of Random Variables 109

On the other hand, if f Y is the p.d.f of Y, then f y f g y g y

Y

d dy

( )= [ − 1( ) ] − 1( )

Therefore the last integral above is equal to ∫−∞∞ yf y dy Y( ) , which is consonantwith the definition of E X( )=∫−∞∞ xf x dx( ) (A justification of the above deriva-tions is given in Theorem 2 of Chapter 9.)

2 For g as above, that is, g(X1, , X k)= X n X

n x

when no confusion is possible, and is called the variance of X Its

positive square root σX or σ(X) or just σ is called the standard deviation (s.d.)

110 5 Moments of Random Variables—Some Moment and Probability Inequalities

is the nth factorial moment of the r.v X j Thus the nth factorial moment of an r.v X with p.d.f f is

From the very definition of E[g(X)], the following properties are immediate.

(E1) E(c) = c, where c is a constant.

(E2) E[cg(X)] = cE[g(X)], and, in particular, E(cX) = cE(X) if X is an

r.v

(E3) E[g(X) + d] = E[g(X)] + d, where d is a constant In particular,

E(X + d) = E(X) + d if X is an r.v.

(E4) Combining (E2) and (E3), we get E[cg(X) + d] = cE[g(X)] + d,

and, in particular, E(cX + d) = cE(X) + d if X is an r.v.

j j n

∑

(E5) If X ≥ 0, then E(X) ≥ 0.

Consequently, by means of (E5) and (E4″), we get that(E5′) If X ≥ Y, then E(X) ≥ E(Y), where X and Y are r.v.’s (with finite

expectations)

(E6) |E[g(X)]| ≤ E|g(X)|.

(E7) If E|X| r < ∞ for some r > 0, where X is an r.v., then E|X| r′< ∞ forall 0 < r′ < r.

This is a consequence of the obvious inequality |X| r′≤ 1 + |X| r

5.1 Moments of Random Variables 111

(E7′) If E(X n

) exists (that is, E|X| n

< ∞) for some n = 2, 3, , then E(X n′)

also exists for all n ′ = 1, 2, with n′ < n.

Regarding the variance, the following properties are easily established bymeans of the definition of the variance

5.1.1 Verify the details of properties (E1)–(E7).

5.1.2 Verify the details of properties (V1)–(V5).

5.1.3 For r ′ < r, show that |X| r′≤ 1 + |X| r

and conclude that if E|X| r< ∞, then

E|X| r′ for all 0 < r′ < r.

Exercises 111

112 5 Moments of Random Variables—Some Moment and Probability Inequalities

5.1.4 Verify the equality ( [ ( )] )E g X = ∫−∞∞ g x f( ) X( )x dx=∫ yf y dy Y( )

5.1.5 For any event A, consider the r.v X = I A , the indicator of A defined by

IA (s) = 1 for s ∈ A and I A (s) = 0 for s ∈ A c

5.1.7 Let X be an r.v with finite EX.

ii) For any constant c, show that E(X − c)2

= E(X − EX)2

+ (EX − c)2

;

ii) Use part (i) to conclude that E(X − c)2

is minimum for c = EX.

5.1.8 Let X be an r.v such that EX4< ∞ Then show that

5.1.10 Let X be the r.v denoting the number of claims filed by a

policy-holder of an insurance company over a specified period of time On the basis

of an extensive study of the claim records, it may be assumed that the

5.1 Moments of Random Variables 113

5.1.11 A roulette wheel has 38 slots of which 18 are red, 18 black, and 2green

ii i) Suppose a gambler is placing a bet of $M on red What is the gambler’s

expected gain or loss and what is the standard deviation?

iii) If the same bet of $M is placed on green and if $kM is the amount

the gambler wins, calculate the expected gain or loss and the standarddeviation

iii) For what value of k do the two expectations in parts (i) and (ii) coincide? iv) Does this value of k depend on M?

i v) How do the respective standard deviations compare?

5.1.12 Let X be an r.v such that P(X = j) = (1

2)j , j= 1, 2,

ii) Compute EX, E[X(X− 1)];

ii) Use (i) in order to compute σ2

5.1.14 Let the r.v X be distributed as U( α, β) Calculate EX n

for any positive

integer n.

5.1.15 Let X be an r.v with p.d.f f symmetric about a constant c (that is, f(c − x) = f(c + x) for every x).

ii) Then if EX exists, show that EX = c;

ii) If c = 0 and EX 2n+1 exists, show that EX 2n+1 = 0 (that is, those moments of X

of odd order which exist are all equal to zero)

5.1.16 Refer to Exercise 3.3.13(iv) in Chapter 3 and find the EX for those α’s

for which this expectation exists, where X is an r.v having the distribution in

114 5 Moments of Random Variables—Some Moment and Probability Inequalities

ii) Use the interpretation of the definite integral as an area in order to give a

geometric interpretation of EX.

5.1.19 Let X be an r.v of the continuous type with finite EX and p.d.f f.

ii) If m is a median of f and c is any constant, show that

integral from −∞ to c and c to ∞ in order to remove the absolute value.

Then the fact that ∫−∞m f x dx( ) =∫m∞f x dx( ) = 1

2 and simple manipulationsprove part (i) For part (ii), observe that ∫m c(c−x f x dx) ( ) ≥0 whether c ≥ m

or c < m.)

5.1.20 If the r.v X is distributed according to the Weibull distribution (see

Exercise 4.1.15 in Chapter 4), then:

x n

x n

1

1 1

11

1 0

5.1 Moments of Random Variables 115

n

x x

2 2

2 0

116 5 Moments of Random Variables—Some Moment and Probability Inequalities

2 1 2 0

2 1 2 0

2 1 2

0

2 2 2 0

2 2 2 0

Multiplying them out, we obtain