Chapter 5 Continuous Random Variables€¦ · 5.1 Introduction 5.2 Expectation and Variance of Continuous Random Variables 5.3 The Uniform Random Variable 5.4 Normal Random Variables

5.1 Introduction 5.2 Expectation and Variance of Continuous Random Variables 5.3 The Uniform Random Variable 5.4 Normal Random Variables 5.5 Exponential Random Variables 5.6 Other Continuous Random Variables 5.7 The Distribution of a Function of a Random Variable

Chapter 5 Continuous Random Variables

Chih-Yuan Hung

School of Economics and ManagementDongguan University of Technology

December 22, 2017


5.1 Introduction:Continuous Random Variables


Continuous R.V.

A continuous random variable is a r.v. of which is valued on auncountable set.For a continuous r.v. X, we define the probability distributionfunction (PDF):

P{X ∈ B} =∫Bf (x)dx


f must satisfy

1 = P{X ∈ (−∞, ∞)} =∫ ∞

−∞f (x)dx

For any interval [a, b] ∈ B,

P{a ≤ X ≤ b} =∫ a

bf (x)dx

If a=b, then P{X = a} =∫ aa f (x)dx = 0

a continuous random variable will assume any fixed value iszero.

For a continuous r.v.,

P{X < a} = P{X ≤ a} = F (a) =∫ a

−∞f (x)dx


Example 1a

Suppose that X is a continuous r.v. whose pdf is given by

f (x) =

{C (4x − 2x2) 0 < x < 2

0 otherwise

a What is the value of C?

b Find P{X > 1}.Sol.

a ∫ 2

0C (4x − 2x2)dx = 1 =⇒ C

[2x2 − 2x3

3

]|20 = 1

orC = 3/8

Hence,

b P{X > 1} =∫ ∞1 f (x)dx = 3

8

∫ 21 (4x − 2x2)dx = 1

2


Example 1b

The amount of time in hours that a computer functions beforebreaking down is a continuous r.v. with pdf given by

f (x) =

{λe−x/100 x ≥ 0

0 x < 0

What is the probability that

a a computer will function between 50 and 150 hours beforebreaking down?

b it will function for fewer than 100 hours?

Sol.

a Since

1 = λ∫ ∞

0e−x/100dx =⇒ 1 = −λ(100)e−x/100|∞0 = 100λ

We have

λ =1

100


Example 1b (Conti.)

Hence, the desired probability is

P{50 < X < 150} =∫ 150

50

1

100e−x/100dx = −e−x/100|15050

=e−1/2 − e13/2 ≈ .383

b Similarly,

P{X < 100} =∫ 100

0

1

100e−x/100dx = −e−x/100|1000 = 1− e−1 ≈ 0.632

Approximately 63.2 percent of the time, a computer will fail beforeregistering 100 hours of use.


Example 1c: Question

The lifetime in hours of a certain kind of radio tube is a r.v. havinga pdf given by

f (x) =

{0 x ≤ 100100x2

x > 100

What is the probability that exactly 2 of 5 such tubes in a radio setwill have to be replaced within the first 150 hours of operation?Assume that the events Ei , i = 1, 2, 3, 4, 5, that the ith such tubewill have to be replaced within this time are independent.


Example 1c: Solution

Solution:

P(Ei ) =∫ 150

0f (x)dx

=100∫ 150

100x−2dx

=1

3

We can treat it as the probability of success in a binomial r.v.,B(5, 1/3). Hence, the desired probability is(

5

2

)(

1

3)2(

2

3)3 =

80

243


The Cumulative Distribution Function of ContinuousRandom Variable

The CDF, F, is

F (a) = P{X ≤ a} =∫ a

−∞f (x)dx

and therefore its derivative

F ′(a) = f (a),

the pdf of the continuous r.v.∗ Note that f (a) is not P{X = a} = 0.


Example 1d

If X is continuous with distribution function FX and densityfunction fX , find the density function of Y = 2X .Solution

FY (a) =P{Y ≤ a}=P{2X ≤ a}=P{X ≤ a/2}=FX (a/2)

Differentiation gives

fY (a) =1

2fX (a/2)

∗ recall the chain rule of differentiation


5.2 Expectation and Variance ofContinuous Random Variables


Definition

If X is a continuous r.v. with pdf f (X ), then expectation of X is

E [X ] =∫ ∞

−∞xf (x)dx

Example 2a Find E [X ] when the density function of X is

f (x) =

{2x if 0 ≤ x ≤ 1

0 otherwise

Solution

E [X ] =∫

xf (x)dx

=∫ 1

02x2dx

=2

3


Example 2b

The density function of X is

f (x) =

{1 if 0 ≤ x ≤ 1

0 otherwise

Find E[eX]Solution Let Y = eX . Since Y is increasing in X ,

X ∈ [0, 1] =⇒ Y ∈ [1, e]

FY (y) =P{Y ≤ y}=P{eX ≤ y}=P{X ≤ lny}

=∫ lny

0f (x)dx

= lny


=⇒ fY (y) =1y for 1 ≤ y ≤ e.

Hence,

E[eX]= E [Y ] =

∫ ∞

−∞yfY (y)dy

=∫ e

1dx

=e − 1

We can take the expectation directly on eX . That is,

E[eX]=∫ 1

0ex f (x)dx

=∫ 1

0exdx

=ex |10 = e − 1


Proposition (2.1)

If X is a continuous random variable with pdf f (x), then, for anyreal-valued function g ,

E [g(X )] =∫ ∞

−∞g(x)f (x)dx

Lemma (2.1)

For a nonnegative random variable Y,

E [Y ] =∫ ∞

0P{Y > y}dy


Proof of Lemma 2.1

Proof.

∫ ∞

0p{Y > y}dy =

∫ ∞

0

∫ ∞

yfY (x)dxdy

=∫ ∞

0(∫ x

0dy)fY (x)dx

=∫

xfY (x)dx

=E [Y ]


Proof of Proposition 2.1

Proof.

By Lemma 2.1, for any function g for which g(x) > 0,

E [g(X )] =∫ ∞

0P{g(X ) > y}dy

=∫ ∞

0

∫x :g (x)>y

f (x)dxdy

=∫x :g (x)>0

(∫ g (x)

0dy)f (x)dx

=∫x :g (x)>0

g(x)f (x)dx


Example 2c

A stick of length 1 is split at a point U having pdff (u) = 1, 0 < u < 1.Determine the expected length of the piece that contains the pointp, 0 ≤ p ≤ 1.Solution Let Lp(U) be the length of the substick that contains thepoint p, and note that

Lp(U) =

{1− U U < p

U U > p


By Proposition 2.1

E [Lp(U)] =∫ 1

0Lp(u)du

=∫ p

0(1− u)du +

∫ 1

pudu

=1

2− (1− p)2

2+

1

2− p2

2

=1

2+ p(1− p)

Since p(1− p) is maximized when p = 12 , maxp E [Lp(U)] will

occur at p = 12 , the middle point.


Example 2d: Question

Suppose that if you are s minutes early for an appointment, thenyou incur the cost cs, and if you are s minutes late, then you incurthe cost ks.Suppose also that the travel time from where you presently are tothe location of your appointment is a continuous random variablehaving pdf f .Determine the time at which you should depart if you want tominimize your expected cost.


Example 2d: Solution

Let X denote the travel time.

You leaves t minutes before the appointment.

the cost function Ct(X ) is

Ct(X ) =

{c(t − X ) X ≤ t

k(X − t) X ≥ t

Hence,

E [Ct(X )] =∫ ∞

0Ct(x)f (x)dx

=∫ t

0c(t − x)f (x)dx +

∫ ∞

tk(x − t)f (x)dx

=ct∫ t

0f (x)dx −

∫ t

0xf (x)dx + k

∫ ∞

txf (x)dx − kt

∫ ∞

tf (x)dx


Example 2d: Solution (conti.)

To minimize E [Ct(X )], we differentiate the expected cost:

d

dtE [Ct(X )] =

d

dt[ct∫ t

0f (x)dx −

∫ t

0xf (x)dx

+ k∫ ∞

txf (x)dx − kt

∫ ∞

tf (x)dx ]

=ctf (t) + cF (t)− ctf (t)

− ktf (t) + ktf (t)− k [1− F (t)]

=(k + c)F (t)− k

=0

=⇒ F (t∗) =k

k + c


Corollary (2.1)

If a and b are constants, then

E [aX + b] = aE [X ] + b

The definition of variance of a continuous random variable withmean µ is

Var(X ) = E[(X − µ)2

]or equvalently

Var(X ) = E[X 2]− (E [X ])2

AlsoVar(aX + b) = a2Var(X )


Example 2e

Find Var(X ) for X as given in Example 2a.

f (x) =

{2x if 0 ≤ x ≤ 1

0 otherwise

Solution

Var(X ) =E[X 2]− (E [X ])2

=∫ 1

0x2(2x)dx − (

2

3)2

=x4

2|10 −

4

9

=1

18


5.3 The Uniform Random Variable


pdf of Uniform r.v.

Definition

A random variable is said to be uniformly distributed over theinterval (α, β) if its pdf is given by

f (x) =

{1

β−α α < x < β

0 otherwise

f (x) is a pdf since f (x) ≥ 0 and∫ ∞−∞

1β−αdx =

∫ βα

1β−αdx = 1

for any α < a < b < β

P{a ≤ X ≤ b} =∫ b

a

1

β− αdx =

b− a

α− β

for any subinterval in (a, b) the probability that X in thatsubinterval equals to the length of that subinterval.


The Graph of f(x) and F(x)

The CDF of uniform r.v. is

F (a) =

0 a ≤ αa−αβ−α α < a < β

1 a ≥ β


Example 3a

Let X be uniformly distributed over (α, β). Find (a) E [X ] and (b)Var(X ).Solution (a)

E [X ] =∫ ∞

−∞xf (x)dx

=∫ β

αx

1

β− αdx

=β2 − α2

2(β− α)

=β + α

2

In words, the expected value of a uniform r.v. over an interval isequal to the midpoint of that interval.


Solution (b) To find Var(X ), we first calculate E[X 2]

E[X 2]=∫ β

α

x2

β− αdx

=β3 − α3

3(β− α)

=β2 + αβα2

3

Hence,

Var(X ) =E[X 2]− (E [X ])

=β2 + αβα2

3− (β + α)2

4

=(β− α)2

12

Therefore, the variance of a uniform r.v. over an interval is thesquare of the length of that interval divided by 12.


Example 3b

If X is uniformly distributed over (0, 10), calculate the probabilitythat (a) X < 3, (b) X > 6, and (c) 3 < X < 8.Solution

(a) P{X < 3} =∫ 30

110dx = 3

10

(b) P{X > 6} =∫ 106

110dx = 4

10

(c) P{3 < X < 8} =∫ 83

110dx = 5

10


Example 3c

Buses arrive at a specified stop at 15-minute intervals starting at 7A.M. That is, they arrive at 7, 7:15, 7:30, 7:45, and so on.If a passenger arrives at the stop at a time that is uniformlydistributed between 7 and 7:30, find the probability that he waits.

(a) less than 5 minutes for a bus;(b) more than 10 minutes for a bus.

Solution Let X denote the number of minutes past 7 thatpassenger arrives at the stop.

X ∼ U(0, 30)

Hence

(a)P{10 < X < 15}+ P{25 < X < 30} =∫ 15

10

1

30dx +

∫ 30

25

1

30dx =

1

3

(b)P{0 < X < 5}+ P{15 < X < 20} =∫ 5

0

1

30dx +

∫ 20

15

1

30dx =

1

3


Supplementary Materials

Definition

Let α ≤ β be integers. Suppose that the value of a randomvariable X is equally likely to be each of the integers α, ..., β. Thenwe say that X has the uniform distribution on the integers α, ..., β.Its pmf is given by

f (x) =

{1

β−α+1 for x = α, ...β

0 otherwise

CDF (P{X ≤ a}):

F (a) =a

∑x=α

1

β− α + 1=

a− α + 1

β− α + 1

;


E [X ] =

β

∑x=α

x

β− α + 1=

1

β− α + 1

β

∑x=α

x

=1

β− α + 1

(β + α)(β− α + 1)

2

=β + α

2

Note that

β

∑x=α

x2

β− α + 1=

1

β− α + 1

β

∑x=α

x2

=1

β− α + 1

(β− α + 1)(β− α + 2)(2β− 2α + 3)

6

=(β− α + 2)(2β− 2α + 3)

6


Variance of Uniform R.V.

Thus the variance of X,

Var(X ) =E[X 2]− (E [X ])2

=(β− α + 2)(2β− 2α + 3)

6− (β + α)2

4

=β2 + α2 − 6αβ + 7(β− α) + 6

12

=(β− α)2 + 4(1− α)(1 + β) + 3(β− α) + 2

12

>(β− α)2

12


Example: Lottery

A popular state lottery game requires participants to select athree-digit number (leading 0s allowed). Then three balls, eachwith one digit, are chosen at random from well-mixed bowls.

The sample space here consists of all triples (i1, i2, i3) whereij ∈ {0, 1, ..., 9} for j = 1, 2, 3

Define X (i1, i2, i3) = 100i1 + 10i2 + i3,. For exampleX (0, 1, 5) = 15

P(X = x) = 0.001 for each x ∈ {0, 1, ...999} is a uniformr.v.( of integers)

E [X ] = 499.5

Var(X ) = 9992+412


5.4 Normal Random Variables


Normal Random Variables

Definition

We say that X is a normal random variable with parameter µ andσ2 if the pdf of X is given by

f (x) =1√2πσ

e−(x−µ)2

2σ2

This density is abell-shapedcurve that issymmetric aboutµ.


f (x) is indeed a pdf

WTS 1√2πσ

∫ ∞−∞ e

−(x−µ)2

2σ2 dx = 1

Let y = X−µσ , =⇒ y ∈ (−∞, ∞) and dy = 1

σdx

1√2πσ

∫ ∞

−∞e−(x−µ)2

2σ2 dx =1√2π

∫ ∞

−∞e−y22 dy

Let I =∫ ∞−∞ e

−y22 dy , I 2 =

∫ ∞−∞

∫ ∞−∞ e

−(y2+x2)2 dydx

change of variables to polar coordinates:let x = rcosθ, y = rsinθ and dydx = rdθdr Thus


f (x) is indeed a pdf

I 2 =∫ ∞

0

∫ 2π

0e−r22 rdθdr

=2π∫ ∞

0re−r22 dr

=− 2πe−r22 |0∞

=2π

Therefore1√2π

∫ ∞

−∞e−y22 dy =

1√2π

√2π = 1


Linear Transformation of a Normal Random Variable

if X ∼ N (µ, σ2) and Y = aX + b, thenY ∼ N (aµ + b, a2σ2)

Let FY denote the CDF of Y

FY (x) =P{Y < x}=P{aX + b < x}

=P{X <x − b

a}

=FX (x − b

a)


Linear Transformation

Differentiating both side

fY (x) =1

afX (

x − b

a)

=1√

2πaσe−( x−ba −µ)2

2σ2

=1√

2πaσe−(x−b−aµ)2

2(aσ)2

=1√

2πaσe−(x−(aµ+b))2

2(aσ)2

This shows that Y ∼ N (aµ + b, a2σ2)


Standard Normal Distribution

A random variable Z ∼ N (0, 1) is called a standard normalrandom variable.

By the result of linear transformation, we can deriveX ∼ N (µ, σ2) from Z .

Let X = σZ + µ (or equivalently let Z = X−µσ ).

The process from X to Z is called standardization


Example 4a

Find E [X ] and Var(X ) when X is a normal random variable withparameter µ and σ2.Solution Starting from Z ∼ N (0, 1), and using the lineartransformation X = σZ + µ. We have

E [Z ] =∫ ∞

−∞z

1√2π

e−z22 dz

=1√2π

∫ ∞

−∞ze−

z2

2 dz

=− 1√2π

e−z22 |∞−∞

=0


Thus

Var(Z ) =E[Z 2]

=1√2π

∫−∞

∞z2e−z2

2 dz

=1√2π

(ze−z2

2 |∞−∞ +∫−∞

∞e−z2

2 dz)

=1√2π

∫−∞

∞e−z2

2 dz

=1

Because X = σZ + µ,

E [X ] = σE [Z ] + µ = µ

andVar(X ) = σ2Var(Z ) = σ2


CDF of Standard Normal Random Variable

The CDF of standard normal r.v. is denoted by

Φ(x) =1√2π

∫ x

−∞e−

y2

2 dy

The value of Φ(x) for nonnegative x are given in Table 5.1

For negative of x can be obtained from the relationship

Φ(−x) = 1−Φ(x) −∞ < x < ∞

This can be prove by the fact of symmetry of normal r.v.

P{Z ≤ −x} = P{Z > x} −∞ < x < ∞

The value of X ∼ N (µ, σ2) can also be obtained by

FX (a) = P{X ≤ a} = P(X − µ

σ≤ a− µ

σ) = Φ(

a− µ

σ)



Example 4b

If X is a normal random variable with parameter µ = 3 andσ2 = 9, find(a) P{2 < X < 5}; (b) P{X > 0}; (c) P{|X − 3| > 6}Solution (a)

P{2 < X < 5} =P{2− 3

3<

X − 3

3<

5− 3

3}

=P{−1

3< Z <

2

3}

=Φ(2

3)−Φ(−1

3)

=Φ(2

3)−

[1−Φ(

1

3)

]≈ 0.3779


(b)

P{X > 0} =P{X − 3

3>

0− 3

3}

=P{Z > −1}=1−Φ(−1)

=1− (1−Φ(1)) ≈ 0.8413

(c)

P{|X − 3| > 6} =P{ |X − 3|3

> 2}

=P{X − 3

3> 2}+ P{−(X − 3)

3> 2}

=2P{Z > 2}=2(1−Φ(2))

≈0.0456


Example 4c: grading on the curve

Grade A: X > µ + σGrade B: µ < X < µ + σGrade C: µ− σ < X < µGrade D: µ− 2σ < X < µ− σGrade F: X < µ− 2σ.

There are 16 percent of the class graded A since

P{X > µ + σ} = P{Z > 1} = 1−Φ(1) ≈ 0.1587

Also check the other grades.


Example 4d

X ∼ N (270, 100): the length of gestation..

Assume that the defendant is the father.

The defendant is out of country 290 days and come back at240 days before the birth of the child.

Thus he is the father if the gestation is shorter than 240 orlonger than 290

The probability will be

P{X > 290 or X < 240} =P{X > 290}+ P{X < 240}

=P{X − 270

10> 2}+ P{X − 270

10< −3}

=1−Φ(2) + 1−Φ(3) ≈ 0.0241


Example 4e

The channel noise is a normal r.v. N ∼ N (0, 1).Two types of error

message 1 was incorrectly determined to be 0 if the messageis 1 and 2 +N < 0.5

message 0 was incorrectly determined to be 1 if the messageis 0 and −2 +N ≥ 0.5

Hence

P{error |message is 1} = P{N < −1.5} = 1−Φ(1.5) ≈ 0.0668

and

P{error |message is 0} = P{N ≥ 2.5} = 1−Φ(2.5) ≈ 0.0062


Example 4f

X ∼ N (µ, σ2): the gain from an investment.

The loss is the negation of the the gain, −XFind a value v such that

0.01 = P{−X > v}

standardize X

0.01 =P{−(X − µ)

σ>

v + µ

σ}

=1−Φ(v + µ

σ)

Φ(2.33) = 0.99 which implies

v + µ

σ= 2.33

v = VAR = 2.33σ− µ


5.4.1 The Normal Approximation to the BinomialDistribution

Theorem (DeMoivre-Laplace)

If Sn denotes the number of successes that occur when nindependent trials, each resulting in a success with probability p,are performed, then, for any a < b,

P{a ≤ Sn − np√np(1− p)

≤ b} → Φ(b)−Φ(a)

as n→ ∞.

When n is large, a binomial random variable, B(n, p), willapproximately a normal random variable, N (np, np(1− p)).

Can standardize it into Z ∼ N (0, 1) by Z = x−np√np(1−p)

.

This is a special case of Central Limit Theorem.


Example 4g

Let X be the number of times that a fair coin that is flipped 40times lands on heads. Find the probability that X = 20. Use thenormal approximation and then compare it with the exact solution.Solution Note that B(n, p) is discrete and N (µ, σ2) is continuous,we shall write P{X = i} as P{i − 1/2 < X < i + 1/2} .So, while approximate X = 20, we have

P{X = 20} =P{19.5 < X < 20.5}

=P{19.5− 20√10

<X − 20√

10<

20.5− 20√10

}

≈Φ(0.16)−Φ(−0.16) ≈ 0.1272

The exact result is

P{X = 20} =(

40

20

)(

1

2)40 ≈ 0.1254


Review and Questions

Documents

Chapter 5 Continuous Random Variables€¦ · 5.1 Introduction 5.2 Expectation and Variance of Continuous Random Variables 5.3 The Uniform Random Variable 5.4 Normal Random Variables