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 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
5.1 Introduction: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 5.5 Exponential Random Variables 5.6 Other Continuous Random Variables 5.7 The Distribution of a Function of a Random Variable
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
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
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
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
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
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
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
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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.
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
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.
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
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
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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.
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
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
5.1 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
5.2 Expectation and Variance ofContinuous 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
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
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
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
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
=⇒ 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
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
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
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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 ]
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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
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
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
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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.
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
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.
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
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
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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
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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 )
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
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
5.1 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
5.3 The Uniform Random Variable
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
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.
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
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
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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.
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
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.
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
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
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
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
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
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
;
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
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
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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
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
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
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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
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.
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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
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
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
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
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)
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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)
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
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
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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
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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
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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)
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
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
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
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
(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
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
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.
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
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
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
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
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
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
5.1 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
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.
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
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
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
Review and Questions