Download Chapter 4 Analysis of Continuous Random Variables 4

Chapter 4
Analysis of Continuous Random Variables
4-1 Continuous Random Variables
y
The range of a random variable X includes all values in an interval of
real numbers; that is, t he range of X can be thought of as a
continuum.
The number of possible values of the random variable X is
uncountably infinite.
y
4-2 Probability Distributions and Probability Density
Functions
Definition: For a continuous random variable X, a probability density
function is a function such that
(1) f ( x ) ≥ 0
∞
(2)
∫
f ( x ) dx = 1
−∞
b
(3) P(a ≤ X ≤ b) =
∫
f ( x ) dx = area under f(x) from a to b for any a
a
and b
y A probability density function provides a simple description of the
probabilities associated with a continuous random variable.
y A histogram is an approximation to a probability density function
(See Fig. 4-3).
Figure 4-3 Histogram approximates a probability density function.
1
y Note: f(x) is used to calculate an area that represents the probability
that X assumes a value in [a, b]
=> For a continuous random variable X and any value x
P ( X = x) = 0
If X is a continuous random variable, for any x1 and x2 ,
P( x1 ≤ X ≤ x 2 ) = P( x1 < X ≤ x 2 ) = P( x1 ≤ X < x2 ) = P( x1 < X < x 2 )
Prove:
P( x1 ≤ X ≤ x2 ) = P( x1 < X ≤ x2 ) + P( x1 ) = P( x1 < x ≤ x2 )
Example 4-1 (Electric Current)
● The continuous random variable X: the current measured in a thin
copper wire in milliamperes (mA).
y The range of X is [0, 20 mA]
y The probability density function of X is f(x) = 0.05 for 0 ≤ x ≤ 20
y What is the probability that a current measurement is less than 10
milliamperes?
y The probability density function is shown in Fig.4-4
10
10
10
P( X < 10) = ∫ f ( x)dx = ∫ 0.05dx = 0.05 x = 0.5
0
0
0
Figure 4-4 Probability density function for Example 4-1
Example 4-2 (Hole Diameter)
y Continuous random variable X: the diameter (直徑) of a hole
2
drilled in a sheet metal component.
y The distribution of X can be modeled by a probability density
−20 ( x −12.5)
function f ( x) = 20e
, x ≥ 12.5.
y If a part with a diameter larger than 12.60 millimeters is scrapped
(丟棄), what proportion of parts is scrapped?
y The density function and the requested probability are shown in
Fig. 4-5
∞
P ( X > 12.60) =
∫
∞
f ( x ) dx =
12.6
− 20 ( x −12.5 )
dx = − e −20 ( x −12.5)
∫ 20e
12.6
∞
12.6
= 0.135
Figure 4-5 Probability density function for Example 4-2
y What proportion of parts is between 12.5 and 12.6
millimeters?
12.6
P(12.5 < X < 12.6) =
∫
f ( x)dx = − e − 20( x −12.5)
12.5
= 1 − P ( X > 12.6) = 1 − 0.135
= 0.865
Exercises 4-2: 4-1, 4-3, 4-5, 4-9
3
12.6
12.5
= 0.865
4-3 Cumulative Distribution Function
Definition: The cumulative distribution function (cdf) of a
continuous random variable X is
x
∫ f (u )du
F ( x) = P( X ≤ x) =
(4-3)
−∞
for –∞ < x < ∞
Example 4-3 (Electric Current)
y See Example 4-1
y The cumulative distribution function of the random variable X
consists of three expressions.
y Case 1:
F ( x) = 0 , for x < 0
y Case 2:
F ( x) =
x
0
x
x
−∞
−∞
0
0
∫ f (u)du = ∫ f (u)du + ∫ f (u )du = ∫ f (u)du = 0.05x,
for 0 ≤ x < 20
y Case 3:
x
F ( x) = ∫ f (u )du = 1 , for 20 ≤ x
0
y Therefore,
x<0
⎧ 0
⎪
F ( x) = ⎨0.05 x 0 ≤ x < 20
⎪ 1
20≤ x
⎩
y The plot of F(x) is shown in Fig. 4-6
Figure 4-6 Cumulative distribution function for Example 4-3.
4
Example 4-4 (Hole Diameter)
y See Example 4-2
y Case 1: F ( x) = 0 for x < 12.5
y Case 2:
x
F ( x) =
∫ 20e
−20 ( u −12.5 )
12.5
du = −e −20(u −12.5) x
12.5
= −e −20 ( x−12.5) − [−e −20 (12.5−12.5) ] = 1 − e −20( x−12.5)
y Therefore,
x < 12.5
0
F ( x) = ⎧⎨
−20 ( x −12.5 )
12.5 ≤ x
⎩1 − e
y Figure 4-7 display a graph of F(x)
Figure 4-7 Cumulative distribution function for Example 4-4
„ The probability density function of a continuous random variable can
be determined from the cumulative distribution function by
differentiating:
d
d x
F ( x) =
f (u ) du = f ( x)
dx
dx ∫−∞
¾
f ( x) =
dF ( x )
as long as the derivative exists.
dx
5
Example 4-5 (Reaction Time)
y
The time until a chemical reaction is complete (in miniseconds) is
approximated by the cumulative distribution function
0
x<0
⎧
F ( x) = ⎨
− 0.01 x
0≤ x
⎩1 − e
y
Determine the probability density function of X
dF ( x) ⎧
0
x<0
=⎨
f ( x) =
−0.01 x
≤x
0
0
.
01
e
dx
⎩
y
What proportion of reactions is complete within 200 miniseconds
P ( X < 200) = F ( 200) = 1 − e −2 = 0.8647
Exercises 4-3: 4-11(a)(b), 4-17,
6
4-4 Mean and Variance of a Continuous Random Variable
Definition: Suppose X is a continuous random variable with
probability density function f(x). The mean or expected value of X,
denoted as μ or E(X), is
∞
μ = E( X ) =
∫ x f ( x) dx
(4-4)
−∞
The variance of X, denoted as V(X) or σ 2, is
∞
σ = V ( X ) = ∫ ( x − μ ) f ( x)dx =
2
∞
2
−∞
∫x
2
f ( x)dx − μ 2
−∞
2
The standard deviation of X is σ = σ
Prove:
∞
∞
σ = ∫ ( x − μ ) f ( x) dx = ∫ ( x 2 f ( x) − 2 xμf ( x) + μ 2 f ( x))dx
2
2
−∞
∞
=
∫x
−∞
∞
=
∫x
2
−∞
∞
2
f ( x )dx − 2 μ ∫ xf ( x)dx + μ
∞
2
−∞
f ( x)dx − 2μ 2 + μ 2 =
−∞
∞
∫x
2
∫ f ( x)dx
−∞
f ( x)dx − μ 2
=1
−∞
Example 4-6 (Electric Current)
y See Example 4-1
y f(x) = 0.05 for 0 ≤ x ≤ 20
y The mean of X is
E( X ) =
∞
20
20
−∞
0
0
∫ xf ( x)dx = ∫ xf ( x)dx = ∫ 0.05x dx = 0.05
x 2 20
= 10
2 0
y The variance of X is
20
V ( X ) = ∫ ( x − 10) 2 f ( x)dx =
0
If X is a continuous random variable with probability density function f(x),
∞
E[h( X )] =
∫ h( x) f ( x)dx
−∞
7
(4-5)
Example 4-7
y In Example 4-1, X is the current measured in milliamperes.
y f(x) = 0.05, 0 ≤ x ≤ 20
y What is the expected value of the squared current?
y h(X) = X2
∞
∫x
E[h( X )] =
2
f ( x) dx =
−∞
Example 4-8 (Hole Diameter)
y For the drilling operation in Example 4-2
y The mean of X
∞
E( X ) =
∫
∞
xf ( x)dx =
12.5
∫ x20e
−20 ( x −12.5 )
12.5
y The variance of X is
∞
V (X ) =
∫ ( x − 12.55)
2
f ( x)dx =
12.5
Exercises 4-4: 4-25, 4-29, 4-31.
8
dx =
4-5 Continuous Uniform Distribution
Definition: A continuous random variable X with probability density
function
f ( x) = 1 /(b − a) , a ≤ x ≤ b
(4-6)
is a continuous uniform random variable.
y
The probability density function of a continuous uniform random
variable is shown in Fig. 4-8.
Figure 4-8 Continuous uniform probability density function
y
The mean of the continuous uniform random variable X is
b
b
x
0.5 x 2
E ( X ) = ∫ xf ( x) dx = ∫
dx =
b
−
a
b−a
a
a
b
=
a
( a + b)
2
The variance of X is
b
b
V ( X ) = ∫ ( x − μ ) 2 f ( x) dx = ∫
a
a
a+b 2
))
2
dx
b−a
(x − (
b
=
a+b 3
)
(b − a ) 2
2
=
3(b − a )
12
(x −
a
If X is a continuous uniform random variable over a ≤ x ≤ b,
a+b
(b − a ) 2
2
and σ = V ( X ) =
μ = E( X ) =
2
12
9
(4-7)
Example 4-9 (Uniform Current)
y Let the continuous random variable X denote the current measured
in a thin copper wire in milliamperes.
y The range of X is [0, 20 mA]
y f(x) = 0.05, 0 ≤ x ≤ 20
y What is the probability that a measurement of current is between 5
and 10 milliamperes?
y See Fig. 4-9
Figure 4-9 Probability for Example 4-9.
10
P(5 < X < 10) = ∫ f ( x)dx = ∫ 0.05dx = 0.05 x 10 = 0.5 − 0.25 = 0.25
5
5
10
5
y The cumulative distribution function of a continuous uniform
random variable is obtained by integration.
y If a < x < b,
x
F ( x) = ∫ 1 /(b − a) du = x /(b − a) − a /(b − a) =
a
0
x<a
⎧
⎪
y F ( x) = ⎨( x − a) /(b − a) a ≤ x < b
⎪
1
b≤x
⎩
Exercises 4-5: 4-33, 4-35, 4-37, 4-39
10
x−a
b−a
4-6 Normal Distribution (Gaussian Distribution)
y Undoubtedly, the most widely used model for distribution of a random
variable is a normal distribution.
y A normal distribution is also referred to as Gaussian distribution.
y The value of E(X) = μ determines the center of the probability density
function and the value of V(X) = σ2 determines the width.
y Figure 4-10 illustrates several normal probability density function with
selected values of μ and σ2.
y Each normal distribution has the characteristic symmetric bell shaped
curve, but the centers and dispersions differ.
Figure 4-10 Normal probability density function for selected values of the
parameters μ and σ2.
Definition: A random variable X with probability density function
− ( x−μ )2
1
2
f ( x) =
e 2σ
–∞ < x < ∞
(4-8)
2π σ
is a normal random variable with parameters μ, where –∞ < μ < ∞,
and σ >0. Also,
E ( X ) = μ and V ( X ) = σ 2
(4-9)
2
and the notation N(μ, σ ) is used to denote the distribution. The mean
and variance of X are shown to equal μ and σ2, respectively, at the end
of this Section 5-6.
11
Example 4-10
y The current measurement in a strip of wire follow a normal
distribution with a mean of 10 milliamperes and a variance of 4
(milliamperes)2. What is the probability that a measurement
exceeds 13 milliamperes?
y P( X > 13) = ?
y See Fig. 4-11.
y There is no closed-form expression for the integration of a normal
probability density function
y The integral of a normal probability density function are found
from a table.
y Some useful results concerning a normal distribution are
summarized below and in Fig. 4-12.
P ( μ − σ < X < μ + σ ) = 0.6827
P ( μ − 2σ < X < μ + 2σ ) = 0.9545
P ( μ − 3σ < X < μ + 3σ ) = 0.9973
Figure 4-11 Probability that X >13 for a normal random variable
with μ = 10 and σ2 = 4.
Figure 4-12 Probabilities associated with a normal distribution.
y From the symmetry of f(x), P( X >μ) = P( X <μ) = 0.5
y Because more than 0.9973 of the probability of a normal
distribution is within the interval (x–3σ, x+3σ), 6σ is often referred
to as the width of a normal distribution.
12
Definition: A normal random variable with
μ = 0 and σ2 =1
is called standard normal random variable and is denoted as Z.
The cumulative distribution function of a standard normal random
variable is denoted as
Φ( z ) = P(Z ≤ z )
y Appendix Table III provides cumulative probability values for Φ(z),
for a standard normal random variable.
Example 4-11 (Standard Normal Distribution)
y Z is a standard normal random variable.
y The use of Table II to find P(Z ≤ 1.5) is illustrated in Fig. 4-13.
Figure 4-13 Standard normal probability density function
y P(Z ≤ 1.5) = 0.93319
y P(Z ≤ 1.53) =
y P(Z ≤ 1.525) =
13
Example 4-12
y The following calculations are shown pictorially in Fig. 4-14
(1) P( Z > 1.26) = 1 − P( Z ≤ 1.26) =
(2) P ( Z < −0.86) =
(3) P( Z > −1.37) = P( Z < 1.37) =
(4) P(−1.25 < Z < 0.37) = P( Z < 0.37) − P( Z < −1.25) =
(5) P( Z ≤ −4.6) cannot be found exactly from Appendix Table III
∵ P ( Z ≤ −3.99) = 0.00003 and P ( Z ≤ −4.6) < P ( Z ≤ −3.99)
∴ P ( Z ≤ −4.6) is nearly zero.
(6) Find the value z such that P ( Z > z ) = 0.05 , P ( Z ≤ z ) = 0.95
⇒ The nearest value is 0.95053 ⇒ z = 1.65
(7) Find the value z such that P ( − z < Z < z ) = 0.99
The value for z corresponds to probability of 0.995 in Table III.
The nearest probability in Table II is 0.99506, when z = 2.58.
14
If X is a normal random variable with E(X) =μ and V(X) =σ2 , the
random variable
X −μ
Z=
(4-10)
σ
is a normal random variable with E(Z) = 0 and V(Z) = 1. That is, Z is a
standard normal random variable.
y Creating a new random variable by this transformation is referred to as
standardizing.
Example 4-13 (Normally Distributed Current)
y Suppose the current measurement in a strip of wire are assumed to
follow a normal distribution with a mean of 10 milliamperes and a
variance of 4 (milliamperes)2.
y What is the probability that a measurement will exceed 13
milliamperes? (P( X > 13 ) = ?)
13 − 10
Z = ( X − 10) / 2 ⇒ Z =
= 1 .5
2
P ( X > 13) = P ( Z > 1.5) = 1 − P ( Z ≤ 1.5) =
y Rather than using Fig. 4-15,
( X − 10) (13 − 10)
P ( X > 13) = P (
>
) = P ( Z > 1.5) =
2
2
Figure 4-15 Standardizing a normal random variable.
Note: 1.5 is referred to as the z-value associated with a probability
15
Suppose X is a normal random variable with mean μ and variance σ2,
then
X −μ x−μ
P( X ≤ x) = P(
≤
) = P(Z ≤ z)
(4-11)
σ
σ
where Z is a standard normal random variable, and z =
(x − μ )
σ
is
the z-value obtained by standardizing X.
The probability is obtained by entering Appendix Table III with
z = (x − μ ) /σ .
Example 4-14 (Normally Distributed Current)
● Continuing the previous example, what is the probability that a
current measurement is between 9 and 11 milliamperes?
P( 9 < X < 11) = P( (9－10)/2 < (X－10)/2 < (11－10)/2 )
= P(－0.5 < Z < 0.5)
= P(Z < 0.5) – P(Z < –0.5)
=
● Determine the value for which the probability that a current
measurement is below this value is 0.98.
● By standardizing
P( X < x) = P(( X − 10) / 2 < ( x − 10) / 2)
= P( Z < ( x − 10) / 2) = 0.98
● The nearest probability from Table III
P ( Z < 2.05) = 0.97982
( x − 10) / 2 = 2.05 ⇒ x = 2(2.05) + 10 = 14.1
16
Example 4-15 (Signal Detection)
y Assume that in the detection of a digital signal the background
noise follows a normal distribution with a mean of 0 volt and
standard deviation of 0.45 volt.
y The system assumes a digital 1 has been transmitted when the
voltage exceeds 0.9
y What is the probability of detecting a digital 1 when none was
sent?
y The random variable N denotes the voltage of noise.
N − 0 0. 9 − 0
P ( N > 0. 9) = P (
>
) = P ( Z > 2) =
0.45
0.45
y The probability can be described as the probability of a false
detection.
y Determine symmetric bounds about 0 that include 99% of all noise
readings: P ( − x < N < x ) = 0.99
P (− x < N < x) = P( − x / 0.45 < N / 0.45 < x / 0.45)
y
y
y
y
y
= P (− x / 0.45 < Z < x / 0.45) = 0.99
From Appendix Table III:
P ( −2.58 < Z < 2.58) = 0.99 ⇒ x / 0.45 = 2.58
⇒ x = 2.58 × 0.45 = 1.16
Suppose a digital 1 is represented as a shift in the mean of the noise
distribution to 1.8 volt.
What is the probability that a digital 1 is not detected?
Let the random variable S denote the voltage when a digital 1 is
transmitted.
S − 1 .8 0 .9 − 1 .8
P ( S < 0 .9 ) = P (
<
) = P ( Z < −2 ) =
0.45
0.45
The probability can be interpreted as the probability of a missed
signal.
17
Example 4-16 (Shaft Diameter)
y The diameter of a shaft (光軸) in an optical storage drive is
normally distributed with mean 0.2508 inch and standard deviation
0.0005 inch.
y The specifications on the shaft are 0.2500 ± 0.0015 inch.
y What proportion of shaft conforms to specification?
y Let X denote the shaft diameter in inches.
P (0.2485 < X < 0.2515)
0.2485 − 0.2508
0.2515 − 0.2508
<Z<
)
0.0005
0.0005
= P (−4.6 < Z < 1.4) = 0.91924
y If the process is centered so that the process mean is equal to the
target value of 0.2500
P (0.2485 < X < 0.2515)
= P(
0.2485 − 0.2500
0.2515 − 0.2500
<Z<
)
0.0005
0.0005
= P (−3 < Z < 3) = 0.9973
= P(
Exercises 4-6: 4-41, 4-43, 4-47, 4-53, 4-55, 4-61
18
4.8 Exponential Distribution
y The Poisson distribution defines a random variable to be the number
of flaws along a length of copper wire.
y The distance between flaws is another random variable that is often of
interest.
y The random variable X: the length from any starting point on the wire
until a flaw is detected.
y The random variable N: the number of flaws in x millimeters of wire.
y The distribution of X can be obtained from the distribution of N
y Key concept: the distance to the first flaw exceeds 3 mm if and only
if there are no flaws within a length of 3 mm
y If the mean number of flaws is λ per millimeter.
y N has a Poisson distribution with mean λx
e − λx (λx)0
P( X > x) = P( N = 0) =
= e − λx
0!
y The cumulative distribution function of X.
F ( x ) = P ( X ≤ x ) = 1 − e − λx , x ≥ 0
y By differentiating F(x), the probability density function of X
f ( x) = λe−λx , x ≥ 0
Note: The deviation of the distribution of X depends only on the
assumption that the flaws in the wire follow a Poisson process
Definition: The random variable X that equals the distance between
successive counts of a Poisson process with mean λ > 0 is an
exponential random variable with parameter λ. The probability
density function of X is
f ( x) = λe −λx for 0 ≤ x < ∞
(4-14)
19
If the random variable X has an exponential distribution with
parameter λ,
1
1
μ = E ( X ) = and σ 2 = V ( X ) = 2
(4-15)
λ
λ
Prove:
∞
∞
0
0
μ = E[ x] = ∫ xf ( x)dx = ∫ xλe −λx dx
⎛
⎞
= ⎜ x ( −e − λx ) −
e − λx ⎟ ∞
λ
⎝
⎠0
1
= x ( − e − λ x ) ∞ − e − λx ∞
0
0
λ
1
x
λe − λx
−1
− e − λx
1
0
λ
e − λx
1
( x)′
1
x
⎛
⎞
= ⎜ lim λx − 0 ⎟ − (0 − ) = lim
+
x
λ
λ x →∞ (−e )′ λ
⎝ x →∞ − e
⎠
1
1
1 1
= lim
+ =0+ =
x
λ
x → ∞ − λe
λ
λ λ
∞
1
2
1
1
σ 2 = E[ x 2 ] − μ 2 = ∫ x 2 f ( x ) dx − ( ) 2 = 2 − 2 = 2
0
∫
∞
0
x 2 λe − λx dx = x 2 ( −e − λx )
∞
0
− 2x
λ
1
λ
e − λx
∞
0
λ
λ
+ 2( −
1
λ2
∞
e − λx )
0
2
x
x
2
2
lim
−
+
x → ∞ − e λx
λ x → ∞ eλx λ2
2x
2
2
= lim
+
=
x → ∞ − λ e λx
λ2 λ2
= lim
λ
x2
λe − λ x
− 2x
− e − λx
1 −λx
e
2
0
20
λ
−
1
λ
2
e −λx
Example 4-21 (Computer Usage)
y In a large corporate computer network, user log-ons to the system
can be modeled as a Poisson process with a mean of 25 log-ons per
hour.
y What is the probability that there are no log-ons in an interval of 6
minutes?
y X: the time in hours from the start of the interval until the first
log-on.
y X has an exponential distribution with λ = 25 log-ons per hour.
y We are interested in the probability that X exceeds 6 minutes.
(6 minutes = 0.1 hour)
∞
∞
P ( X > 0.1) = ∫ 25e −25 x dx = −e − 25 x
0 .1
0 .1
= 0 − ( −e − 25 ( 0.1) ) = e − 2.5 = 0.082
y Another point of view – from cumulative distribution function:
P ( X > 0.1) = 1 − P ( X ≤ 0.1) = 1 − F (0.1)
= 1 − (1 − e − 25 ( 0.1) ) = e − 2.5 = 0.082
y What is the probability that the time until the next log-on is
between 2 and 3 minutes? (0.033 and 0.05 hour)
P(0.033 < X < 0.05) = ∫
0.05
25e −25 x dx
0.033
or
P(0.033 < X < 0.05) = F (0.05) − F (0.033)
= (1 − e −25×0.05 ) − (1 − e −25×0.033 )
y Determine the interval of time such that the probability that no
log-on occurs in the interval is 0.90.
P( X > x) = e −25 x = 0.90
⇒ ln e −25 x = ln 0.9
⇒ − 25x = ln 0.9 = −0.1054
⇒ x = 0.00421 hour = 0.25 minute
y The mean time until the next log-on is
μ = 1 / λ = 1 / 25 = 0.04 hour = 2.4 minutes
y The standard deviation of the time until the next log-on is
μ = 1 / λ = 1 / 25 = 0.04 hour = 2.4 minutes
Note: The probability that there are no log-ons in a 6-minute interval is
0.082 regardless of the starting time of the interval
● An even more interesting property of an exponential random variable
is concerned with conditional probabilities.
21
Example 4-22
y X denote the time between detection of a particle with a Geiger
counter
y X has an exponential distribution with λ = 1.4 minutes.
y The probability that we detect a particle within 30 seconds (0.5
minutes) of starting the counter is
−0 .5 / 1 .4
= 0.30
P(X < 0.5 minute) = F(0.5) = 1 − e
y Note: λ’ = 1/1.4 count/minute
y Now, suppose we turn on the Geiger counter and wait 3 minutes
without detecting a particle. What is the probability that a particle
is detected in the next 30 seconds?
y The requested probability can be expressed as the conditional
probability that
P( X < 3.5 | X > 3) = P (3 < X < 3.5) / P( X > 3)
where
P(3 < X < 3.5) = F (3.5) − F (3)
= [1 − e−3.5 / 1.4 ] − [1 − e−3 / 1.4 ] = 0.035
P( X > 3) = 1 − F (3) = e −3 / 1.4 = 0.117
P ( X < 3.5 | X > 3) = 0.035 / 0.117 = 0.30
y After waiting for 3 minutes without a detection, the probability of
a detection in the next 30 seconds is the same as the probability
of a detection in the 30 seconds immediately after starting the
counter.
Lack of Memory Property
For an exponential random variable X,
P( X < t1 + t 2 | X > t1 ) = P( X < t 2 )
A=
C
C+D
Exercises 4-8: 4-77, 4-79, 4-81, 4-83, 4-85, 4-91
22
(4-16)
4.9 Erlang and Gamma Distributions
4.9.1 Erlang Distribution
Recall:
‹ The Poisson random variable is defined to be the number of
counts in a length of wire.
‹ The exponential random variable describes the length until the
first count is obtained.
y The random variable that equals the interval length until r counts
occur in a Poisson process has an Erlang random variable.
Example 4-23 (Processor Failure)
y The failures of CPU of large computer system are often modeled as
a Poisson process.
y The mean number of failures per hour is 0.0001.
y X: the time until four failures occur in a system.
y Determine the probability that X exceeds 40,000 hours.
y N: the number of failures in 40,000 hours of operation.
y The time until four failures occur exceeds 40,000 hours if and only
if the number of failures in 40,000 hours is three or less.
P( X > 40,000) = P( N ≤ 3)
1
y The assumption that the failures follow a Poisson process implies
that N has a Poisson distribution with
E ( N ) = λ = 40,000(0.0001) = 4 failures per 40,000 hours
3
e − λ λk
e −4 4 k
=∑
= 0.433
!
!
k
k
k =0
k =0
3
y P ( X > 40,000) = P ( N ≤ 3) = ∑
y X: the time until the r-th event in a Poisson process, then
r −1 − λx
e (λ x ) k
P( X > x) = P( N ≤ r − 1) = ∑
k!
k =0
dF ( x) d
d ⎡ r −1 e − λx (λx) k ⎤ λr x r −1e − λx
=
¾ f ( x) = dx = dx [1 − P( X > x)] = dx ⎢1 − ∑
k! ⎥⎦
(r − 1)!
⎣ k =0
3
X
X
X
x
23
r-1
r
X
2
X
1
Note:
F ( x) = P ( X ≤ x) = 1 − P ( X > x)
e − λx (λx) k
P ( X > x) = P ( N ≤ r − 1) = ∑
k!
k =0
r −1 − λx
r −1 − λx k k
e ( λx ) k
e λ x
⇒ F ( x) = 1 − ∑
=1− ∑
k!
k!
k =0
k =0
r −1
⎡ r −1 (−λ )e − λx λk x k + e − λx λk kx k −1 ⎤
dF ( x)
= − ⎢∑
f ( x) =
⎥
dx
k!
⎣ k =0
⎦
k −1
r −1 k
r −1 k +1 k
λ x
λ kx
= e −λx ∑
− e −λx ∑
k!
k!
k =0
k =0
r −1
λk +1 x k
k =0
k!
r −1
λk +1 x k
= e − λx ∑
k =1
r −2
λk x k −1
(k − 1)!
λm+1 x m
−e ∑
k!
m!
m =0
r −1 k +1 k
r − 2 k +1 k
λ x
λ x
= e −λx ∑
− e −λx ∑
k!
k!
k =0
k =0
=e
− λx
∑
r −1
− e − λx ∑
− λx
k =0
=e
− λx
λ( r −1)+1 x r −1
( r − 1)!
=
λr x r −1e − λx
(r − 1)!
Definition: The random variable X that equals the interval length
until r counts occur in a Poisson process with mean λ > 0 has an
Erlang random variable with parameters λ and r. The probability
density function of X is
λr x r −1e − λx
f ( x) =
, for x > 0 and r = 1, 2, ...
(r − 1)!
y An Erlang random variable can be thought of as the continuous
analogy of a negative binomial random variable.
y A negative binomial random variable can be expressed as the sum of r
geometric random variables.
¾ An Erlang random variable can be represented as the sum of r
exponential random variables.
24
If X is an Erlang random variable with parameters λ and r,
μ = E[ X ] = r / λ and σ 2 = V ( X ) = r / λ2
Prove:
∞
μ = E[ X ] = ∫ x
λr x r −1e − λx
(r − 1)!
0
=
λr
(r − 1)! ∫
∞
0
x r e −λx dx
⎡ r 1 − λx ∞
r! −λx ∞ ⎤
r −1 1
− λx ∞
−
+
−
(
)
(
)
x
e
rx
e
L
e
0
0
0 ⎥⎦
( r − 1)! ⎢⎣
λ
λ2
λr +1
λr ⎡ r! −∞ 0 ⎤ + x r
e − λx
=
−
−
e
e
⎥⎦
(r − 1)! ⎢⎣ λr +1
1 −λx
r −1
−
rx
−
e
r
λ
=
λ
1
+ r (r − 1) x r −2
+ 2 e −λx
λr
=
[
+ x r +1
dx
− ( r + 1) x r
+ ( r + 1)( r ) x r −1
M
− ( r + 1)( r )( r − 1) L1
0
]
λ
e − λx
1
− e − λx
λ
1
+ 2 e − λx
λ
M
1
− r +1 e − λx
λ
1
+ r + 2 e − λx
λ
M
M
M
M
− r (r − 1)(r − 2) L1
0
−
1
λ
+
r
e −λx
1
λ
r +1
e −λx
兩邊正負號同步
σ 2 = E[ x 2 ] − μ 2
r r −1 − λx
∞
∞
e
λr
2
2 λ x
E[ X ] = ∫ x
dx =
x r +1e −λx dx
∫
0
(r − 1)!
(r − 1)! 0
∞
λr ⎡ r +1 1 −λx ∞
1
(r + 1)! −λx ∞ ⎤
=
x (− e ) − (r + 1) x r ( 2 e −λx ) + L −
e
⎢
0
0
0 ⎥⎦
λ
(r − 1)! ⎣
λ
λr + 2
(
r + 1)r
r2 + r
−∞
0
[− e + e ] =
=
λ2
λ2
2
r2 + r ⎛ r ⎞
r
σ = E[ X ] − μ = 2 − ⎜ ⎟ = 2
λ
λ
⎝λ⎠
2
2
2
25
4-9-2 Gamma Distribution
y The Erlang distribution is a special case of the gamma distribution.
y If the parameter r of an Erlang random variable is not an integer, but
r > 0, the random variable has a gamma distribution.
y How to define (r – 1)! for r ∈ R?
Definition: The gamma function is
Γ(r ) = ∫ x r −1e − x dx, for r > 0
(4-17)
Note:
„ It can be shown that the integral in the definition of Γ(r) is finite
and Γ(r) = (r – 1) Γ(r – 1)
„ If r is a positive integer, then Γ(r) = (r – 1)!
„ Γ(1) = 0! = 1 and it can be shown that Γ(1/2) = π1/2
„ Γ(r) can be interpreted as a generalization to non-integer values
of r of the term (r – 1)! that is used in the Erlang probability
density function.
Definition: The random variable X with probability density function
λr x r −1e − λx
f ( x) =
, for x > 0
(4-18)
Γ(r )
has a gamma random variable with parametersλ> 0 and r > 0. If r
is an integer, X has an Erlang distribution.
Figure 4-26 Gamma probability density functions for selected values of
λ and r.
26
If X is a gamma random variable with parameters λ and r,
μ = E[ X ] = r / λ and σ 2 = V ( X ) = r / λ2 (4-19)
Example 4-24 (Processor Failure)
y The time to prepare a micro-array slide fro high-throughput
genomics is a Poisson process with a mean of two hours per slide.
y What is the probability that 10 slides require more than 25 hours to
prepare?
y X: the time to prepare 10 slides.
y The assumption of a Poisson process implies that X has a gamma
distribution with λ = 1/2 and r = 10
y The request probability is P(X > 25)
9
e − λx (λ x ) k
e −12.512.5k
=∑
= 0.2014
P ( X > 25) = ∑
k!
k!
k =0
k =0
y What is the mean and standard deviation of the time to prepare 10
slides?
E(X) = r/λ = 10/(1/2) = 20
V(X) = r/λ2 = 10/(1/2)2 = 40
y The slides will be completed by what length of time with
probability equal to 0.95?
y Find x such that P(X ≤ x) = 0.95
9
9
e − λ x (λ x ) k
e −0.5 x (0.5 x) k
=∑
= 0.05
P( X > x) = ∑
!
!
k
k
k =0
k =0
9
y The chi-squared distribution is a special case of the gamma
distribution in which λ = 1/2 and r equals one of the values 1/2, 1, 3/2,
2, …
y The distribution is used extensively in interval estimation and tests of
hypotheses
Exercises 4-9: 4-97, 4-99, 4-101, 4-103
27
y Appendix A Table III
28
29