7-1 chapter 7 special continuous distributions 7.1 uniform random variable def: a continuous rv x is...
TRANSCRIPT
7-1
Chapter 7 Special Continuous Distributions 7.1 Uniform Random Variable
Def: A continuous RV X is said to have a uniform distribution if the pdf of X is
We symbolize the distribution as X~U(a, b).
Example:
If X~U(a, b), evaluate (a) MX(t) (b) X (c)X2
others , 0
),( ,1
)( baxabxf X
7-2
Sol:
12
)(][][][
3
1
3
1
)0(''][(c)2
1
2
1
)0('][(b))(
1
][)((a)
222
2232
2
2
abXEXEXVar
baba
a
b
ab
xdx
abx
MXE
ba
a
b
ab
xdx
abx
MXEabt
eedx
abe
eEtM
b
a
X
b
a
X
b
a
atbttx
tXX
7-3
Ex:r.v. X 之分配為 U(-3,3) 。試問以下
方程式 有實根之機率? Sol: 有實根,則
0)2x(xt4t4 2
2
1
6
1
6
2
13
1
0)2)(1(
0)2(16
0)2(44)4(2
2
有實根之機率=
3x2 or x
2x or x
xx
xx
xx
7-4
Ex: Let X be a uniformly distributed continuous random variable with E [X] =1 and E [X2] =2 , what is the probability density function of X .
Sol: 令 r.v. X 之 p.d.f. 為
23
aabbdx
ab
xXE
12
abdx
ab
xXE
others , 0
bxa , ab
1)x(f
22b
a
22
b
a
7-5
others , 0
31x31, 32
1)x(f
31b,31a
故
聯立可得
7-6
7.2 Normal (Gaussian) Random Variable1.Def: A continuous RV X is said to have a normal
distribution with parameters and (or and 2), where - < < and > 0, if the pdf of X is
We symbolize the distribution as X~N(, 2).
xexf
x
X ,2
1)(
2
2
2
)(
7-7
Note: (1)
(2) fX(x+) = fX(-x+) (3) P[X ] = P[X ] =
X
fX(x)
2
1
7-8
(4) Remark:(a) If RV X~B(n, p) and = np, then X~P() when n and p 0 ( = np is constant).(b) If RV X~B(n, p), then X~N(, 2) when n (p 0), where = np and 2 = np(1- p) (c) If RV X~P(), then X~N(, 2) when .
7-9
Example:
If X~N(, 2), evaluate (a) MX(t) (b) X (c)X2
Sol:
):(
22
1
2
1
2
1
][)( (a)
2
2222
2
2222
2
2
2
1
2
1
2
)(
2
1
2
)(
adxeNote
ee
dxee
dxee
eEtM
ax
tttt
txtt
xtx
tXX
7-10
2. Proposition:
If X~N(0, 2), then
Proof:
2
222
1
)0(''][ (c))0('][
2
1]ln[
)](ln[)( (b)22
kXVarkXE
tte
tMtktt
X
!2
)!2(][
22
n
nXE
n
nn
7-11
!2
)!2(][
!2)!2(
][,2 let (2), (1) Compare
!
][]
!
)([][)((2)
!2!
)2
1(
)((1)
22
22
00
2
0
2
0
22
2
1 22
n
nXE
nn
XE nk
tk
XE
k
tXEeEtM
tnn
tetM
n
nn
n
nn
k
kk
k
ktX
X
n
nn
n
n
n
t
X
7-12
Ex: Let X have the normal density
Find the mean and variance of
X2
Sol:
),0(N 2
etM
NX
XEXE
XEXEXVar
XEXE
XEXEXVar
XE
t
X
,0~
)2(
)1(
0
22
2
1
2
224
22222
2222
222
7-13
4
224
2242
4
0t4
X4
2
)(3
XEXE XVar
3
tMXE
7-14
Example:
Let X(t) be a signal defined by X(t) = Acos(wt + ), where A and are independent random variables, and w is a positive constant. Suppose that A is a zero-mean gaussian random variable with unity variance and is uniformly distributed on [0, 2]. Calculate the mean and variance of X(t).
Sol: RV A~N(0, 1), RV ~U(0, 2)
(1) E[X(t)] = E[Acos(wt + )]
= E[A]E[cos(wt + )]
= 00 = 0
7-15
(2)
2
1
))]2sin()42(sin(4
1[
2
1
2
)22cos(1
2
1
2
1)(cos1
)]([cos][ )](cos[)]([
)]([)]([)]([
2
0
2
0
2
22
222
22
wtwt
dwt
dwt
wtEAEwtAEtXEtXEtXEtXVar
7-16
3. Thm: RV X~N(, 2)(1) If RV Z = aX + b Z~N(a + b, a22).
(2) If RV Z = Z~N(0, 1).
Proof:
(1)
(2)
X
),(~ RV][
)()(
22
)(2
1)()(
2
1)( 22222
abaNZeee
atMetMtatbaatat
bt
Xbt
Z
)1,0(~),(~ RV
,1
Let
22 NabaNZ
ba
7-17
4. Standardizing
When X~N(, 2). The standardized variable
is . Let Z = ,
Then Z is a standard normal RV.
(E[Z] = 0, Var[Z] = 1.)
X
X
7-18
5. Function (x):
Def: dtexx
t
2
2
2
1)(
Xx
(x)
Standard normal
density function
Note:(a) (-x)=1- (x)
(b) When RV X~N(0, 1), P[X x] = (x)
-x
7-19
Note:(1) If RV Z~N(0, 1),
(2) If RV X~N(, 2),
)()( )()(][
abaFbFbZaP ZZ
)()(
][
][
ab
bXaP
bXaP
7-20
(3) Table of Function:
7-21
Example: A fair coin is drawn 1000 times. Let RV X be the
number that the head appears. Evaluate P[460 X 535].
Sol: Method 1: RV X~B(1000, 0.5)
535
460
10001000
10001000
)2
1(]535460[
)2
1()(
xx
xX
C
C
XP
xf
7-22
Method 2: RV X~B(1000, 0.5)
9807.00057.09864.0
)53.2()21.2()5
104()
10
107(
]250
500535
250
500
250
500460[
]535460[)250,500(~
2502
1
2
11000
5002
11000 where
),(~
2
2
XP
XPNX
NX
7-23
Ex: 某輪胎公司宣稱該公司所製的輪胎有90% 機率在 25000 哩和 35000 哩之間損壞 ,假設其為常態分佈 ,試求和。
Sol:
645.1X
645.1r
9.0rZP
1,0N~ZX
Z
300002
2500035000
由查表得
令
7-24
5.3039645.1
5000
3500025000
645.1645.1
X
X
又
7-25
Ex: 某次選舉有甲乙參加。若已知有 55%民眾支持乙候選人 ,試問在抽樣調查 100人中。至少半數支持甲候選人之機率。
Sol:
1.005ZP
975.4
4550ZP50XP
1,0N~X
Z
4.975p-1np , 45
X(r.v. 45.0p,100n
則
令
不適用卜瓦松分配)使用常態分配來模擬(
支持甲之人數)可得
7-26
0.16
由查表可知)(84.01
1.005-1
1.005ZP-1
7-27
Ex:Suppose that the weight of a person
selected at random from some
population is normally distributed with
parameter and . Suppose also that
and2
1]160[ XP
4
1]140[ XP
7-28
(1)Find and , and
(2)find
(3)Of all the people in population weighting
at least 200 pounds, what percentage
will weigh over 220 pounds?
]200[ XP
7-29
Sol:
4
1201
2020
160140160]140[
.
160
160 2
1)1(
,~..
ZP
XP XP
160xZ vr
XPXP
NXvr
令
7-30
%69.80869.0
9131.01
1.36-1
1.36ZP
41.29
160200
41.29
160200)2(
41.2968.0
2068.0
20
75.04
320
function error
XPXP
z
查表得
為其中
7-31
%82.23
0869.0
0207.0200220
0207.0
9793.01
04.21
04.2
41.29
160220
41.29
160220
200
220200220)3(
XXP
ZP
XPXP
XP
XPXXP
7-32
Ex: (a) A random variable X has mean =15 and variance 2=9, and an unknown probability distribution .Find the value of c using the Chebyshev’s theorem such that (b) If the random variable X in (a) has the binomial distribution, compute the value of ( c is the same as in (a), and assume
, where Z is a random variable having the standard normal distribution.)
75.0)|15(| cXP
)|15(| cXP
)()( aaZP
7-32-A
Thm: Chebyshev’s Inequality
Let X be a RV with mean and variance 2.
Then
proof:
So the theorem results.
2
2
] |-XP[|
dxxfxXE X
)()(])[( 222
dxxfxdxxfx XX
)()()()( 22
)22)(
22)((
xxx
xxx
dxxfdxxf XX
)()( 22
]|[|2 XP
7-32-B
Note:
(1)
(2)
x
fX(x)
- +
2
2
] |-XP[|:figure
2
1] |-XP[|
kk
2
11] |-XP[|
kk
7-33
Sol:( 接 p7-32) (a)
625.0
9
925.075.0115
75.0]15[
15115
:theorem sChebyshev'
9XVar
15..
2
22
2
2
2
2
cc
cccXP
cXP
cXPcXP
XP
XEXvr
由
7-34
ondistributi Normal : ,,
small very not is 4.0
large very
0.4p2
75n
91
15
,...,2,1,0 , 1
,~..)(
2NpnB
p
n
pnpXVar
npXE
nxppCxf
pnBXvrb
n
xnxnx
7-35
122
212
22
22 615
ondistributi Normal Standard : 1,0~3
15..
23
152615
ZPXP
NZ
XZvr
XPXP
令
7-36
7.3 Exponential Random Variable7.4 Gamma Distribution 1. Gamma Function:
Def: For > 0, the gamma function () is defined by
Note:
(1) For any > 1, () = ( - 1)(- 1). (2) For any possible integer n, (n) = (n - 1)!
(3)
(請同學自證!)
0
1)( dxex x
.)2
1(
7-37
2. Gamma distribution: Def: A continuous RV X is said to have a gamm
a distribution if the pdf of x is
We symbolize the distribution as X~(,) or X~Gamma(,).Note: The standard gamma distribution has = 1, so the pdf of a standard gamma RV is
.0,0 where
0,)(
1)( 1
xexxf
x
X
0,)(
)(1
xex
xfx
X
7-38
3. Negative Exponential Distribution: Def: A continuous RV X is said to have a negati
ve exponential distribution if the pdf of x is
We symbolize the distribution as X~NE().
Note:
(1) (1, ) = NE().
0 where
0,)(
xexf x
X
1
7-39
(2) Let the failure rate of a system be per unit time. (a) If the system can not used again after it was failed, then the pdf of it’s life is ~NE().
(b) If the system has fault tolerable protection that allows -1 failures, then the pdf of it’s life is ~Gamma(,), where = 1/.
Example:
If X~Gamma(,), evaluate (a) MX(t) (b) X (c)X2
7-40
Sol: (a)
)1(
1
)1)((
1
)/1(])/1[()(
1)(
)/1(,
)/1()
1(Let
)(
1
)(][)(
)(
0
1
0 1
1
)1
(1
0
1
tdueu
t
t
due
t
utM
t
dudx
t
uxxtu
dxex
dxex
eeEtM
u
uX
o
xt
x
txtXX
7-41
(b) Let k(t) = ln[MX(t)] = -ln(1 - t) E[X] = k’(0) = (c) Var[X] = k’’(0) = 2
Note:If RV X~NE(), then
2
1][(c)
1][(b)
1
1)((a)
XVar
XE
ttM X
7-42
Example: Suppose that the number of miles that a car can
run before its battery wears out is exponentially distributes with an average value of 10,000 miles. If a person desires to take a 5000-mile trip, what is the probability that he or she will be able to complete the trip without having to replace the car battery?
Sol: Let RV X be the number of miles that a car can
run before the battery wears out. X~NE(), E[X] = 10000 = 1/10000
7-43
607.0)(]5000[
0,10000
1)(
2
1
5000
10000
edxxfXP
xexf
X
x
X
7-44
4. Thm: If the RV X~NE(), then RV X is memoryless. That is:
P[X > s + t | X > s] = P[X > t], s and t.
Proof:
][
][
][]|[
][
)(
tXPee
e
sXP
tsXPsXtsXP
et
edxetXP
ts
ts
tx
t
x
7-45
5. Thm: Assume X(t) means the number of events occur in [t1, t2] and X(t)~P(t),where t = t2 - t1.(1) If RV T means the time interval of length bet
ween two events occurred. Then T~NE().(2) If RV T means the time interval of length bet
ween + 1 events occurred. Then T~Gamma(, ), where = 1/.
Proof: (1)
0,)(
)(
1]0)([1 ][1][)(
4)-ch4 Process,(Poisson !
)()(
tedt
tdFtf
etXPtTPtTPtF
x
texf
tTT
tT
xt
X
7-46
(2)
),(Gamma~
0,)1)((
)()(
!
)(1
]1)([1 ][1][)(
)(~
1
1
0
T
tet
dt
tdFtf
x
tetXP
tTPtTPtFNET
tT
T
x
xt
T
1 where !
)(1 )(
then, and ),,(Gamma If:1
0
x
xt
T x
tetF
NX~Note
7-47
Proof:
!
)(1)(
!
)(1
)!1(
)!1(])!1(...))(1()[(
)()(
1)(
Let
)1)(()()(
1
0
1
0
21
0
1
0
1
0
k
kt
T
k
kx
x
x
X
xt
x
XX
k
tetF
k
xe
xxe
dexF
t
dtet
dttfxF
7-48
Note: If RV Xi~NE(), i = 1, 2, …, , then Z~Gamma
(, 1/ ) if Z = X1 + X2 + … + X.6. Def: A continuous RV X is said to have a chi-squ
ared distribution with parameter v if the pdf of x is the gamma density with = v/2 and = 2. The pdf of a chi-squared RV is thus
The parameter v is called the number of degrees
of freedom (df) of X. We symbolize the distribution as X~2(v).
0,2)
2(
1)( 2
1)2
(
2
xexv
xfxv
vX
7-49
Ex: 已知卡車通過地磅站符合卜瓦松試驗 ,且 每小時平均有六部卡車通過 ,試問 : (1) 以指數分配來模擬卡車之通過間隔時間 ? (2) 已知 10:15 有一輛卡車通過 ,試問在 10:45 以
後才有一輛車通過之機率 ? (3) 自某一輛卡車通過後開始計算 ,並取其後 第十輛卡車之通過時間為隨機變數 X。求 E[X],Var[X], 及 P[X2] 。 Sol:
tetf tTvr
λ
66:..
6)1(
小時 隔時間表示兩部卡車通過之間
小時卡車
7-50
)(18
5
36
103
5
6
10
!9
6
10
6
6
11,10~ )3(
62
1
)2(
22
2
69106910
3
2
16
小時
(小時)
屬於伽傌分配
由於指數分配無記憶性
XVar
XE
exex
xf
XX
edteTp
xx
t
7-51
9
0
12
7896
610
67
63
7
62
869
2
0
6910
!
121
0
2]1
!1
6...
!7
)6(
!8
)6(
!9
)6([
0
2)]
6
1)(!9()
6
1)(!9(
)6
1)(89(
)6
1)(9()
6
1)([(
!9
!9
62
k
k
x
xx
x
xx10
x
ek
xxxxe
eex
... ex
exex6
dxexXP
7-52
Ex: 已知單位時間顧客前往櫃臺結帳之人數符合 卜瓦松程序 ,且平均每二分鐘一人 ,試問 :
(a) 十分鐘內結帳人數之方差為 ? (b) 第三位結帳的客人和第四位 ,其間隔時間大於三分鐘之機率 ?
(c) 在二十分鐘內有多於 20 人結帳之機率 ? (d) 第三位和第五位結帳客人之時間間隔方差為何 ?
Sol:
)(5!
5
)(10),/(5.0
10..)(
25
分
分分人分鐘內結帳人數表示令
tXVarx
exf
t
Xvra
x
7-53
!
10xf
)(20),/(5.0
..)(2
13
5.0)(
..)(
10
3
2
3
3
2
5.0
x
xe
t
Xvrc
edtedttfTP
etf
)/0.5(
e~T
Tvrb
t
t
t
分分人數表示二十分鐘內結帳人令
分人其中
間間隔表示每位客人結帳的時令
7-54
)(822
42)2(
21
,2,~
)(!20
10...
!2
101011
!
101
!
1020
222
2
2
2
2210
20
0
10
21
10
分
其中
時間間隔表示每兩位客人結帳的令
βαTVar
te
Γ
tetf
λβ α βαGammaT
r.v.Td
e
x
xe
x
xeXP
tt
xx
7-55
Ex:Seven light bulbs are turned on at t=0. The lifetime of any particular bulb is independent of the lifetimes of all other bulbs and is decreased by the probability density function.
Determine the mean and variance of random variable Y, the time until the third failure .
otherwise , 0
0 if , )(
tetf
t
7-56
Sol:
)5()6()(
21,0
..
yU yUy7
yyYP
Yvr
個燈泡壽命大於個燈泡壽命大於個燈泡壽命大於
個燈泡壞掉個或 個時間,只有經過了
止之時間表示直到第三個壞掉為
7-57
yFeyYP
yYPe
e
eCeC
ppCppCp
yYP
edteytPp
y
Yy
y
y
yy
y
yt
y2-y-5
y2-y-5
2y-y-y-y2-5
2y-572
y-671
7y-
2572
671
7
e1535e-211
1e1535e-21
e121e1e7e
e1e1e
)1()1(
][
][
的機率一個燈泡壽命大於
7-58
49
1
18
1
25
1
!2!4
!7
e1e!2!4
!7)1(
0 , e1e!2!4
!7
e105e210-105
e30e35e75e175-105
e30e35
e1535e-215
0
2y-y5-
2y-y5-
y2-y-5
y2-y-y2-y-5
y2-y-5
y2-y-5
dyyYE
y
e
e
e
e
dy
ydFyf
y
y
y
y
YY
7-59
22
222
0
2y-y5-22
343
1
54
1
125
1
!2!4
!7
e1e!2!4
!7)2(
YEYEYVar
dyyYE
7-60
7.5 Beta Distribution 1. Beta function:
Note: For any positive r and s, 2. Def: A RV X is said to have a Beta distribution
with parameters , (both positive) if the pdf of X is
We symbolize the distribution as X~Be(,).
1
0
11 )1(),( dxxxsrB sr
)(
)()(),(
sr
srsrB
10,)1()()(
)(
),(
)1()( 11
11
xxxB
xxxf X
7-61
Example: If X~Be(,), evaluate (a) X (b)X
2.Sol: (a)
(b)
)1(
)()1(
)()(
)(
),1()()(
)(
)1()()(
)(][
1
0
11
B
dxxxxXE
)1()(][ ][ ][
(a)) as same (The )1)((
)1(][
222
2
XEXEXVar
XE
7-62
7.6 Weibull Distribution1. Def: A RV X is said to have a Weibull distributi
on with parameters and if the pdf of X is
We symbolize the distribution as X~W(,). Note:
In general, there are two forms of pdf of Weibull distribution. One is shown above, and the other form is
(5.4.1) 0,)( 1 xexxf xX
(5.4.2) 0,)()(
1 xex
a
bxf
b
a
xb
bX
7-63
Note:
(1) Consider of (5.4.1), if we let = 1/ab and
= b, then it will become the form of
(5.4.2).
(2) W(, 1)~NE().
(3) If X~W(, ), then
0
)1(
!)( (a)
nn
nn
n
ttM
7-64
2. Failure Rate:
Def: If the pdf and cdf of a RV X is fX(x) and FX
(x), then the failure rate R(x) is defined by
2
2)]11([)21(][ (c)
XVar
)(1
)()(
xF
xfxR
X
X
1
)11(][ (b)
XE
7-65
Note:
If RV X means the life time and the pdf of X is fX
(x). Then the probability that X fails in the time interval (x, x + dx) is fX(x)dx.Hence the failure rate of X is
)(1
)(
][1
)(1lim
][
][1lim
]|[lim)(
0
0
0
xF
xf
xXP
xxf
x
xXP
xxXxP
x
x
xXxxXxPxR
X
XX
x
x
x
7-66
Note: (1) If RV X~NE(),
(2)
Proof:
]1[1)(1
)()(
x
x
X
X
e
e
xF
xfxR (constant)
0,)()( 0)(
xexRxfx
dR
X
)](1ln[0
)](1ln[
)(1
)()(
00
xFx
F
dF
fdR
XX
x
X
Xx
7-67
x
x
x
dRXX
dRX
dRX
exRxFdx
dxf
exF
exF
0
0
0
)(
)(
)(
)()()(
1)(
)(1
7-68
Ex: 某電氣用具的壽命的機率密度函數為 t0, 其中 t以年為單位 ,若公司欲使不多於 5% 的用具要求保固服務 (warranty services), 則其保固其應至多設為幾個月 ?
Sol:
ttetf 5.025.0)(
2,2),(~
2)2(
1
25.0)()1(
22
5.0
其中 gammaT
te
tetft
t
7-69
05.01
2
4
122
1
421
05.01
)2(
2
1
2
1
222
2
et
te
tR
et
te
tet
dtte
tTPtF
tF
tftR
ttt
t
t
7-70
(月)即
(年)
3
89
205.0
42
t
tt
t
7-71
7.7 Lognormal Distribution Def: A nonnegative RV X is said to have a logno
rmal distribution if the RV Y = ln(X) has a normal distribution. The resulting pdf of a lognormal RV when ln(X) is normally distributed with parameter and is
0,2
1)(
2
2
2
])[ln(
xe
xxf
x
X
xexf
NXComparex
X ,2
1)(
then,),(~ If :
2
2
2
)(
7-72
Note: If ln(X)~N(, 2), then
)(not )1(][
)(not ][22
2
22
2
eeXVar
eXE