2.4 continuous r.v. suppose that f(x) is the distribution function of r.v. x , if there exists a...
TRANSCRIPT
![Page 1: 2.4 Continuous r.v. Suppose that F(x) is the distribution function of r.v. X , if there exists a nonnegative function f(x) , (-](https://reader035.vdocuments.us/reader035/viewer/2022081519/56649f145503460f94c28c11/html5/thumbnails/1.jpg)
2.4 Continuous r.v.
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Suppose that F(x) is the distribution function of r.v. X , if there exists a no
nnegative function f(x) , (-<x<+) , such that for any x , we have
( ) ( ) ( )x
F x P X x f t dt
= =
Definition2.8---P35
The function f(x) is called the Probability density function ( pdf ) of X, i.e. X ~ f(x) , (-<x<+)
The geometric interpretation of density function
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xo
)(xf
11d)(
xxfS
(1) and (2) are the sufficient and necessary properties of a de
nsity function
Note:
Properties of f(x)-----P35
;0)()1( xf;1d)()2(
xxf
, 0 3,
( ) 2 , 3 4,2
0,
kx x
xf x x
otherwi es
Suppose that the density function of X is specified by
Try to determine the value of K.
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( 3 ) For any a , if X ~ f(x) , (-<x<) , then P{X=a} = 0 。Proof
0 { }P X a P a x X a F a F a x
0,x then X a a x X a Therefore
Assume that
0 { } 0x F a F a x P X a
continuousF x is right
P36
}{ bXaP }{ bXaP }{ bXaP }.{ bXaP
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1 2 2 14 { } ( ) ( )P x X x F x F x ;d)(2
1
xxfx
x
xo
)(xf
1 1S
1x
2x
xxfx
d)(2
Proof
.d)(2
1
xxfx
x1 2 1 2 2 1{ } { } ( ) ( )P x X x P x X x F x F x
xxfx
d)(1
}{ bXaP }{ bXaP }{ bXaP { } ( )d .b
aP a X b f x x
P35
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(5) If x is the continuous points of f(x), then
)()(
xfdx
xdF
P35
. . ( ) ( )i e F x f x
Note:P36---(1)
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1, 0 3,
6
( ) 2 , 3 4,2
0,
x x
xf x x
otherwi es
Example1 Suppose that the density function of X is specified by
Try to determine 1)the value of K
2)the d.f. F(x),
3)P(1<X≤3.5)
4)P(x=3)
5)P(x>3.5 x>3)∣
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Suppose that the distribution function of X is specified by
0 1
( ) ln 1
1
x
F x x x e
x e
Try to determine
(1) P{X<2},P{0<X<3},P{2<X<e-0.1}.
(2)Density function f(x)
Example2
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1. Uniformly distribution
if X ~ f(x) = 1
,
0
a x bb a
, el se
。 。
0 a b
ab
cddxab
dxxfdXcPd
c
d
c
=== 1
)(}{
)x(f
x
It is said that X are uniformly distributed in
interval (a, b) and denote it by X~U(a, b)
For any c, d (a<c<d<b) , we have
Several Important continuous r.v.
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Example 2.14-P38
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2. Exponential distribution
If X ~
0x,0
0x,e)x(f
x
=
It is said that X follows an exponential distribution with parameter >0, the d.f. of exponential distribution is
)x(f
x
0
0,0
0,1)(
x
xexF
x
=
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Example Suppose the age of a electronic instrument 电子仪器 is X (year), which follows an exponential distribution with para
meter 0.5, try to determine
(1)The probability that the age of the instrument is more than 2
years.
(2)If the instrument has already been used for 1 year and a half,
then try to determine the probability that it can be use 2 more ye
ars.
,00
05.0)(
x
xexf
0.5x
37.0)1(
1
2
0.5x edx0.5e2}P{X
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}5.1|5.3{)2( XXP
37.0
1
1.5
0.5x
3.5
0.5x
e
dx0.5e
dx0.5e
}5.1{
}5.1,5.3{
XP
XXP
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The normal distribution are one the most important
distribution in probability theory, which is widely applied
In management, statistics, finance and some other areas.
3. Normal distribution
A
B
Suppose that the distance between A , B is , the observed value of is X, then what is the density function of X ?
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where is a constant and >0 , then, X is said to follow
s a normal distribution with parameters and 2 and rep
resent it by X ~ N(, 2).
Suppose that the density function of X is specified by
2221
~ ( )2
x
X f x e x
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(1) symmetry
the curve of density function is symmetry with respect to x= and
f() = max f(x) = .21
Two important characteristics of Normal distribution
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(2) influences the distribution
, the curve tends to be flat ,
, the curve tends to be sharp ,
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4.Standard normal distribution A normal distribution with parameters = 0 an
d 2 = 1 is said to follow standard normal distributi
on and represented by X~N(0, 1) 。
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.,2
1)( 2
2
xexx
and the d.f. is given by
xdte
xXPx
xt
,
}{)(
221
2
the density function of normal distribution is
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The value of (x) usually is not so easy to compute
directly, so how to use the normal distribution table
is important. The following two rules are essential
for attaining this purpose.
Note : (1) (x) = 1 - ( - x) ;
(2) If X ~ N(, 2) , then
).(}{)(
x
xXPxF
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1 X~N(-1,22), P{-2.45<X<2.45}=?
2. XN(,2), P{-3<X<+3}?
EX 2 tells us the important 3 rules, which are widely
applied in real world. Sometimes we take
P{|X- |≤3} ≈1 and ignore the probability of
{|X- |>3}
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Example The blood pressure of women at age 18 are
normally distributed with N(110,122).Now, choose a
women from the population, then try to determine (1)
P{X<105},P{100<X<120};(2)find the minimal x such that
P{X>x}<0.05
105 110Answer 1 { 105} 0.42 1 0.6628 0.3371
12P X
:()
120 110 100 110{100 120}
12 12
0.83 0.83 2 0.7967 1 0.5934
P X
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{ } 0.05P X x (2) Let
1101 0.05
12
x
1100.95
12
x
1101.645
12
x
129.74x
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Example 2.15,2.16,2.17,2.18-P40-42
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Homework:
P50--- Q15 , 18
P51: 17,19,