Transcript (1) (x)

§4. Continuous random variables
and their density functions
Definition2.8---P35
Suppose that F(x) is the distribution function of r.v. X，if there exists a
nonnegative function f(x)，(-<x<+)，such that for any x，we have
F ( x)＝P( X  x)＝
x

f (t )dt
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
Properties of f(x)-----P35
(1) f ( x )  0 ;
( 2)

 f ( x ) d x  1;
f ( x)
Note:
S


f ( x)d x  1
1
o
x
(1) and (2) are the sufficient and necessary properties of a
density function
Suppose that the density function of X is specified by
0  x  3,
kx,

x

f ( x)  2  , 3  x  4,
 2
ot her wi es
0,
Try to determine
the value of K.
P36
（3） For any a ，if X～ f(x)， (<x<)，then P{X=a}＝0。
Proof Assume that x  0, then X  a  a  x  X  a
Therefore
0  P{X  a}  P a  x  X  a  F  a   F  a  x 
F  x  is right continuous
x  0  F  a   F  a  x   P{X  a}  0
P{a  X  b} P{a  X  b}  P{a  X  b}  P{a  X  b}.
P35
 4
P{x1  X  x2}  F ( x2 )  F ( x1 ) 
x2
x
f ( x)d x ;
1
f ( x)
S1
1
Proof
 
x1 x2
o
x
P{x1  X  x2 }  P{x1  X  x2 }  F ( x2 )  F ( x1 )
x2
x1
x2


x1
  f ( x) d x   f ( x) d x  
f ( x ) d x.
b
P{a  X  b}  P{a  X  b}  P{a  X  b}  P{a  X  b}  a f ( x)d x.
P35
(5) If x is the continuous points of f(x), then
dF ( x )
 f ( x) i.e.F ( x)  f ( x)
dx
Note:P36---(1)
Example1
Suppose that the density function of X is specified by
1
0  x  3,
 6 x,

x

f ( x)  2  , 3  x  4,
 2
ot her wi es
0,


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)
Example2
Suppose that the distribution
function of X is specified by
x 1
 0

F ( x)  ln x 1  x  e
 1
xe

Try to determine
(1) P{X<2},P{0<X<3},P{2<X<e-0.1}.
(2)Density function f(x)
Several Important continuous r.v.
f (x)
1. Uniformly distribution
 1
if X～f(x)＝ b  a , a  x  b


 0， el se
0
。
。
a
b
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
1
d c
P{c  X  d }＝ f ( x)dx＝
dx＝
c
c ba
ba
d
d
x
Example 2.14-P38
2. Exponential distribution
f (x)
e  x , x  0
If X～ f ( x )＝
 0, x  0
x
0
It is said that X follows an exponential
distribution with parameter >0, the d.f. of
exponential distribution is
1  e  x , x  0
F ( x)＝
 0, x  0
Example Suppose the age of a electronic instrument电子仪器
is X (year), which follows an exponential distribution with
parameter 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
years.
0.5e0.5x
f ( x)  
 0

x0
x  0,
(1)P{X  2}  0.5e0.5xdx e 1  0.37
2
( 2) P{ X  3.5 | X  1.5}
P { X  3. 5, X  1 . 5 }

P { X  1.5}


0.5x
0.5e
dx

3.5

0.5x
0.5e
dx

1.5
e
1
 0.37
3. Normal distribution
The normal distribution are one the most important
distribution in probability theory, which is widely applied
In management, statistics, finance and some other areas.
B
A
Suppose that the distance between A，B is ，the
observed value of  is X, then what is the density
function of X ？

Suppose that the density function of X is specified by

1
X ~ f ( x) 
e
2
 x 
2 2
2
  x  
where  is a constant and >0 ，then, X is said to follows
a normal distribution with parameters  and 2 and
represent it by X～N(, 2).
Two important characteristics of Normal distribution
(1) symmetry
the curve of density function is symmetry
with respect to x= and
f()＝max f(x)＝

1
.
2 
(2)  influences the distribution
 ，the curve tends to be flat，
 ，the curve tends to be sharp，


4.Standard normal distribution
A normal distribution with parameters ＝0 and
2＝1 is said to follow standard normal distribution
and represented by X~N(0, 1)。
the density function of normal distribution is
1
e
2
 ( x) 
x2

2
,  x  .
and the d.f. is given by
( x )  P { X  x }

1
2

x
e

t2

2
dt ,  x  
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
F ( x )  P{ X  x }  (
x

).
1 X~N(-1,22), P{-2.45<X<2.45}=?
2. XN(,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}
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  110 
Answer: （）
1 P{ X  105}   
    0.42   1  0.6628  0.3371
 12

 120  110 
 100  110 
P{100  X  120}   


12
12




   0.83    0.83  2  0.7967  1  0.5934
( 2) Let P{X  x}  0.05
 x  110 
1  
  0.05
 12 
 x  110 

  0.95
 12 
x  110
 1.645
12
x  129.74
Example 2.15,2.16,2.17,2.18-P40-42
Homework:
P50--- Q15，18
P51: 17,19,