Transcript 投影片 1

Chap. 4
Continuous Distributions
Examples of Continuous
Random Variable

If we randomly pick up a
real number between 0 and 1,
then we define a continuous
uniform random variable V
with ProbV  t   t , for any real
number 0≤t≤1.
Distribution Function

The distribution function of
a continuous random variable
X is defined as same as that
of a discrete random
variable, i.e.
FX t   Prob( X  t ).
Probability Density Function

The probability density function
（p.d.f.）of a continuous random
variable is defined as
dFX (t )
f X t  
.
dt

For example, the p.d.f. of an
uniform random variable defined
in [0,2] is
1
f X t   for t  [0,2].
2

The p.d.f. of a continuous
random variable with space S
satisfies the following
properties:
a  f X x   0 for all x  S .
b  S f X x dx  1 .
c  The probability of event A is A f X x dx .
Uniform Distribution

Let random variable X
correspond to randomly
selecting a number in [a,b].
Then,
0 if x  a
 x-a

FX ( x)  Prob ( X  x)  
if a  x  b
 b-a
if x  b
1
Uniform Distribution


 1
if a  x  b
dFX ( x) 
f X ( x) 
 b  a
dx
0
otherwise
X has a uniform distribution.
Some Important
Observations

The p.d.f. of a continuous random
variable does not have to be
bounded. For example, the p.d.f.
of a uniform random variable with
space [0,1/m] is

 1 
m if x  0,
f X x   
 m 
0 otherwise


The p.d.f. of a continuous random
variable may not be continuous,
as the above example demonstrates.
Expected Value and Variance

The expected value of a
continuous random variable X
is

EX    xf x dx.


The variance of X is
VarX   


x   X  f x dx.

2
Expected Value of a Function of
a Random Variable

Let X be a continuous random
variable and Y=G(x). Then,

E[Y ]   G( x) f X ( x)dx,

where f X () is the p.d.f. of X.

In the following, We will only
present the proof for the cases,
in which G(.) is a one-to-one
monotonic function.
Expected Value of a Function of
a Random Variable

proof :

E[Y ] 
 yf
Y
( y )dy


dFY ( y )
  G ( x)
dy , where x  G 1 ( y )
dy

dFY ( y ) dFX (G 1 ( y )) dx
dx

  f ( x) 
dy
dx
dy
dy
Therefore,


dFY ( y )
E[Y ]   G ( x)
dy   G ( x) f ( x)dx
dy


Moment-Generating Function
and Characteristic Function

The moment-generating function of
a continuous random variable X is

M X t   E[e ]   e f x dx.
tX


tx
Note that the moment-generating
function, if it is finite for
-h<t<h for some h>0, completely
determines the distribution. In
other words, if two continuous
random variables have identical
m.g.f., then they have identical
probability distribution function.
Moment-Generating Function
and Characteristic Function

The characteristics function
of X is defined to be the
Fourier transformation of
its p.d.f.

 X w   e

iwx
f x dx
Illustration of the Normal
Distribution


Assume that we want to model
the time taken to drive a
car from the NTU main campus
to the NTU hospital.
According to our experience,
on average, it takes 20
minutes or 1200 seconds.
Illustration of the Normal
Distribution


If we left the NTU main campus
and drove to the NTU hospital now,
then the probability that we
would arrive in 1200.333…seconds
would be 0.
In addition, the probability that
we would arrive in
3600.333…seconds would also be 0.
Illustration of the Normal
Distribution

However, our intuition tells
us that it would be more
likely that we would arrive
within 600 seconds and 1800
seconds than within 3000
seconds and 4200 seconds.
Illustration of the Normal
Distribution
Therefore, the likelyhood
function of this experiment
should be of the following form:
Likelihood

0
600
1200
Time
1800
Illustration of the Normal
Distribution


In fact, the probability density
function models the likelihood
of taking a specific amount of
time to drive from the main
campus to the hospital.
By the p.d.f., the probability
that we would arrive within a
time interval would be

FT ( )  FT ( )   fT (t )dt

Illustration of the Normal
Distribution

In the real world, many
distributions can be well
modeled by the normal
distributions. In other words,
the profile of the p.d.f. of a
normal distribution provides a
good approximation of the exact
p.d.f. of the distribution just
like our example above.
The Standard Normal
Distribution


A normal distribution with
μ=0 and σ=1 is said to be a
standard normal distribution.
The p.d.f. of a standard
normal distribution is
1
e
2

1 2
x
2
.
1
1
 y2
 z2






1
1
2
2
2
c   
e
dy  
e
dz 


2
2



1

2

 

 
e


1 2 2
y z
2

dydz
 is a circularly
Since e 
symmetric function on the Y-Z plane,


 

 
 2e
e


1
y2  z2
2
1 2
y z2
2
1
 2
0
2



dydz   2e
0
1
 2
2
d
 2 .
Therefore,c2=1 and c=1.
Linear Transformation of the
Normal Distribution


Assume that random variable
X has the distribution N  , 2  .
X 
Y

Then,
has the standard

normal distribution.
Proof：
 X 

FY  y   ProbY  y   Prob
 y
 

 Prob X    y   
 y

1
e
2 
1  x  2
 

2  
dx
Linear Transformation of the
Normal Distribution

By substituting
we have
1
FY  y   
y


x
t

,
1  2t 2
e dt.
2
Therefore, Y is N 0,1.
Accordingly, if we want to
compute F X w , we can do
that by the following
procedure.
w 
 X  w 

 w 
FX w  Prob X  w  Prob

  Prob Y 
  FY 
.
 
 
 

  
Expected Value and Variance
of a Normal Distribution

Let X be a N(0,1).
1
2
E[ X ] 


x2

2
xe

1
e
2
dx 
x2

2

 0.

Var[ X ]  E[ X 2 ]  E 2 [ X ]  E[ X 2 ]

1
2


x 2e


Since
1
2
dxe
dx




x2
2
x 2e
x2
2
dx.
e
x2

2

x2
2
dx 
 x 2e

x2
2
,
   x2
x2

1 
2
2
e
dx

xe

2 
 



 1.


 



Therefore, the expected
value and variance of X are
0 and 1, respectively.
The expected value and
variance of a N  ,  2 .
distribution are μ and σ2,
respectively,
Y 
since  is N(0,1),
if Y is N(μ, σ2).
The Table for N（0,1）
The Chi-Square Distribution



Assume that X is N（0,1）.
In statistics, it is common
that we are interested in
Prob X  x  or Prob X  x .
Therefore, we define Z=X2.
The distribution function of
F  z   Prob Z  z   Prob X  z 
Z is
2
Z


1  12 x 2
e
dx, for z  0.
2
z
 z
 2
0
z
1  12 x 2
e
dx 
2
2
 
0
z
e
1
 x2
2
dx
The Chi-Square Distribution

The p.d.f. of Z is
dFZ  z 

dz
2


z
d
0




1 2
x
2
dx
dz
t
2
e

d e

1 2
x
2
0
dt
dx

dt
, where t 
dz
1
1

 z
1
z 2 e 2 , for z  0.
2
z
The Chi-Square Distribution

Z is typically said to have
the chi-square distribution
of 1 degree of freedom and
2
1.

denoted by
Chi-Square Distribution with
High Degree of Freedom


Assume that X1, X2, ……, Xk are k
independent random variables
and each Xi is N(0, 1).
Then, the random variable Z
defined by Z   X is called
a chi-square random variable
with degree of freedom = k
and is denoted by  2 k 
.
k
i 1
2
i
Addition of Chi-Square
Distributions

2
2
2
A
 r    s    r  s 

An


 ki      ki 

 i 1 
i 1
2
2
n

Example of Chi-Square
Distribution with Degree of
Freedom = 2
Assume that a computercontrolled machine is
commanded to drill a hole at
coordinate (10,20). The
machine moves the drill along
the x-axis first followed by
the y-axis.
Example of Chi-Square
Distribution with Degree of
Freedom = 2



According to the calibration
process, the positioning accuracy
of the machine in terms of
millimeters along each axis is
governed by a normal distribution
N(u,0.0625).
The engineer in charge of quality
assurance determines that the
center of the hole must not
deviate from the expected center
by more than 1.0 millimeters.
What is the defect rate of this
task.
The Table of the χ2
Distribution
The distribution Function and
2
p.d.f. of  k  is
F 2 k  t  
t
r2

k 1
2
1

k
 2
2
where w  r 2 .
0
k
1
2
r e
t
k
w
1 
1
2
2
dr  
w
e
dw
k
k 2
0
 2
2
k
t
1 
1
f 2 t  
t 2 e 2 where 0  t  
k
 k 
k 2
  2
2

m    x m 1e  x dx   x
0
m 1  x 
e
0

 m  1 x m 2e  x dx
0

 m  1 x m2e  x dx  m  1(m  1).
0