Download 2.1 One-dimensional random variable and distribution

Probability
Theory
School of Mathematical Science
and
Computing Technology in CSU
Course groups of Probability and Statistics
Chapter 2
Discrete type random variable
§2.1 One-dimensional random variable and
distribution
§2.2 Multiple random variables and distribution
§2.3 The distribution of the function of random
variable
§2.4 Numerical characteristic of random variable
§2.5 Conditional distribution and conditional
mathematical expectation
In order to better reveal the regularity of random
phenomenon, and to use mathematical tools to describe its
rule，we introduce stochastic variable to describe the
different results of randomized trials.
E.g. The receiving phone numbers of telephone
switchboard over a long period can be described by
variable X
E.g. Two results of throwing a coin can also be
described by a variable
1, Positive upward
X ( )  
0, Reverse upward
§2.1 One-dimensional random variable and
distribution
1、The concept of random variables
E.g. (1) Randomly throw a dice，ω said
all the sample points,
ω:
One point Two point Three point
Four point
X(ω): 1
2
Five point
3
4
Six point
5
6
(2) Someone continuously shoot on the same target, until
really shot and then stop. ω is the number of shots，so
ω
One shooting Two shooting.... n shooting......
X(ω)
1
2
....
n
......
(3) Every 10 minutes a bus is driven from one station ,
passengers at any time get to the station, and ωis the
passenger waiting time,
ω
X(ω)
Waiting time
[0, 10]
1、The concept of discrete random variables
Define If the possible value of the random variable
X is limited or infinite list，say X for discrete
random variables, generally use X, Y , Z ,or
lowercase letters Greece , ,  to show.
Description of probability characteristic of discrete
random variables commonly use its probability
distribution or distribution law,
P( X  xk )  pk , k  1,2,
The characteristic of probability distribution


pk  0, k  1,2,
nonnegative

 pk  1
k 1
standard
Random variables is a
mapping，  R
The characteristic of this mapping as follows：
Field of definitions : 
Randomicity: Random variable X possible value is
more than one，before the test can only predict its
possible values but cannot predict which value is
Probability characteristic: X to take a certain
probability to take someone value or some values
After introducing random variables, using random
variable equation or inequality to express random
events
For example，If use X to depict the receiving
phone numbers of telephone switchboard at
9:00-10:00,
{ X  100} or ( X  100)
—— is “Some days at 9:00 ~ 10:00 the
receiving telephone numbers is more than
100 times ”
Such as，use random variables
1, Positive upward
X ( )  
0, Reverse upward
to describe the possible results of throwing a
coin ,so
( X ( )  1) — Positive upward
also
0, Positive upward
Y ( )  
1, Reverse upward
describe the randomized trial results
Such as，To study the development on children's
growing often require multiple indexes，the height、
weight、head circumference etc.
 = {the development on children's growing  }
X (  ) — height Y (  ) — weight
Z (  ) — head circumference
The random variables may have a certain relationship，or
may have not a certain relationship——be Independent to
each other
note：
The probability distribution of the discrete random
variables can be gotten as following steps:
(1) Determine all possible values of random
variables;
(2) Try to（Using the classical probability）
calculate the probability to take each value.
(3) List the probability distribution of random
variables（or write probability function）.
Example 1 From 1 ~ 10 this 10 numbers
randomly take out of five figures，and make
X：remove the maximum number of five
figures. Try to get the distribution law for
X．
Getting the distribution
Solution: The possible value
rate must explain k
for X is 5,6,7,8,9,10. And
value range!
C
4
k 1
k  5， 6， ， 10 .
PX  k =——
5
C 10
Concretely writing，and get the distribution
law for X：
x
5
6
7
8
9
10
P
1
252
5
252
15
252
35
252
70
252
126
252
Example 2 5 black balls and 3 white balls within a bag,
each one taken does't return every time, until the ball
for black. Remember X as the numbers of taking white
balls, remember Y as the numbers of taking balls ，Get
the probability distribution for X, Y and the probability
for taking at least three times.
Solution (1) The possible value for X is 0,1,2,3,
P(X=0)=5/8, P(X=1)=(3×5)/(8×7)=15/56
P(X=2)=(3×2×5)/(8 ×7 ×6)=5/56,
P(X=3)=1/56, so, The probability distribution of X is
X
0
1
2
3
P
5/8
15/56
5/56
1/56
(2) The possible value for Y is 1,2,3,4,
P(Y=1)=5/8, P(Y=2)=P(X=1)=15/56, similarly
P(Y=3)=P(X=2)=5/56, P(Y=4)=P(X=3)=1/56,
So the probability distribution of Y is
Y
1
2
3
4
P
5/8
15/56
5/56
1/56
(3) P(Y≥3)=P(Y=3)+P(Y=4)=6/56
Third. Common discrete random variables
distribution
(1) 0 – 1 distribution
X = xk
1
0
Pk
p
1-p
0<p<1
Application
All randomized trials only have two possible
results，usually described by 0-1 distribution，such
as whether the product is qualified, the population
gender statistic, whether the system is normal, and
whether the consumption of power is overloaded etc.
Note The distribution law can be written as
P( X  k )  p k (1  p)1k , k  0,1
(2) Discrete even distribution
X
pk
x1
x2
1
n
1
n


xn
1
n
As in “dice” experiment，use { X  i }to
depict｛i appear｝，random variables X
is even distribution.
X
pk
1
2
3
4
5
6
1
6
1
6
1
6
1
6
1
6
1
6
(3) Binomial distribution
B(n, p)
Background：In N heavy Bernoulli trial，
the times of A interested in each test of n
tests—— X is a discrete random variable
If P ( A ) = p , so
Pn ( k )  P ( X  k )  C p (1  p)
k
n
k
nk
,
k  0,1,, n
Say X is the binomial distribution and parameter is
n, p ，remember as
X ~ B(n, p)
0 – 1 distribution is binomial distribution (n = 1)
Example 2 The defective product rate in a large number
of product is 0.1, now one is scoopped out. Try to get
the the probability as following events ：
B={Just 2 defective products from removing 15
products }
C={At least 2 detective products from removing 15
products }
Solution：Because of removing 15 products in a
large number products，so it can be similarly
regarded as a 15 heavy Bernoulli test．
A   Take a product is defective  ,
so P  A  0.1.
So，
P B   C 152  0.12  0.913
P C   1  P C 
0
1
 1  C15
 0.10  0.915  C15
 0.1  0.914
Example 3: A person who don't know
English at all attend a English exam.
Supposed this exam has 5 multiple-choice,
every problem has n choices, only one answer
is correct. Try to: the probability of the event
that he can answer three questions above and
then pass.
Solution: Because he completely
didn't know，So every answer in
every question is the same for him，
and whether he correctly answered
the question also is independent. So,
the process of his answering is the
Bernoulli test.
"
"
amount on correct answers m random variables ~ B(5,1/ n)
 n k
pk  P ( m  k )    p (1  p)n k , ( k  0,1, ,5)
k
1
p  ,so when n  4, the passing probability :
n
 5  1   3   5  1   3   5  1 
p3  p4  p5                  0.10
 3  4   4   4  4   4   5  4 
3
2
4
5
(4) Poisson Distribution  ( ) or P ( )
k


If P( X  k )  e
, k  0,1,2,
k!
  0 is constant，Say the parameters for X is
for Poisson Distribution ,as  ( ) or P ( )

Application
In a certain time intervals：
The receiving numbers of phone from telephone
switchboard；
The numbers of defects on the cloth；
The numbers of customers in shopping malls；
The numbers of patinents in Municipal hospital
emergency；
The numbers of bacteria in a container；
The numbers of traffic accidents in one area;
The numbers of particles from radioactive materials；
The numbers of printing mistakes from every page
in a book；
……
Example 4 Set random variable X obey the
Poisson distribution and the the
parameters isλ，and
PX  1  PX  2
Try to get P  X  4 ．
Solution：
The distribution law for random
variable X is
PX  k  
Because

k
k!
e 
 k  0， 1， 2，  
PX  1  PX  2
Example 5 If the distribution law of the
random variable X is
PX  k   c

k
k  1,2,
k!
.
   0 and is constant  Try to get the constant c
unknown .
Soulation:
The nature of the
distribution law

and
k
 k!
k 1

k
k 0
k!


c
k 1

k
k!
c
1  e  1
1
so c   .
e 1

k 1

k
k!
 1,
(5) Geometric distribution
Set the probability of a machine gun shotting
down a aircraft for p,Unlimited firing，so the
numbers needed of first shotting down plane ,
the distribution of X obey geometric
distribution ,and the parameters is p，written
as X ~ G( p) ,so
P ( X  k )  (1  p)
k 1
p,
k  1,2, 
Easily prove,if the airplanes didn't shotted
down before m times ,so,in the condition，In
order to make the waitting time, before plane
was shotted down, also obey the same
geometric distribution，the distribution is
irrelevant with m，This is the so-called no
memory.
(6) Super geometry distribution
With S products,there are N good
products，and M defective
products( s  M  N )，randomly take n
products and don't put back，( n  N )
write X as the numbers of the good
products taken，get the distribution
law for X.
Now the probability of taking k good
products：
 N  M 
 

k  n  k 

P( X  k ) 
 s
 
 n
k = 0，1，… ，n
so X obey super geometry distribution.as
X ~ H ( M , N , n)
It can be proved that limit distribution of
super geometry distribution was binomial
distribution,in practical application, when
all S，M，N, are large，super geometry
distribution can be approximately depicted
as follow
 N  M 
 

k  n  k 

P( X  k ) 
 s
 
 n
 n  N k M n k
  ( ) ( ) ,
s
k s
(7) Negative binomial distribution
（Pascal distribution) (Self-study)
(8) Zipf distribution (Self-study)
Have a break and then go on