Prof. Xiaorong Gan
Kunming University of Science & Technology
ch0
An Introduction to Set Theory
Sets
Subsets
Operations on sets
Further results
--- Some Basic Definitions and Results
A set is any well-defined collection of distinct objects.
Some special sets are denoted by symbols.
For example,
R denotes the set of all real numbers,
Z denotes the set
of all integers while Z+ denotes the set of all positive
integers.
The objects in a set are called the elements or members
of the set.
We write
to denote the fact that the
fact that the object x is an elements of the set S.
If x is
not a member of S then we write
sets
(i) Listing the elements of a set :
we denote the set by
enclosing all its elements in a list within curly brackets.
.
is the statements that S is the set
whose four elements are 1, 3, 5 and 7.
(ii) The conditional definition of a set :
a set may be
described by specifying some condition which
determines whether or not an object is an element of
the set.
.
means that S is the set of
Describing a set
all integers n which satisfy
(ii) The conditional definition of a set :
a set may be
described by specifying some condition which
determines whether or not an object is an element of
the set.
.
means that S is the set of
We could have described this particular set by listing
the elements.
(iii) The constructive definition of a set:
formula for constructing the elements of the set.
.
here we give a
is the set of integer squares which
.
may be listed as
(iii) The constructive definition of a set:
formula for constructing the elements of the set.
here we give a
is the set of integer squares which
.
A set S is said to be finite if it contains only finitely many
different members;
otherwise it is said to be an infinite
set.
If S is a finite set then the number of different objects
in S is called the order (or cardinality) of S and is
denoted by
D is an element of C.
Two sets are said to be equal if they have precisely the
same elements.
.
C=D Means if and only if (iff)
To show that two sets C and D are equal it is necessary to
prove two things (although this can often be done
together in simple cases):
prove that every element of C
is an element of D and conversely that every element of
Equality pf sets
If we have sets A, B, C with
Given two sets A and B,
we say that A is subset of B,
written ,
when every element of A is also an
element of B.
.
If ,
Thus,
Then
A=B if and only if
If A and B are in fact unequal so that B contains at least
some element not contained in A,
then we say that A is a
proper subset of B and this is denoted by
Subsets
If we have sets A, B, C with
The empty set is the unique set which has no elements at
all and is denoted by the symbol Ø.
.
the set
is clearly empty.
The complement of a subset:
Let A be a subset of a
given set S.
The complement of A in S is the subset
and is denoted by
Note that when all sets being considered are subsets of
a particular set S then S is called the universal set.
Let A and B be two subsets of a universal set S.
(i) The intersection of A and B is the subset of S
Thus
A and B are said to be disjoint if
.
A and B have no elements in common at all.
comprising those elements which belong to both A
and B.
those elements which belong to A or to B or to both.
(ii) The union of A and B is the subset of S comprising
Thus
Operations on sets
those elements which belong to A or to B or to both.
(ii) The union of A and B is the subset of S comprising
Thus
The above two definitions are easily extended to the
case when we are considering the intersection or union
of n>2 sets.
.
let
be n subsets of S.
Then,
the
intersection
comprises those elements in S
that belong to all of
.
.
while the union comprises those elements in S that
belong to at least one of the subsets
.
(iii) The difference of A and B, denoted A – B,
consists of
elements which lie in A but not in B.
Thus
(v) Double complement:
Let A, B and C be subsets of some universal set S ,
then
we have the following identities:
(i) Commutatively:
(ii) Associativity:
(iii) Distributivity:
(iv) Complementation:
Further results
(v) Double complement:
(i) Commutatively:
(ii) Associativity:
(iii) Distributivity:
(iv) Complementation:
(vi) de Morgan’s laws:
The dice is six-sided and
Now this dice is rolled.
Example :Consider a fair dice.
is numbered 1 through to 6.
The dice is six-sided and
Now this dice is rolled.
Example :Consider a fair dice.
is numbered 1 through to 6.
B= number less than 5 and C=even
Let A=odd number ,
number but less than 5 .
Find the following sets:
(a) S=
(b) A=
(c) B=
(d) C=
(e)
(f)
(g)
(h)
(i)
(j)
A
B
A
B
ch1
The Axioms of Probability
Basic Ideas
Relative frequency definition of
probability
The Axioms of Probability
Conditional Probability
Introduction
Random Experiment
It is generally believed that the mathematical
Theory of probability was started by the French
Mathematicians Blaise Pascal(1623-1662) and
Pierre Fermat(1601-1665) when they succeeded
in deriving exact probability for certain gambling
Problems involving dice.
The theory of probability has been developed
steadily since the seventeenth century and has been
widely applied in diverse fields of study.
Today, probability theory is an
important tool in most areas of engineering,
science, and management.
Certainty and uncertainty
How to deal with this uncertainty and its effects?
Probability theory deals with situations in which there
is a degree of randomness or chance in the outcome of
some experiment.
In such a situation it is desirable to know quantitativel
what the chances of such an outcome are of occurring.
Here are a few illustrative examples:
Random Experiment
E2:A coin is tossed three times.
E1:Choose two people at random from a group of five
What is the all possible outcomes that a
people.
particular person is in the selected pair?
What is the all possible
Outcomes that exactly two heads are obtained?
the following experiment:
E3:Choose a positive integer
by means of
Toss a fair coin repeatedly until you get heads and
let n be the number of tosses up to and including the
first toss resulting in heads.
What is the possible outcomes that n is an odd num-
ber?
E4:Select at random some point in the range [0, 4],
the interval between 0 and 4.
What is the possible
outcomes that the point is
(i) 2 ?
大量抛掷硬币
正面出现频率
(ii) inside the range [0, 2] ?
(iii) inside [3, 4] ?
6
possible outcomes is.
The first thing to do before answering any question
involving random outcomes is decide what the set of all
(1) You can repeat the experiment with the same
(2) You can get either one of all possible outcomes;
the experiment.
A property common to all random experiment:
conditions;
(3) You can not tell what outcome is before you do
1. 概率论是研究随机现象规律性的一门数学学科.
2. 随机现象是通过随机试验来研究的.
随机现象的特征:
条件不能完全决定结果.
随
机
试
验
(1) 可以在相同的条件下重复地进行;
(2) 每次试验的可能结果不止一个, 并且能事先明
现.
小 结
确试验的所有可能结果;
(3) 进行一次试验之前不能确定哪一个结果会出
Basic ideas
Sample space
Event
The set of all possible outcomes of a random experiment
is called the sample space of the experiment and is
denoted by S.
The possible outcomes themselves are called sample
points or elements and are denoted by etc.
Running our random experiment should result in
exactly one of these outcomes.
E1:If we denote the five people as A, B, C, D, E and
E3: Let S3=the event that the outcome n is an odd
disregard the order in which they were chosen. Let
S1= the event that person E is chosen.
E2: Let S2=the event that two heads are obtained.
number.
er of elements or an infinite number which can be listed
If the sample space S contains either a finite numb-
in
Then
Then
Then
(E1) and (E2) are finite whereas(E3) is
succession (countably infinite) then we call S a discrete
sample space.
Examples (E1),(E2) and (E3) each have a discrete
sample space;
Countably infinite.
If S consists of an uncountable set of outcomes,
example all possible values in the range [a , b] where
for
a<b,
Example (E4) has a continuous sample space.
then we call S a continuous sample space.
Give out the sample spaces of the following random
Record the sum of all numbers for a roll of three
products which meet the standard.
Answer
Class Exercise
experiments:
dice ;
Record the sum of the products until there are 10
in A then we say that A has occurred.
If the actual outcome of the experiment is contained
Definition
An event is a subset A, of the sample space. .
set of some of the possible outcomes of the experiment.
it is a
life in years of a certain electronic component,
then the
event A that the component fails before the end of the
fifth year is the subset
Example
where t is the
Given the sample space
the null set and denoted by the symbol Ø,
It is conceivable that an event may a subset that inc-
ludes the entire sample space S,
or a subset of S called
Which conta-
ins no elements at all.
of 7 are the odd numbers 1 and 7.
For instance,
if
then B
must be the null set,
the only possible Factors
Definition
The complement of an event A with respect to S is the
subset of all elements of S that are not in A.
We denote
the complement of A by the symbol AC
A
B
S
the deck is not a red but a black card.
Example
Let R be the event that a red card is selected from an
ordinary deck of 52 playing cards,
and let S be the entire
deck.
Example
Consider the sample space
S={book, catalyst, cigarette, precipitate, engineer, river}
Let A ={catalyst, river, book, cigarette}.
Then is the event that the card selected from
Then ={precipitate, engineer}.
Definition
The intersection of two events A and B,
denoted by the
symbol
are common to A and B.
AB
A
B
S
is the event containing all elements that
Example
Let P be the event that a person selected at random
while dining at a popular cafeteria is a taxpayer,
and let
Q be the event that the person is over 65 years of age.
cafeteria who are over 65 years of age.
Then the event
is the set of all taxpayers in the
Example
Let
That is,
M and N have no elements in common and,
therefore,
cannot both occur simultaneously.
then
For certain statistical experiments it is by no means
unusual to define two events,
A and B,
that cannot both
occur simultaneously.
The events A and B are then said
to be mutually exclusive. Stated more formally,
we have
the following definition:
Definition
Two events A and B are said to be disjoint or
mutuall yexclusive if
More generally the events A1, A2, A3,…are said to be
S
A
B
pairwise disjoint or mutually exclusive if
whenever
Example
A cable television company offers programs of eight
Different channels,
three of which are affiliated with
ABC,
two with NBC,
The other two
are an educational channel and the ESPN sports channel.
and one with CBS.
Solution:
Suppose that a person subscribing to this
Service turns on a television set without first selecting
to the NBC network and B the event that it belongs to
the CBS network.
the channel.
Let A be the event that the program belongs
Since a television program cannot belong to more
than one network,
the events A and B
have no programs in common.
ally exclusive.
Therefore,
grams,
the intersection
contains no
pro-
and consequently the events A
and B are mutu-
Definition
The union of the two events A and B,
denoted by the
that belong to A or B or both.
symbol
is the event containing all the elements
Example
S
B
A
Let
then
Example
Let P be event that an employee selected at random
from an oil drilling company smokes cigarettes.
Let Q be the event that the employee selected drinks
alcoholic beverages.
either drink or smoke,
or do both.
Example
Then the event
is the set of all employees who
then
If
Definition
An event A is said to imply an event B if
S
B
A
This means that if A occurs
then B necessarily occurs since
the outcomes of the experiment
is also an element of B.
We denoted by A=B.
Definition
An event A is said equal to event B if
disjoint
complement
S
A
B
S
A
B
A、B disjoint
A、B complement
The difference of disjoint and complement
(ii) Associativity:
(iii) Distributivity:
(vi) de Morgan’s laws:
(i) commutatively:
Let A, B and C be subsets of some universal set S:
Class Exercise
B and C, where A, B and C are three random events:
(1) A happens, but B and C do not;
(5) Three events do not happen at all;
(2) A and B happen but C does not;
(3) All three events happen;
(4) At least one of the three events happens;
Not more than one event happens;
. 1 Try to represent the following events with A,
Solution:
(7) Not more than two of them happen;
(8) At least two of them happen;
(9) At least one of A and B happen but C does not;
(10) Just two of them happen.
Solution:
Conclusion
ple space and random event.
Random experiment
Sample space
subset
Random event
Basic/Simple event
Determined event
Impossible event
Complicated event
Random event
The relationship among random experiment, sam-
theory and Set theory:
Sample space, determined event
Whole set
Impossible event
Empty set
Basic or simple event
Element
Event or random event
Subset
The opposite event of A
Complement set of A
Event A implies event B
A is a subset of B
Event A equals event B
A is B
Sign
Probability theory
Set theory
2. The signs & its relationship between Probability
The difference of event a
And event B
The difference of set A and B
Event A and event B are
Mutually exclusive (disjointed)
Set a and B are disjointed
The union of event A and
event B
The union of set A and B
The intersection of event A
and event B
The intersection of set A and B
(1) A happens, but B and C do not:
(5) Three events do not happen at all:
(2) A and B happen but C does not:
(3) All three events happen:
(4) At least one of the three events happens:
Not more than one event happens:
Solution:
Solution:
(7) Not more than two of them happen:
(8) At least two of them happen:
(9) At least one of A and B happen but C does not:
(10) Just two of them happen:
definition of probability
Frequency definition of probability
The relative frequency
When n(A) is the number of times that A occurs.
Consider performing our experiment a large numb-
er n,
The relative frequency of A is then defined to be
A occurs.
times and counting the number of those times when
Properties:
nts then
are mutually exclusive (disjoint) eve-
if
(3) Finite Additivity
numbers and frequencies of the head.
. Toss a coin 5, 50 and 500 times respectively in
each 7 repeating observation and then look and count the
Exp.
Order
1 2 3 4 5 6 7
2
3
1 5 1 2 4
22
25
21
25
24
18
27
251
249
256
247
251
262
258
As n increases the frequency f turns to stable
More stable here
Most stable here
Unstable here
Experimenter
De Morgan
Buffon
2048
1061
4040
2048
12000
6019
24000
12012
with n increased
PLAY ESC
Galton Experiment
Born: 25 Apr. 1903 in
Tambov, Tambov
province,Russia Died: 20 Oct. 1987 in
Moscow, Russia
Andrey Nikolaevich Kolmogorov
Frequency definition of probability
The Axioms of Probability
Equally Likely Outcomes
then
Definition
A Probability Measure on a sample space S is a function
P which assigns a number P(A) to every event A in S in
such a way that the following three axioms are satisfied:
Axiom 1. P(A)≥0 for every event A.
Axiom 2. P(S) =1
Axiom 3. Countable Additivity
. if A1,A2,…is an
infinite sequence of mutually exclusive (disjoint) events
Example
A coin is tossed twice.
What is the probability that at least one head occurs?
Solution:
Example
A die is loaded in such a way that an even number is
twice as likely to occur as an odd number.
If E is the event that a umber less than 4 occurs on a
single toss of the die,
find P(E).
Solution:
Example
In Example ,
ible by 3 occurs.
let A be the event that an even nu-
mber turns up and let B be the event that a number divis-
Find
and
Solution:
Properties of Probability
1) P(Ø)=0.
2) Finite Additivity
.
if are mutually exclusive (disjoint) events
then
3)
4)
5) P(A)≤1 for all events A.
6) If A and B are arbitrary events then
and hence
where |A| denotes the number of elements in A.
Equally Likely Outcomes
Theorem
Let S be the sample space of an experiment.
If S has n elements which are all equally likely,
then for any event
Model One: Drawing ball
(1) Drawing without replacement
find the probability when the two balls are white.
Problem 1 There are 4 white balls and 2 black balls
in a bag. Draw two balls consecutively out of the bag. Now
Solution:
(2) Drawing with replacement
black and the third ball drawn is red.
Problem 2 There are 4 red balls and 6 black balls in
a bag. Draw consecutively 3 balls with replacement. Find
Find the probability when the first two balls drawn are
Solution:
2o Dice 3 die are cast, find the probability that the numbers you get after you throw the die add up to 4.
Class Exercise
1o Phone number Given that a phone number is made up of 7 digits, while the first digit can not be 0, find the probability that the number 0 appears exactly 3 times.
Model Two: Ball placed in cylinder
(1) No restriction on the capacity of cylinder
Problem 1 Four balls are to be placed inside three cylinders. Assuming that there is no restriction on the number of balls each cylinder can contain, find the probability that cylinders 1 and 2 each contains 2 balls.
Solution:
(2) Each cylinder can only have 1 ball
Problem 2 Four balls are to be placed into ten cylinders. Knowing that each cylinder can only contain one ball, find the probability that for the first 4 cylinders, each one contains a ball.
Solution:
2o Birthday A class has a total of 20 students, who are all born in the same year. Find the probability that out of the 20 students, 10 were born on January 1st and 10 were born on December 31st.
Class Exercise
1o Room arrangement There are three people Mr. A, B and C who are going to be arranged into three rooms. Find the probability that each room has just one people in it.
The sample space for this experiment is
we assign a probability of w to each
Therefore,
point.
Then 4w=1,or w=1/4.
If A represents the event of at least one head occuring
then
Solution:
sample
The sample space is S={1,2,3,4,5,6}.
We assign a probability of w to each odd number and
Solution:
a probability of 2w to each even number.
Since the sum of the probabilities must be 1,
9w=1 or w=1/9.
Hence probabilities of 1/9 and 2/9 are assigned to
each odd and even number,
we have
respectively.
Therefore,
For the events A={2,4,6} and B={3,6},
By assigning a probability of 1/9 to each odd number
and 2/9 to each even number,
we have
and
Solution:
we have
The number of all events in the sample space is
The number of basic events in A is
Solution:
,So
Solution:
First draw
10
second draw
10
Third draw
10
6
Black on the first draw
6种
Black on the second draw
4
Red on the third draw
6
Suppose
Total number of basic/simple events
A contains
events,
So
According to this type of placement, there
ers.
Solution:
are
ways to place the 4 balls into 3 cylind-
Therefore, the proba-
bility that cylinders 1 and
2 will each have 2 balls is:
Solution:
Problem 2: Four balls are to be placed into ten cylinders. Knowing that each cylinder can only contain one ball, find the probability that for the first 4 cylinders, each one contains a ball.
Conditional Probability
Conditional Probability
The multiplication rule
Suppose that we roll a fair six-sided die and note the score obtained. Let A = the event that the outcome is > 3 and B= the event that the outcome is an even number. What is the conditional probability that B occurs given that A has occurred?
Solution:
provided that P(B)>0
Definition
The conditional probability of an event A given than an
event B has already occurred is given by
given
that it departed on time,
Example
The probability that a regularly scheduled flight dep-
time is P(A)=;
the probability that it arrives on
and the probability that it departs
that it has arrived on time.
given
and arrives on time is
on time is P(D)=;
arts
Find the probability that a plane (a) arrives on time
and (b) departed on time
Solution:
In other words, it does not matter which event is referred
Thus the probability that both A and B occur is equal to
If in an experiment the events A and B can both
Theorem
the probability that A occurs multiplied by the probability
that B occurs,
given that A occurs. Since the events
are equivalent,
it follows from Theorem
that we can also write
to as A and which event is referred to as B.
occur, then
Following theorem generalizes these results to n events
Theorem (The multiplication rule)
If
then
proof:
Example
Suppose that we have a fuse box containing 20 fuses,
which 5 are defective,
if 2 fuses are selected at random
and removed from the box in succession without replacing
the first,
what is the probability that both fuses are defe-
ctive?
of
Solution:
Example
One bag contains 4 white balls and 3 black balls,
and a
second bag contains 3 white balls and 5 black balls.
One
ball is drawn from the first bag and placed unseen in
the second bag.
What is the probability that a ball now
drawn from the second bag is black?
Solution:
We denote the conditional probability that A occurs
Solution:
then
give that B has occurred by
time
given that it departed on time is
Solution:
a. The probability that a plane arrives on
b. The probability that a plane departed on time
given that it has arrived on time is
Proof:
fuse is defective and B the event that the second fuse is
Solution:
We shall let A be the event that the first
defective;
The probability of first removing a
then we interpret as the event
that A occ-
urs, and then B occurs after A has
occurred.
defective fuse is
then the probability of
1/4;
fuse from the
remaining 4 is 4/19.
Hence
removing a second defective
Solution:
events
possibilities and
are illustrated in
their probabilities
Let B1, B2, and W1 represent,
respectively,
drawing of a black ball from bag 1,
the
bag 2,
and a white ball from bag 1.
We are interested in
the union of the
mutually exclusive
The various
Figure .
61
a black
ball from
The Law of Total Probability
The Law of Total Probability
(Partition Theorem)
Let A and B be two events in a sample space.
We have
Therefore
, So ,
if
Definition
We say that events B1,B2,…, Bn represent a partition
0f the sample space S if
(i)
for all
(ii)
(iii)
In other words,
when the experiment is performed one
and only one of the events Bi occurs.
Theorem (Partition Theorem)
Let A be some event in S and B1,B2,……, Bn be a
partition of S.
Then
The Law of Total Probability
Proof:
说明 全概率公式的主要用处在于它可以将一个复杂事件的概率计算问题,分解为若干个简单事件的概率计算问题,最后应用概率的可加性求出最终结果.
Example
In a certain assembly plant,
three machines,
B1, B2，and
B3,
make 30%,
45%,
and 25%,
respectively,
of the
products.
It is known from past experience that 2%,
3% and 2% of the products made by each machine,
respectively,
are defective.
Now,
suppose that a finished
product is randomly selected.
What is the probability
that it is defective?
Solution:
Proof:
Consider the Venn diagram of Figure .
The
event A is seen to be the union of the mutually
exclusive events
that is,
Using Properties 2) and then ,
we have
化整为零,各个击破
Solution:
Consider the following events:
A: the product is defective,
B1: the product is made by machine B1,
B2: the product is made by machine B2,
B3: the product is made by machine B3.
Applying the rule of elimination,
we can write
Bayes Theorem
Bayes Theorem
Independent
Theorem (Bayes’Rule)
If the events B1,B2,…, Bn constitute a partition of the
sample space S,
where then for
any event A in S such that
----Bayes’Rule
Proof:
probability that it was made by machine B3?
Example
With reference to Example ,
if a product were
chosen randomly and found to be defective,
what is the
Solution:
小 结
1.条件概率
全概率公式
贝叶斯公式
乘法定理
Now consider an experiment in which 2 cards are
drawn in succession
from an ordinary deck, with
replacement.
The events are defined as
A: the second card is a spade,
B: the first card is an ace.
Since the first card is replaced, our sample
space for both the first and second draws consists of 52
Independent
That is,
P(A|B)=P(A).
When this is true,
B are said to be independent.
the events A and
Consider
Solution:
cards, containing 4 aces and 13 spades. Hence
It show that B has occurred does not change the
Definition
Two events A and B are independent if and only if
P(A|B)=P(A) or P(B|A)=P(B)
A and B are dependent.
Otherwise,
probability of A occurring.
We have
we must have
So
Therefore B is independent of A,
.
knowledge that A
has occurred does not change the probability of B
occurring.
Hence,
independence is a symmetric relation
on the set of all events in a sample space.
We have
we must have
So
otherwise they are called dependent events.
Theorem
We say that A and B are independent events if
and the fire engine will be available.
Example
A small town has one fire engine and one ambulance
available for emergencies.
The probability that the fire
engine is available when needed is ,
probability that the ambulance is available when called
and the
is .
In the event of an injury resulting from a burning
building,
find the probability that both the ambulance
Solution:
Example
An electrical system consists of four components.
The
system works if components A and B work and either
of the components C or D work.
The reliability of each
component is also shown in Figure .
Find the
probability that
(a) the entire system works,
And (b) the
component C does not work,
given that the entire system
works.
Assume that four components work independently.
A
B
C
D
Fig.
Solution:
We can generalize this notion to n events as follows.
Definition
The n events A1, A2,…An are mutually independent if
for every subset
without replacement,
from an ordinary deck of playing cards.
Three cards are drawn in succession,
Example
Find the
probability that the event occurs,
where A1
is the event that the first card is a red ace,
A2 is the event
that the second card is a 10 or a jack,
and A3 is the event
that the third card is greater than 3 but less than 7.
Solution:
Example
A coin is biased so that a head is twice as likely to occur
as a tail.
If the coin is tossed 3 times,
what is the
probability of getting 2 tails and 1 head ?
Solution:
Notes:
(ii) Do not confuse independence with disjointness .
if A and B are disjoint and P(A)>0 and
In fact,
P(B)>0 then they are dependent because the
occurrence of one precludes the occurrence of the
other.
(i) Any event A with P(A)=0 or 1 is independent of
every other event B.
两事件相互独立
两事件互斥
例如
由此可见两事件相互独立，但两事件不互斥.
两事件相互独立与两事件互斥的关系.
请同学们思考
二者之间没
有必然联系
由此可见两事件互斥但不独立.
小 结
Proof:
Thomas Bayes(1702---1761)
Solution:
Using Bayes’ rule to write
and then substituting the probabilities calculated in
example , we have
In view of the fact that a defective product was selected,
this result suggests that it probably was not made by
machine B3.
Solution:
Let A and B represent the respective events
that the fire engine and the ambulance are available.
Then
and the fire engine will be available.
A small town has one fire engine and one ambulance
available for emergencies.
The probability that the fire
engine is available when needed is ,
probability that the ambulance is available when called
and the
is .
In the event of an injury resulting from a burning
building,
find the probability that both the ambulance
Example
b. To calculate the conditional probability in this case,
A
B
C
D
Solution:
a. Clearly the probability that the entire system
works can be calculated as the following
notice that
b. To calculate the conditional probability in this case,
notice that
Solution:
First we define the events
A1: the first card is a red ace,
A2: the second card is a 10 or jack,
A3: the third card is greater than 3 but less than 7.
Now
Solution:
The sample space
Assigning probabilities of w and 2w for getting a tail
and a head,
respectively.
Then we have 3w=1 or w=1/3.
Hence
P(H)=2/3 and P(T) =1/3.
Let A=the event of getting 2 tails and 1 head in the 3
tosses of the coin.
Then, A={TTH, THT, HTT}.
Similarly,
ch2
their distributions
Random variables and
Random variables
Distributions
Random Variables
Example
The classification of .
Example 1 Toss coin:
H
T
Random variables
R
S
Example 2 Test the life in years of light bulbs:
Definition
Let S be the sample space associated with a particular
experiment.
A single-valued function X assigning to
every element a real number,
X(ω) ,
is called a
random variable.
Denoted by X .
In general,
Definition
and x, y, z…represent a real number.
we use X,Y,Z….represent a random variable
Notice that RX is always a set of real numbers.
Definition
The set of all possible values of X is called the range
space of X and is denoted by RX.
Let S be the sample space associated with a particular
experiment. A single-valued function X assigning to
every element a real number, X(ω) , is called a
random variable. Denoted by X .
Definition
The set of all possible values of X is called the range
space of X and is denoted by RX.
Notice that RX is always a set of real numbers.
For above example,
For above example,
随机变量随着试验的结果不同而取不同的值, 由于试验的各个结果的出现具有一定的概率, 因此随机变量的取值也有一定的概率规律.
(2)随机变量的取值具有一定的概率规律
随机变量是一个函数 , 但它与普通的函数有着本质的差别 ,普通函数是定义在实数轴上的,而随机变量是定义在样本空间上的 (样本空间的元素不一定是实数).
2.说明
(1)随机变量与普通的函数不同
随机事件包容在随机变量这个范围更广的概念之内.或者说 : 随机事件是从静态的观点来研究随机现象,而随机变量则是从动态的观点来研究随机现象.
(3)随机变量与随机事件的关系
随机变量的分类
离散型
(1)离散型 随机变量所取的可能值是有限多个或
无限可列个, 叫做离散型随机变量.
观察掷一个骰子出现的点数.
随机变量 X 的可能值是 :
随机变量
连续型
实例1
1, 2, 3, 4, 5, 6.
非离散型
其它
实例2 若随机变量 X 记为 “连续射击, 直至命中时的射击次数”, 则 X 的可能值是:
实例3 设某射手每次射击打中目标的概率是,
现该射手射了30次,则随机变量 X 记为“击中目标
的次数”,
则 X 的所有可能取值为:
实例2 随机变量 X 为“测量某零件尺寸时的测量
误差”.
则 X 的取值范围为 (a, b) .
实例1 随机变量 X 为“灯泡的寿命”.
(2)连续型 随机变量所取的可能值可以连续地充
满某个区间,叫做连续型随机变量.
则 X 的取值范围为
小 结
2. 随机变量的分类: 离散型、连续型.
1. 概率论是从数量上来研究随机现象内在规律性的，因此为了方便有力的研究随机现象， 就需将随机事件数量化，把一些非数量表示的随机事件用数字表示时， 就建立起了随机变量的概念． 因此随机变量是定义在样本空间上的一种特殊的函数．
ch2
their distributions
Random variables and
Random variables
Distributions
Discrete random variables
The definitions
Examples
Definition
With each possible outcome
, we associate a
Number
called the probability of xi ·
The numbers
must satisfy
(i)
(ii)
A random variable X is said to be a discrete random
variable if its range space is either finite or countably
infinite,
.
The numbers
must satisfy
(i)
(ii)
Definition
The function p is called the probability mass function
called the probability distribution of X.
(pmf) and the collection of pairs
Example
Solution:
Let X be a random variable whose values x
are the possible numbers of defective
computers Purchased by the school.
Then x can be any of the numbers 0,1, and 2.
A shipment of 8 similar microcomputers to a retail
outlet contains 3 that are defective.
a random purchase of 2 of these computers,
find the
probability distribution for the number of defectives.
If a school makes
Solution:
Let X be a random variable whose values x
are the possible numbers of defective
computers Purchased by the school.
Then x can be any of the numbers 0,1, and 2.
Thus the probability distribution of X is
x 0 1 2
p(x)
ch2
their distributions
Random variables and
Random variables
Distributions
Some Important Discrete
Discrete Probability Distributions
Probability Distributions
Uniform
We have a finite set of outcomes
which has the same probability of occurring (equally
likely outcomes).
X is said to have a Uniform distribution and we
each of
Write
So
bulb,
Example
When a light bulb is selected at random from a box
that contains a 40-watt bulb,
a 60-watt bulb,
a 75-watt
and a 100-watt bulb.
Find
Solution:
Example
When a die is tossed,
S={1,2,3,4,5,6}.
P(each element of the sample space) = 1/6.
Therefore,
we have a uniform distribution,
with
Bernoulli trial
A Bernoulli trial is an experiment which has two
Let p = P(success), q = P (failure ) (q=1-p).
‘success’ and ‘failure’.
possible outcomes:
The pmf of X is
or
伯努利资料
Binomial
each of which must result in either a ‘success’ with
Consider a sequence of n independent Bernoulli trials
probability of p or a ‘failure’ with probability q=1-p.
Let X= the total number of successes in these n trial
so that
X is said to have a Binomial distribution with parameters
P( the total number of x successes ) =
n and p and we write X~Bin(n, p) or X~b(x;n, p)
X is said to have a Binomial distribution with parameters
n and p and we write X~Bin(n, p) or X~b(x;n, p)
Special case,
when n=1,we have
We write
B(n,p)
b(1, p)
二项分布的图形
The probability that a certain kind of component will survive a given shock test is 3/4. Find the probability that exactly 2 of the next 4 components tested survive.
Example
Example
The probability that a patient recovers from a rare blood
disease is .
this disease,
survive,
If 15 people are known to have contracted
what is the probability that
(a) at least 10
(b) from 3 to 8 survive,
and (c) exactly 5 survive?
Solution:
Solution:
from a shipment.
(b) Suppose that the retailer receives 10 shipments in a
A large chain retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%.
Example
(a) The inspector of the retailer randomly picks 20 items
will be at least one defective item among these 20?
What is the probability that there
shipments containing at least one defective device?
shipment.
month and the inspector randomly tests 20 devices per
What is the probability that there will be 3
Solution：
Poisson
The pmf of a random variable X which has a Poisson
distribution with parameter is given by
and we write
泊松资料
电话呼唤次数
交通事故次数
商场接待的顾客数
地震
火山爆发
特大洪水
泊松分布的图形
单击图形播放/暂停 ESC键退出
二项分布 泊松分布
During a laboratory experiment the average number of radioactive particles passing through a counter in 1 millisecond is 4. What is the probability that 6 articles enter the counter in a given millisecond?
Example
Solution：
handle at most 15 tankers per day.
Example
Ten is the average number of oil tankers arriving each day a certain port city. The facilities at the port can
What is the probability that on a given day tankers
have to be turned away?
Solution：
Solution:
each element of the sample space occurs with
probability 1/4.
Therefore, we have a uniform distribution:
Assuming that the tests are independent and
Solution:
p=3/4 for each of the 4 tests, we obtain
Solution：
Let X = the number of people that survive.
(a)
(b)
Solution：
(a) Denote by X the number of defective
Devices among the 20.
Then this X follows a b (x; 20,).
Hence
(b) Assuming the independence from shipment to
shipment and denoting by Y .
Y=the number of shipments containing at least one
defective.
Then Y~b(y;10,).
28
Therefore,
Solution：
Using the Poisson distribution with x=6 and
, We find from Table 1 that
Solution：
Let X be the number of tankers arriving each
day.
Then,
using Table,
we have
伯努利资料
Jacob Bernoulli
Born: 27 Dec 1654 in Basel, Switzerland Died: 16 Aug 1705 in Basel, Switzerland
泊松资料
Born: 21 June 1781 in Pithiviers, France Died: 25 April 1840 in Sceaux (near Paris), France
Siméon Poisson
ch2
their distributions
Random variables and
Random variables
Distributions
Cumulative Distribution
Functions
Cumulative DistributionFunctions
Definition
The cumulative distribution function (cdf) of the
random variable X is defined to be
and is denoted by F(x).
Properties of F(x):
(i) F is non-decreasing.
(ii)
.
. if
Properties of F(x):
(i) F is non-decreasing. . if
(ii)
(iii) F is right continuous.
(iv) F(x) is defined for all real numbers x.
The cdf of a discrete random variable X is a step
function with jumps at the
.
The cdf of a discrete random variable X is a step
function with jumps at the
.
The pmf of X is
Find: 1)The Cumulative distribution function of X.
2)
X -1 2 3
Example
Solution:
X -1 2 3
Solution:
-1 2 3
2)
-1 0 1 2 3
1
ch2
their distributions
Random variables and
Random variables
Distributions
Continuous Random
Variables
Continuous Random Variables
Definition
X is a continuous random variable if there exists a
The function f is called the probability density function
with the property that for every subset of real
nonnegative function f defined for all real x
numbers B
(pdf) of X.
Properties of the pdf
(i)
Properties of the pdf
(i)
This follows by setting B=
(ii)
1
This follows from
(iii) If we let B=[a, b] then
This follows from
(iv)
This follows from
1
(iv)
This follows from
Notes
(a) If X is continuous then F(x)
is continuous.
Also,
(b) P(a≤X≤b) represents the
between x = a and x = b.
area under the graph of f
f(x)=F’(x) at all point where F is continuous.
f(x)=F’(x)
(c) The meaning of density function:
. The probability that X is in a small interval
is approximately equal to f(x) times the width of the
interval
(d) For any specified value of X, say x0, we have .
(b) P(a≤X≤b) represents the
between x = a and x = b.
area under the graph of f
. The probability that X is in a small interval
is approximately equal to f(x) times the width of the
interval
(d) For any specified value of X, say x0, we have .
Hence,
then
if X is continuous then the probabilities
Example
Let X be a continuous . with. pdf
Find
Solution:
Solution:
ch2
their distributions
Random variables and
Random variables
Distributions
Some Continuous Prob-
ability Distributions
Continuous Probability Distributions
Uniform (or rectangular) Distribution
A uniform random variable X on the interval (a,b) has
probability density function (pdf)
We write X~U (a,b).
Cdf:
Example
Suppose that a large conference room for a certain company can be reserved for no more than 4 hours, However, the use of the conference room is such that
both long and short conferences occur quite often.
In fact,
it can be assumed that length X of a conference has a
uniform distribution on the interval [0, 4].
(a) What is the probability density function?
(b) What is the probability that any given conference
lasts at least 3 hours?
Normal Distribution
The pdf of a Normal random variable, X, with
is given by
parameters
We write
正态概率密度函数的几何特征
Cdf:
正态分布是最常见最重要的一种分布,例如
测量误差, 人的生理特征尺寸如身高、体重等 ;
正常情况下生产的产品尺寸:直径、长度、重量
高度等都近似服从正态分布.
正态分布的应用与背景
高斯资料
In the special case when and the
distribution is called the standard Normal distribution.
.
We write
For a standard Normal random variable X
distribution is called the standard Normal distribution.
.
We write
Example
Given a standard normal distribution，
find the area
under the curve that lies
(a) to the right of z=
(b) between z = and z =.
Figure Areas for
Example
Given a standard normal distribution，
find the value
of k such that
(a) P(Z>k)=，
and (b) P(k<Z<)=.
For a Normal random variable X
A Theorem:
If
Then
Example
Given a random variable X having a normal
distribution with
，Find the
probability that X assumes a value between 45 and 62.
Example
Given that X has a normal distribution with
and
，find the probability that X assumes a
value greater than 362.
Exponential Distribution
The pdf of an Exponential random variable with
Parameter is given by
We write
某些元件或设备的寿命服从指数分布.例如无线电元件的寿命 、电力设备的寿命、动物的寿命等都服从指数分布.
应用与背景
Cdf:
If 5 of these components are installed in different
Suppose that a system contains a certain type of
component whose time in years to failure is given by T.
The random variable T is modeled nicely by the
Example
exponential distribution with mean time to failure
Systems what is the probability that at least 2 are still
functioning at the end of 8 years?
高斯资料
Born: 30 Apr. 1777 in Brunswick, Duchy of Brunswick (now Germany)
Died: 23 Feb. 1855 in Göttingen, Hanover (now Germany)
Carl Friedrich Gauss
ch2
their distributions
Random variables and
Random variables
Distributions
Functions
Functions of Random Variables
of Random Variables
a) when the is discrete
The pmf of Y takes on a given value, say yj, is
Functions of Random Variables
Suppose the has the probability distribution below
Example
xi 0 1 2 3 4 5
P(xi) p(0) p(1) p(2) p(3) p(4) p(5)
Let Y=(X-2)2, Find the pmf of the .
Solution:
0 1 4 9
p(2)
p(1)+p(3)
p(0)+p(4)
p(5)
calculated by taking
The cdf of Y is
b) when the is continuous
Once the cdf FY(y) is obtained,
the pdf fY(y) can be
Example
Suppose X~U(0,1) and let Y=X2,
Find the cdf and pdf of Y.
suppose X is continuous with pdf fX(x). Assesses
(i) y=g(x) defines a one-to–one transformation.
General result:
(ii) the derivative is continuous and non-zero,
Where
is the inverse function of g(x).
Then the pdf of the continuous . Y=g(x) is give
Proof:
Example
Suppose X~N( ), Let Y=aX+b where
Find the pdf and cdf of Y.
（另解）
小 结
1. 离散型随机变量的函数的分布
2. 连续型随机变量的函数的分布
方法1
方法2
其中h(y)=g-1(x)是g(x)的反函数.
Couse exercice
1. Let X be a continuous random variable with probability distribution
Find the probability distribution of random
variable Y=2X-3
2. Given the random variable X with probability
distribution
Find the probability distribution of Y,where
(i) y=g(x) defines a one-to–one transformation.
is continuous and non-zero
(ii) the derivative
then
.
. So that
Proof:
First, suppose g(x) is a monotone increasing function
consider the cdf of Y.
then the pdf of Y is
Second,
.
then
if g(x) is a monotone decreasing function,
(i) y=g(x) defines a one-to–one transformation.
is continuous and non-zero
(ii) the derivative
Second, if g(x) is a monotone decreasing function,
.
then
另解:
ch3
Multionariate
Multionariate Random Variables
Random Variables
Cumulative
distribution function
Cumulative distribution function
Definition
Let S be the sample space associated with a particular
experiment.
X and Y be two . assigning to
a real number vector, (X, Y) , are called
two-dimensional random variable.
Denoted by (X,Y)
a) Joint cdf
Definition
Let X, Y be two random
cumulative distribution
r. v. (X, Y) is defined as
function (cdf) of bivariate
variables.
The joint
(2)
Properties of bivariate cdf F(x,y)
(1) F(x,y) is non-decreasing about x and y. .
(3) F(x,y) is right continuous in each argument, .
(4)
0
Obviously
the marginal cdf can be determined by the joint cdf.
.
b) marginal cdf
Definition
If FX,Y (x,y) is the joint cdf of the X and Y, then
the cdfs FX (x) and FY (y) of X and Y are called
marginal cdfs of X and Y, respectively.
ch3
Multionariate
Multionariate Random Variables
Random Variables
Probability
mass/density functions
Probability mass/density functions
a) Discrete Multivariate
Definition
Suppose the . X and Y are discrete.
is defined by
The joint pmf
The Joint pmf of X and Y
Example
A bag contains 3 black, 2 white and 1 red balls. 2 balls
Find the joint probability mass function
are chosen at random without replacement. Let
Solution:
The Joint cdf of X and Y
Example
The joint pmf of X and Y is given by
Find the cdf of X and Y
Solution:
The marginal pmf of X and Y
Definition
Suppose the . X and Y are discrete.
pmf is defined by
The marginal
The marginal cdf of X and Y
Example (Example )
A bag contains 3 black, 2 white and 1 red balls. 2 balls
Find the joint pmf and marginal pmf of
are chosen at random without replacement. Let
Solution:
b) Continuous Multivariate .
1) Double integrals
How do we calculate the double integral above?
① R is a region bounded by the curves
where
② R is a region bounded by the curves
where
Example
Let R is the triangle bounded by
Define
Find the volume
2) Joint probability density functions
Definition
The nonnegative function is a joint pdf of
the continuous random variable X and Y if
for all
The properties of f (x , y):
(2) The relationship between the joint cdf and pdf is
. the joint pdf is the derivative of
with respect to x and y.
(3) We can calculate the probability that (X,Y) falls in
a rectangle as below
(3) We can calculate the probability that (X,Y) falls in
a rectangle as below
More generally
The marginal cdf of X and Y
The marginal pdf of X and Y
Let X and Y denote the proportion of time out of one
Example
working day that two employees,
performing their assigned tasks.
Y is given by
(3) Find the marginal pdf of X and Y.
A and B,
spend
The joint pdf of X and
小 结
联合分布
边缘分布
Solution:
The joint pmf of X and Y is given by
0 1 2
0
1
2
0 3/15 3/15
2/15 6/15 0
1/15 0 0
A bag contains 3 black, 2 white and 1 red balls. 2 balls
Find the joint probability mass function
are chosen at random without replacement. Let
Solution:
So,the cdf of ( X ,Y ) is given by
0 1 2
0
1
2
0 3/15 3/15
2/15 6/15 0
1/15 0 0
The joint pmf of X and Y is given by
Solution:
A bag contains 3 black, 2 white and 1 red balls. 2 balls
Find the joint pmf and marginal pmf of
are chosen at random without replacement. Let
0 1 2
0
1
2
0 3/15 3/15
2/15 6/15 0
1/15 0 0
The marginal distributions of X and Y are
0 1 2
0
1
2
0
3/15
3/15
0
6/15
2/15
0
0
1/15
The joint pmf and marginal pmf of X and Y is given by
Joint
Marginal
+
+
+
+
+
+
+
+
+
+
+
+
ch3
Multionariate
Multionariate Random Variables
Random Variables
Independence
Independence
a) Definition
Definition
The and Y are said to be independent if
for all
Y, while are the marginal cdf of X and Y.
Discrete case:
is the joint cdf of X and
Discrete case:
Continuous case:
Example
0 1 2
0
1
0 3/15 3/15
2/15 6/15 0
2
1/15
0
0
(look at Example )
Example
(look at Example )
Example
Suppose the joint pdf of (X,Y) is
Show that X and Y are independent.
b) Properties of independence
1) The definition of independence is equivalent to
2) If are independent .
then are independent
ch3
Multionariate
Multionariate Random Variables
Random Variables
Two particular
Uniform distribution
multivariate distribution
Bivariate normal distribution
a) Uniform distribution
S is area of D
Example
Suppose
(1) Find joint pdf and marginal pdf of X and Y
(2) Show that X and Y are not independent.
b) Bivariate Normal distribution
Graph of bivariate Normal distribution.
Properties of bivariate Normal distribution.
Suppose
, then
(1) The marginal distribution are
(2)
Example
Assume
show that (1)
(2)
二维正态分布和其边缘分布的关系
单击图形播放/暂停 ESC键退出
ch3
Multionariate
Multionariate Random Variables
Random Variables
Distribution
Uniform distribution
of special functions
Bivariate normal distribution
(1) Discrete Multivariate
Example
Suppose the joint pmf of (X,Y) is
Find the joint pmf of X+Y and︱X-Y︱
Solution:
结论
a) sum and difference
Let X and Y be continuous . with joint pdf
Let Z=X+Y and V=X-Y , Find the pdfs of Z and V.
(2)Continuous Multivariate .
1. Z=X+Y
So
Which is called the convolution of and
In particular,
if X and Y are independent,
then
2. V=X-Y
Similarly,
we have
Let
Example
Find
(look at )
Example
Suppose X,Y are independent random variables.
X ~ N ( 0, 1 ), Y ~ N ( 0, 1 )
Let Z=X+Y.
Find
In general,
Let are independent random variables.
Then
Let X and Y be jointly distributed continuous
b)Product and quotient
Let X and Y be jointly distributed continuous
b)Product and quotient
with density ,
Find the pdfs of Z and U
and let
1. Z=XY
In particular,
if X and Y are independent then
So the pdf of Z is
2. U=X/Y
In particular,
if X and Y are independent then
So the pdf of U is
Example
Let X and Y be independent each uniformly
distributed over the interval (0, 1).
Let Z=XY and U=X/Y .
Find
THANKS FOR
YOUR
LISTENING!
Solution:
P
P
So, the joint pmf of X+Y and︱X-Y︱are
P
ch4
The Expectation and Variance
The Expectation of
The variance of random variables
Some Important Expectations
Covariance & Correlation
and Variances
Expectations
Expectations of random variables
of random variables
Consider the score of ten student :
50,65,65,70,70,70,80,80,90,100
分赌本问题
A, B 两人赌技相同, 各出
赌金100元,并约定先胜三局者为
胜, 取得全部 200 元.由于出现意
外情况 ,在 A 胜 2 局 B 胜1 局时,
不得不终止赌博, 如果要分赌金,
该如何分配才算公平?
A 胜 2 局 B 胜 1 局
前三局:
后二局:
把已赌过的三局(A 胜2局B 胜1局)与上述结果
相结合,
即 A、B 赌完五局,
A A
A B
B A
B B
A 胜
B 胜
分析
假设继续赌两局,则结果有以下四种情况:
A A
A B
B A
B B
A胜B负
A胜B负
A胜B负
B胜A负
B胜A负
A胜B负
B胜A负
B胜A负
因此, A 能“期望”得到的数目应为
而B 能“期望”得到的数目, 则为
故有, 在赌技相同的情况下,
A, B 最终获胜的
可能性大小之比为
即A 应获得赌金的 而 B 只能获得赌金的
因而A期望所得的赌金即为X的 “期望”值,
等于
X 的可能值与其概率之积的累加.
即为
若设随机变量 X 为:在 A 胜2局B 胜1局的前提
下, 继续赌下去 A 最终所得的赌金.
则X 所取可能值为:
其概率分别为:
Definition
If the is discrete with mass points x1, x2,… and
pmf ,
, then the Expectation of X,
denoted by E(X),
If X is continuous with pdf fX(x),
of X is defined by
Example
A lot containing 7 components is sampled by a quality
inspector;
is defined as
then the Expectation
the lot contains 4 good components and 3
Example
A lot containing 7 components is sampled by a quality
inspector; the lot contains 4 good components and 3
defective components.
inspector.
good components in this sample.
What is his expected gain?
Example
In a gambling game a man is paid $5 if he gets all
Heads or all tails when three coins are tossed,
he will pay out $3 if either one or two heads show.
A sample of 3 is taken by the
Find the expected value of the number of
and
Example
Let X be the random variable that denotes the life in
house of certain electronic device.
density function is
Find the expected life of this type of devise.
Theorem
Let X be a random variable with probability
distribution f(x).
random variable g(X) is
if X is discrete.
if X is continuous.
The mean or expected value of the
In general,
if X is discrete.
Let Z= g(X,Y), then
Theorem
Let X be a random variable with probability
distribution f(x).
random variable g(X) is
if X is discrete.
if X is continuous.
The mean or expected value of the
if X is continuous.
when g ( X,Y )=X, then
especially,
when g ( X,Y )=Y, then
The Properties of expectation
Example
Suppose that the number of cars X that pass through a
car wash between 4:00 . and 5:00 . on any sunny
Friday has the following probability distribution:
x
P(X=x)
4 5 6 7 8 9
Let g(X)=2X-1 represent the amount of money in dollars,
paid to the attendant by the manager.
expected earnings for this particular time period.
Find the attendant’s
Example
Let X be a random variable with density function
Find the expected value of g(X) = 4X+3
小 结
数学期望是一个实数, 而非变量,它是一种加权平均, 与一般的平均值不同,它从本质上体现了随机变量 X 取可能值的真正的平均值.
2. 数学期望的性质
ch4
The Expectation and Variance
The Expectation of
The variance of random variables
Some Important Expectations
Covariance & Correlation
and Variances
The variance
The variance of random variables
of random variables
Example: the life of bulbs.
Batch Ⅰ (X):
700 hours for one half, 1300 hours for another half
950 hours for one half, 1050 hours for another half
Batch Ⅱ (Y):
950
1050
700
1300
Batch Ⅰ (X):
700 hours for one half, 1300 hours for another half
950 hours for one half, 1050 hours for another half
Batch Ⅱ (Y):
Average life: E(X)=E(Y)=1000 h.
We use to measure the average value
of the squared deviation of X from its centre E(X).
Definition
Let X be a . and
denoted by or Var(X),
if X is discrete with the pmf P(xi).
if X is continuous with the pdf fX(x).
is defined by
OR, it is defined by
The variance of X,
方差是一个常用来体现随机变量 X 取值分散程度的量. 如果 V(X) 值大, 表示 X 取值分散程度大, E(X) 的代表性差; 而如果 V(X) 值小, 则表示X 的取值比较集中, 以 E(X) 作为随机变量的代表性好.
方差的意义
Notes:
iii) The positive square root,
i) If Var(X) is not finite,
not exist.
ii) Var(X) is the average value of the squared deviation
of X from its centre .
obviously,
standard deviation of the .
by
we say the variance of X does
So, it is a measure its means.
, is called the
It is usually denoted
Example
Let the random variable X represent the number of
automobiles that are used for official business purposes
on any given workday.
company A is
and for company B is
x 1 2 3
f(x)
x 0 1 2 3 4
f(x)
Show that the variance of the probability distribution
for company B is greater than that of company A.
The probability distribution for
Theorem
The variance of a random variable X is
Using above Theorem,
probability distribution of X.
Example
Let the random variable X represent the number of
defective parts for a machine when 3 parts are sampled
from a production line and tested.
x 0 1 2 3
f(x)
The following is the
calculate
Example
The weekly demand for Pepsi,
from a local chain of efficiency stores,
random variable X having the probability density
Find the variance of X .
in thousands of liters,
is a continuous
If X is discrete ,
If X is continuous
Theorem
Let X be a random with probability distribution f(x).
The variance of the random variable g(X) is
and
Example
Calculate the variance of g(X)=2X+3, where X is a
random variable with probability distribution
x 0 1 2 3
f(x)
Example
Let X be a random variable with density function
Find the variance of the random variable g(X)=4X+3.
The Properties of variance:
(1)
(2)
(3)
ch4
The Expectation and Variance
The Expectation of
The variance of random variables
Some Important Expectations
Covariance & Correlation
and Variances
Some Important
Expectations & Variance
Some Important Expectations &
Variance
1. (0-1)
The pmf of is given by
then
2. Binomial
then
The pmf of which has a Binomial distribution with parameter n, p is given by
3. Poisson
then
The pmf of which has a Poisson distribution
with parameter λ is given by
So
4. Uniform
then
5. Exponential
then
6. Normal Distribution
then
分 布
参数
数学期望
方差
两点分布
二项分布
泊松分布
均匀分布
指数分布
正态分布
ch4
The Expectation and Variance
The Expectation of
The variance of random variables
Some Important Expectations
Covariance & Correlation
and Variances
Covariance & Correlation
Covariance
Correlation
Question
If X and Y are independent, then
If X and Y are dependent, then
a) Covariance
Definition
Suppose X and Y are jointly distributed with
expectations ,
of X and Y is defined as
provided the expectation exists.
Note :
1) Cov (X,Y) is a measure of association between X and
respectively.
The covariance
Note :
1) Cov (X,Y) is a measure of association between X and
Y, It may take any values is R .
2)
3) If X and Y are independent Cov(X,Y)=0.
Properties:
1) Cov (aX+b, cY+d)=ac Cov (X,Y) where a,b,c and d
are constants with
2)
Example ( Balls in Bag in Example )
Find E(X+Y), Var(X+Y).
0
1
2
0 3/15 3/15
0 1 2
2/15 6/15 0
1/15 0 0
The pmf of X and Y are
b) Correlation coefficient
Definition
Let X and Y be jointly distribution r. v. s with
var(X)>0,
The correlaltian coefficient of X and Y is defined as
Example
Var(Y)>0 and ,
(Balls in the bag, Example )
Example
Solution:
结论
单击图形播放/暂停 ESC键退出
(1) 不相关与相互独立的关系
注意
相互独立
不相关
(2) 不相关的充要条件
Solution:
ch5
The Law of Large Numbers
Chebyshev's Inequality
Law of large numbers
The central limit theorem
and the Central Limit Theorem
Chebyshev’s Inequality
Chebyshev’s Inequality
(Chebyshev’s Inequality) Let X be a random
or equivalently,
Theorem
variable for which and exist.
Then for any positive number ,
Proof:
In words, this result states that the probability that
any given random variable will differ from its mean
Because of its generality, Chebyshev Inequality is
Chebyshev inequality yields
by more than 3 standard deviations canot exceed 1/9.
very example, let , then the
Similarly, let
,we have
Example
Suppose that a random variable sample is to be
taken from a distribution for which the value of
determine how large the sample size must be in order
will be less than 1 unit.
to make the probability at least that
that the standard deviation is 2 units. We shall
the mean is not known, but for which it is known
Proof:
For convenience, we shall assume X has a continuous
distribution with pdf f(x) . The proof for a discrete
distribution or more general type of distribution
is similar.
ch5
The Law of Large Numbers
Chebyshev's Inequality
Law of large numbers
The central limit theorem
and the Central Limit Theorem
Law of Large numbers
Law of Large numbers
Theorem
Then for any, we have
be a sequence of independent random variables,
(Chebyshev’s Law of Large Numbers) Let
with finite expected value
and finite
variance
(Bernoulli’s Law of Large Numbers) Let n(A) be the
number of successes in n Bernoulli trials with
probability p for success on each trail. Then for any
,we have
Theorem
Example
is the sum of the first n rolls. This is a case of
Thus, by the Chebyshev’s Law of Large Numbers,
,
.
of the i-th roll. Then
Consider n rolls of a die. Let
be the outcome
independent trials with
for any
ch5
The Law of Large Numbers
Chebyshev's Inequality
Law of large numbers
The central limit theorem
and the Central Limit Theorem
The Central Limit Theorem
The Central Limit Theorem
Theorem
of independent identically distributed random
Then
and variance
variables with mean
(Lindeberg and Levy Theorem) Let
be a sequence
Theorem
the number of successes in n Bernoulli trials with
probability p for success on each trail, .
Then for any a<b, we have
(The De Moivre-Laplace Theorem) Let
be
Example
sample of size 100 house holds. Find the
probability that the mean household size exceeds .
and standard deviation
. Take a random
Household size X in the . has mean
ch6
Random Sampling
Random Sampling
Some Important Statistics
Sampling Distributions
Random Sampling
Random Sampling
Definition
A population consists of the totality of the observations
with which we are concerned.
Definition
A sample is a subset of a population.
Definition
Let be n independent random variables,
each having the same probability distribution f(x).
Definition
Let be n independent random variables,
each having the same probability distribution f(x).
We then define to be a random sample of
size n from the population f(x) and write its joint
probability distribution as
小 结
个体 总体
有限总体
无限总体
基本概念:
说明1 一个总体对应一个随机变量X, 我们将不区分总体和相应的随机变量, 统称为总体X.
说明2 在实际中遇到的总体往往是有限总体, 它对应一个离散型随机变量; 当总体中包含的个体的个数很大时, 在理论上可认为它是一个无限总体.
随机样本
ch6
Random Sampling
Random Sampling
Some Important Statistics
Sampling Distributions
sample mean
Some Important Statistics
sample variance
sample standard deviation
Definition
Any function of the random variables constituting
a random sample is called a statistic.
1. sample mean
2. sample variance
3. sample standard deviation S.
1. sample mean
2. sample variance
Example
A comparison of coffee prices at 4 randomly selected
grocery stores in San Diego showed increases from the
previous month of 12, 15, 17, and 20 cents for a 1 pound
bag.
Find the variance of this random sample of price
increases.
Solution:
Example
Find the variance of the data 3, 4, 5, 6, 6 and 7,
representing the number of trout caught by a random
sample of 6 fishermen on June 19, 1996,
at Lake
Muskoka.
Solution:
Solution:
Find the variance of the data 3, 4, 5, 6, 6 and 7
Solution:
according to
ch6
Random Sampling
Random Sampling
Some Important Statistics
Sampling Distributions
Chi-squared Distribution
Sampling Distributions
t-Distribution
F-Distribution
Definition
The probability distribution of a statistic is called a
sampling distribution.
Chi-squared Distribution
Let ,
Theorem
the distribution of the
random variable ,
where
is given
by the density function
Let ,
Theorem
the distribution of the
random variable ,
where
is given
by the density function
This is known as the chi-squared distribution with
degrees of freedom.
Denoted by
then
.
2. If
then
1. If
Properties:
Notes:
Notes:
represent the t-value above
1.
which we find an area equal to
leaves an area of to the right.
If then
2.
Example
Assume
. Find
where
Example
Assume
what is the value
of n such that
Solution:
Example
Assume
Find
where is the sample variance.
Solution:
Theorem
Let Z ~N(0,1)，V~ .
If Z and V are independent,
then the distribution of the random variable T,
where
is given by the density function
This is known as the t-distribution with degrees of
freedom.
denoted by
t-Distribution
Corollary
Let
Then the random variable
has a t-distribution with n-1 degrees of freedom.
.
Notes:
1. represent the t-value above which we find an
area equal to
leaves an area of to the right.
2.
3.
to the right,
is
Example
area of to the left,
and therefore an area of
The t-value with degrees of freedom that leaves an
Example
Example
Find k such that
for a random
sample of size 15 selected from a normal distribution
and
Solution:
Theorem
Then the distribution of the
random variable is given by the density
This is known as the F-distribution with and
degrees of freedom.
Denoted by
F-Distribution
Note:
represent the t-value above which we find an
area equal to
leaves an area of to the right.
Theorem
Writing for with and degrees of
freedom,
we obtain
Theorem
Writing for with and degrees of
freedom,
we obtain
Proof:
Example :
Find
Solution:
第六章 样本及抽样分布
二、主要内容
三、练习习题
一、重点与难点
一、重点与难点
1.重点
(1) 正态总体某些常用统计量的分布.
2.难点
(1) 几个常用统计量的构造.
(2) 临界值的查表计算.
(2) 标准正态分布和F分布临界值的查表计算.
二、主要内容
总 体
个 体
样本
常用统计量的分布
分位点
概率密度函数
统计量
常用统计量
性质
关于样本和方差的定理
t 分布
F 分布
分布
关于样本和方差的定理
试验的全部可能的观察值称为总体.
个体
总体中的每个可能观察值称为个体.
总体
样 本
统计量
常用统计量
(1)样本平均值:
(2)样本方差:
(3)样本标准差:
常用统计量
(4)样本 k 阶(原点)矩:
(5)样本 k 阶中心矩:
常用统计量的分布(一)
分布的性质
性质1
性质2
常用统计量的分布(二)
t 分布又称学生氏(Student)分布.
常用统计量的分布(三)
常用统计量的概率密度函数
常用统计量的概率密度函数
常用统计量的概率密度函数
常用统计量的分布的分位点
常用统计量的分布的分位点
常用统计量的分布的分位点
关于正态总体的样本和方差的定理
定理一
定理二
定理三
关于正态总体的样本和方差的定理
Solution:
Solution:
Example
Find k such that
for a random
sample of size 15 selected from a normal distribution
and
Solution:
Theorem
Writing for with and degrees of
freedom,
we obtain
Proof:
Example :
Find
Solution:
ch7
Estimation Problems
Point Estimation
The particular properties ofestimators
Interval Estimation
The method of moment
Point Estimation
The method of maximum likelihood
(a) The method of moment
Definition
The kth sample moment
The kth population moment
Steps of the method of moment:
Steps of the method of moment:
Then
is the estimator of
Example
Given a random sample of size n from a uniform
uniform population U (a , b) with ,
use the method
of moments to obtain a formula for estimating the
Parameter a.
Solution:
(b) The method of maximum likelihood
Definition
If are the values of a random sample
from a population with the parameter ,
the
likelihood function of the sample is give by
for values of within a given domain.
Here
is the value of the joint probability distribution.
Steps of find maximum likelihood estimate:
(1) Write likelihood function
(2) Calculate
(3) Find
(4) Decide the value of that maximizes as the
maximum likelihood estimate of .
likelihood estimate of the parameter of the
corresponding binomial distribution.
Example
Let X ~ b (n , p ),
find the maximum
Solution:
Example
If are the values of a random sample from
find the maximum likelihood
an exponential population,
estimator of its parameter .
Solution:
Example
If constitute a random sample of size n
from a normal population with the mean and the
find joint maximum likelihood estimates
of these two parameters.
variance ,
Solution:
Example
Given a random sample of size n from a uniform
uniform population U (a , b) with ,
use the method
of moments to obtain a formula for estimating the
Parameter a.
Solution:
likelihood estimate of the parameter of the
corresponding binomial distribution.
Example
Let X ~ b (n , p ),
find the maximum
Solution:
Hence, the maximum likelihood estimator is
Example
If are the values of a random sample from
find the maximum likelihood
an exponential population,
estimator of its parameter .
Solution:
Hence, the maximum likelihood estimator is
Example
If constitute a random sample of size n
from a normal population with the mean and the
find joint maximum likelihood estimates
of these two parameters.
variance ,
Solution:
Hence, the maximum likelihood estimators are
ch7
Estimation Problems
Point Estimation
The particular properties ofestimators
Interval Estimation
unbiasedness
The particular properties of
estimators
efficiency
Consistency
(a) unbiasedness
Definition
A statistic is an unbiased estimator of the parameter
if and only if
unbiased estimator of .
Example
If X has the binomial distribution with the parameters
show that the sample proportion,
,
n and ,
is an
Solution:
Example
population give by
show that is a biased estimator of .
If constitute a random sample from the
Solution:
Theorem
If S2 is the variance of a random sample from an
infinite population with the finite variance ,
then
(b) efficiency
Theorem
If and are two unbiased estimators of the
parameter of a given population If
then we say that is relatively more efficient than .
Example
If constitute a random sample from a
normal population with
(1) Show that are unbiased
estimators of .
(2) Compare the efficiency of these two estimators of .
Solution:
Definition
The statistic is a consistent estimator of the parameter
if and only if for each c>0
(c) Consistency
unbiased estimator of .
Example
If X has the binomial distribution with the parameters
show that the sample proportion,
,
n and ,
is an
Solution:
Hence,
is an unbiased estimator of
show that is a biased estimator of .
Solution:
is a biased estimator of
Example
(1) Show that are unbiased
estimators of .
(2) Compare the efficiency of these two estimators of .
Solution:
1)
2)
is more efficient than
Example
(1) Show that are unbiased
estimators of .
(2) Compare the efficiency of these two estimators of .
ch7
Estimation Problems
Point Estimation
The particular properties ofestimators
Interval Estimation
The estimation of means
Interval Estimation
The estimation of variances
Definition
An interval estimate of is an interval of the form
,where and are values of appropriate
random variables and .
If
for
the interval is then called a
confidence interval.
Also, is called the degree
and the endpoints,
and ,
called the lower and
upper confidence limits.
of confidence,
are
(a) The estimation of means
Theorem ( know)
If is the value of the mean of a random sample of
size n from a normal population with the known
variance ,
then
is a ( )100% confidence interval for the mean of
Theorem ( know)
If is the value of the mean of a random sample of
size n from a normal population with the known
variance ,
then
the population.
Proof:
, construct a 95% confidence interval for the
Example
If a random sample of size n=20 from a normal
population with the variance has the mean
population mean .
Solution:
Theorem ( unknow)
If and s are the values of the mean and the standard
deviation of a random sample of size n from a normal
population,
then
is a ( )100% confidence interval for the mean of the
population.
Proof:
Example
A paint manufacturer wants to determine the average
drying time of a new interior wall paint.
If for 12 test
areas of equal size he obtained a mean drying time of
minutes and a standard deviation of minutes,
construct a 95% confidence interval for the true mean .
Solution:
is a ( )100% confidence interval for .
(b) The estimation of variances
Theorems
If S2 is the value of the variance of a random sample
of size n from a normal population,
then
gallons.
Construct a 99% confidence interval for ,
Example
In 16 test runs the gasoline consumption of an
experimental engine has a standard deviation of
which measures the true variability of the gasoline
consumption of the engine.
Solution:
第六章 参数估计
一、重点与难点
三、练习习题
二、主要内容
一、重点与难点
1.重点
最大似然估计.
正态总体参数的区间估计.
2.难点
显著性水平 与置信区间.
二、主要内容
矩估计量
估计量的评选
最大似然估计量
估 计 量 的 性 质
似然函数
无偏性
正态总体均值方差的置信区间与上下限
有效性
相合性
置信区间和上下限
矩估计量
用样本矩来估计总体矩,用样本矩的连续函数来估计总体矩的连续函数,这种估计法称为矩估计法.
矩估计法的具体做法:
最大似然估计量
似然函数
正态总体均值方差的置信区间与上下限
无偏性
有效性
由于方差是随机变量取值与其数学期望的偏离程度, 所以无偏估计以方差小者为好.
相合性 (一致性)
置信区间和置信上限、置信下限
求置信区间的一般步骤
Proof:
So, The confidence interval of
is
Solution:
, construct a 95% confidence interval for the
Example
population mean .
So, The confidence interval of
is
Proof:
Solution:
gallons.
Construct a 99% confidence interval for ,
Example
In 16 test runs the gasoline consumption of an
experimental engine has a standard deviation of
which measures the true variability of the gasoline
consumption of the engine.
Solution:
