Transcript Lect1 - Brandeis

Probability & Statistics
Lecture 1
Michael Partensky
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Topics of Discussion (1)
 The Laws of Chance… Are they possible?


Gambling – it’s serious
On the shoulders of Giants
 Collecting the data
 Random experiments and events.



Sample space, sample sets.
Operations on sample sets.
Frequencies of events
 The first definition of probability (P)



Axioms of probability
Frequency interpretation of P
General properties of P (derivation from Axioms).
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Topics of Discussion (2)
 Random variables and distribution functions.
Discrete and continuous variables. Examples.
Distribution function for discrete variables.
Continuous distributions.
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On the shoulders of Giants
Pierre de Fermat,
French mathematician
(1601-1655)
Blaise Pascal,
Frenh mathematician
and philospher (16231662)
Christiaan Huygens,
Dutch astronomert,
mathematician,
physicist
(1629-1695)
Jacob Bernoulli,
Swiss mathematician
(1654-1705)
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Not only only from their achievements, but also from errors …
D'Alembert argued, that if two coins are tossed,
there are three possible cases, namely:
(1) both heads, (2) both tails, (3) a head and a tail.
So he concluded that the probability of " a head
and a tail " is 1/3. If he had figured that this
probability has something to do with the
experimental frequency of the occurrence of the
event, he might have changed his mind after
tossing two coins more than few times. Apparently,
he never did so. Why? We do not know.
Was he right?
Think!
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A famous statistician would never travel by airplane, because she had
studied air travel and estimated the probability of there being a bomb
on any given flight was 1 in a million, and she was not prepared to
accept these odds.
One day a colleague met her at a conference far from home.
"How did you get here, by train?"
"No, I flew"
"What about the possibility of a bomb?"
"Well, I began thinking that if the odds of one bomb are 1:million,
then the odds of TWO bombs are (1/1,000,000) x (1/1,000,000) =
10-12. This is a very, very small probability, which I can accept. So,
now I bring my own bomb along!"
We will be dealing a lot with the paradoxes of probability.
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Here is a cute brain-stretcher.
Adventures on the Moscow Subway*
Boris commutes to college by the Moscow subway which runs in a circle. The
school happens to be at the point on the circle that is exactly opposite to
where Boris boards the train. The trains run in both directions, and the
schedule is very regular.
The time intervals for trains in both directions are equal; for instance, if there
is an hour between the arrival of clockwise (cw) trains, there is also an hour
between the arrival of counterclockwise trains. Boris observed, however, that
he caught the cw trains
much more often than the ccw trains.,
despite the fact that his schedule was
irregular and he arrived at the
Station at random times.
Can you explain this?
T-station
*From “The chicken from Minks” by Y.B. Chernyak and R.M. Rose
School
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Random experiments
Term "random experiment" is used to describe any action whose outcome is not
known in advance. Here are some examples of experiments dealing with
statistical data:
• Tossing a coin
•Counting how many times a certain word or a combination of words appears
in the text of the “King Lear” or in a text of Confucius
• Counting how many Japanese sedans passed the Washington Bridge between
12 and 12.30 p.m.
•counting occurrences of a certain combination of amino acids in a protein
database.
•pulling a card from the deck
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Sample spaces, sample sets and events
The sample space of a random experiment is a set  that includes all possible
outcomes of the experiment. For example, if the experiment is to throw a die and
record the outcome, the sample space is
  {1, 2, 3, 4, 5, 6}
the set of possible outcomes.  describes an event that always occurs.
Each outcome is represented by a sample point in the sample space.
There is more than one way to view and experiment, so an experiment may have more
than one associated sample space.
For example, suppose you draw one card from a deck. Here are some sample spaces.
Sample space 1 (the most common) The space consists of 52 outcomes, 1 for each card.
Sample space 2 The space consists of just two outcomes, black and red,
Sample space 3 This space consists of 13 outcomes, namely 2,3,4,… 10,J,Q,K,A.
Sample space 4. This space consists of two outcomes, picture and non-picture.
In tossing a die, one sample space is {1,2,3,4,5,6}, while two others are {odd, even}
and {less then 3.5, more then 3.5}
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The examples above describe the discrete sample spaces.
Intuitively, you all understand what it means although a
precise mathematical definition would be cumbersome.
We will discuss the continuous sample spaces a little later.
The examples can be represented by various physical
variables describing a population, such as weight, height,
concentration etc.
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Race between seven horses,
1,2,3,4,5,6,7.
•
Experiment 1 : determining the winner.
•
Experiment 2: determining the order of finish
Describe the sample spaces
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Events
Certain subsets of the sample space of an experiment are referred to as events. An event
is a set of outcomes of the experiment. This includes the null (empty) set of outcomes
and the set of all outcomes. Each time the experiment is run, a given event A either
occurs, if the outcome of the experiment is an element of A, or does not occur, if the
outcome of the experiment is not an element of A.
What to consider an event is decided by the experimentalist.
Examples.
In the Sample space 1 (see above) of 52 cards the following events can be considered
among others: A= “drawing a king”, B= “drawing Q or A”, C = “drawing a nonpicture”, D= “drawing a black”.
For example, event D consists of 26 points.
D is also an event in sample space 2 (consisting of 1 point).
It is not an event in sample spaces 3 and 4.
Similarly, A (“king) is an event in sample spaces 1 and 3 but not in 2 and 4.
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The examples of sample spaces and events
Example 1.1
Flip two coins. Try to figure out what is the sample space for this experiment How
many simple events does it contain?
The answer could be described by a following table:
1\2
H
T
H
HH
HT
T
TH
TT
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The examples of sample spaces and events
Example 1.2
Role two dice. For convenience we assume that one is red and the other is green
(we often use this trick which hints that we can tell between two objects which
is which). The following table describes 
1
2
3
1
11
12
2
21
3
4
5
6
13 14
15
16
22
23 24
25
26
31
32
33 34
35
36
4
41
42
43 44
45
46
5
51
52
53 54
55
56
6
61
62
63 64
65
66
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Example 1.2 (continued) . Any event corresponds to some collection of the
cells of this table. We showed here four different events (colored blue, red,
green and gray) . Try to describe them in plain English
1
2
3
1
11
12
2
21
3
4
5
6
13 14
15
16
22
23 24
25
26
31
32
33 34
35
36
4
41
42
43 44
45
46
5
51
52
53 54
55
56
6
61
62
63 64
65
66
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Example 1.2 (continued) . And what about the following events?
Which of these event (red, orange, purple, etc) occurs more often?
1
2
3
1
11
12
2
21
3
4
5
6
13 14
15
16
22
23 24
25
26
31
32
33 34
35
36
4
41
42
43 44
45
46
5
51
52
53 54
55
56
6
61
62
63 64
65
66
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Sample spaces and events
Example 1.3 An experiment consists of drawing two numbered balls from the
box of balls numbered from 1 to 5. Describe the sample space if
(a) The first ball is not replaced before the second is drawn.
(b) The first ball is replaced before the second is drawn.
Example 1.4 Flip three coins. Show the sample space. What is the total number
of all possible outcomes?
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Composite events
Quite often we are dealing with composite events. Example. We
study a group of students, picking them at random and considering
the following events: A = “A student is female”, B=“A student is
male”, C= “A student has blue eyes”, D=“A student was born in
California”.
After this information was collected and the probabilities of A. B, C
and D were determined, we decided to find the probabilities of some
other events:
U= “Student is female with blue eyes”, V=“Student is male, or
has blue eyes and was not born in California”, etc.
The latter questions are dealing with the events that are composed
from the “atomic” events A, B, C and D. To describe them, it is
convenient to use a language of the Set Theory.
Warning: please, don’t be scared. We are not going to be too
theoretical. The new symbols will be introduced for our
convenience.
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Inclusion: A is a subset of B
Occurrence of A implies the occurrence of B.
Example: B = {2,3,5,6}, A={2,6}
B
AB
A
Intersection: A B belongs to A and B
Examples:
(1) B = {2,3,4,6}, A={1,2,5,6}
A B ={2,6};
(2) A=“Female students”, B=“Students having blue
eyes”, A B = “Female students with blue eyes”.
A
B
A
B
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Operations on sample sets (1)
The empty set  is the event with no
outcomes. The events are disjoined if
they do not have outcomes in common:
A B
Example: A={1,3,4}, B= {2,6}
The union of events A and B, A  B is the
event that A or B or both occur. Example: A
= “Male student”, B= “Having blue eyes”,
 Bare either male or have
=“StudentsAwho
blue eyes (or both)”
Complementary (opposite) events
Ac is the compliment to A if A AC  
and A
AC  
A
B
B
A
AB
A

Ac
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The first definition of probability
We have introduced some important concepts:




Experiments,
Outcomes,
Sample space,
Random variables.
These concepts will come to life after we introduce another crucial
concept- probability, which is a way of measuring the chance.
A probability is a way of assigning numbers to events that satisfies
following conditions or
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The Axioms of Probability
(1) For any event A, 0  P(A) 1.
(2) If  is the sample space, then P( )=1
(1.1)
(1.2)
(3) For a sequence of disjoined (incompatible) events Ai (finite or
infinite),
P ( i Ai )   P ( Ai )
(1.3)
i
In other words, the probability for a set of disjoined events equals the sum of
individual probabilities
We will introduce here another important property, although it is not an
Axiom and will be justified later in the lecture devoted to the analysis of
conditional probability:
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(4) If A and B are independent, then
P( A  B)  P( A)P(B)
(1.4)
In other words, for any number of independent
events, N,
P( A j )   j P( A j ) , j  1,2,..., N
(1.5)
Incompatible (disjoined) and independent events
These new concepts introduced above require some explanation.
•Two events are said to be incompatible or disjoined if they can not occur
together.
•Two events are said to be independent if they have “nothing to do” with each
other.
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Examples:
• (Disjoined) Jack can arrive to Boston either on Monday (“M”) or on Wednesday
(“W”). These events can not occur together (he can not arrive on Monday and on
Wednesday). Therefore the events are disjoined. If P(M)=0.72 and P(W)=0.2, then
P(M or W)=0.92 (there is still a chance that Jack won’t come to Boston at all).
• (Independent) A= “It will be sunny today”, B=“Celtic will win the game
tomorrow”
. What si the probability that both events will occur simultaneously?
P(A and B)=P(A) P(B).
Please, offer some more examples of both kinds.
You can find a very useful and provocative discussion of these concepts and of
the probability in general in the book “Chance and Chaos” by David Ruelle
(Princeton Sci. Lib.)
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Frequency interpretation of probability
If we repeat an experiment a large number of times, then
the fraction of times the event A occurs will be close to P(A).
In other words, if N(A,n) is the number of times that event occurs
in the first n trials, than
P ( A)  limn 
N ( A, n )
n
(1.6)
It's easy to prove that defined this way, P(A) satisfies conditions
(1) and (2) (Try to prove it).
Hint: the property (3) follows from N (
i
Ai , n )   N ( Ai , n )
i
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Some other properties of probability. Deduction from the
axioms 1-3:
(a). Given that A +Ac  we find from (3):
P(A)=1 - P(Ac )
(1.7).
(b) if A  B, then P(A)  P (B )
(1.8)
(c) P(A  B )  P ( A)  P (B )  P ( AB )
(1.9)
Try to prove (a) and (b). The property (c) is harder to prove
formally, although intuitively it is quite clear. Summing up the
areas of A and B, one counts their intersection twice. The
probability is kind-a proportional to the area. Therefore, one of
the occurrences of AB should be removed.
A AB
B
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Comment
In many cases it is easier to find a probability of a compliment event Ac
rather then A itself. It such cases, the rule (1.7) can be very helpful.
Example: Toss two dice. Find the probability that they show different
values, for example (4,6) and (2,3) but not (2,2).
You can count favorable outcomes directly, or better still, by (3)
P(non-matching dice) = 1 –P(matching dice) = 1 – 6/36 = = 5/6.
We will be using this trick quite often.
Example: the classical “birthday problem”.
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Find a probability that in a group of n people at least two will
have the same birthday.
You will solve a version of this problem at home.
Let’s discuss here a simpler problem.
N people arrive to a vacation place during the same week. Each one chooses
randomly his/her arrival day. What is the probability that at least two people
arrive on the same day?
Let’s discuss it together (working in groups)
What is the complimentary event?
•Consider first N=2,
•N=8.
•Now let us check N=3.
•The general case
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Random Variables and the Distribution Function.
The simplest experiments are flipping coins and throwing dice. Before we
can say anything about the probabilities of their various outcomes (such as
"Getting an even number" on the die, or "Getting 3 heads in 5 consecutive
experiments with a coin") we need to make a reasonable guess about the
probabilities of the elementary events (getting H or T for a coin , or one of
six faces for a die). For instance, we can assume (as we usually do) that a
die is perfectly balanced and all faces are equivalent. Then , the probability
of any number equals 1/6.
As we will see, it can be described in terms of distribution (because it
distributes probabilities between different outcomes) function.
The term “function”, however, implies some arguments (or variables). It
would be inconvenient to use a “function of faces”, or a “functions of Heads
or Tails”. That’s why we will introduce a general term for describing various
random outcomes.
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Random Variables
We now introduce a new term
Instead of saying that the possible outcomes are 1,2,3,4,5 or 6, we say
that random variable X can take values {1,2,3,4,5,6}.
A random variable is an expression whose value is the
outcome of a particular experiment.
The random variables can be either discrete or continuous.
It’s a convention to use the upper case letters (X,Y) for the names of the
random variables and the lower case letters (x,y) for their possible
particular values.
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Examples of random variables
Discrete: (you name !)
Continuous:
For instance, weights or a heights of people chosen randomly, amount
of water or electricity used during a day, speed of cars passing an
intersection.
Please, add a few more examples.
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The Probability Function for discrete random variables
We assigned a probability 1/6 to each face of the dice. In the same manner, we
should assign a probability 1/2 to the sides of a coin.
What we did could be described as distributing the values of probability between
different elementary events:
P(X=xk)=p(xk), k=1,2,…
(1.9)
It is convenient to introduce the probability function p(x) :
P(X=x)=p(x)
(1.10)
In other words, the probability of a random variable X taking a particular value x
Is called the probability function.
For x=xk (1.10) reduces to (1.9) while for other values of x, p(x)=0.
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The probability function should satisfy the following equations :
p( x )  0
 p( xi )  1
i
P (E ) 
 p( xi )
(1.11)
(1.12)
(1.13)
i E
Example:
Suppose that a coin is tossed twice, so that the sample space is
={HH,HT,TH,TT}.
Let X represent a number of heads that can come up. Find the
probability function p(x).
Assuming that the coin is fair, we have P(HH)=1/4, P(HT)=P(TH)=1/4,
P(TT)=1/4;
Then, P(X=0)=P(TT)=1/4; P(X=1)=P(HTTH)=1/4+1/4=1/2.
P(X=2)= ¼.
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The probability function is thus given by the table:
x
0
1
2
p(x)
1/4
1/2
1/4
The probability function p(x) is related to the probability density function
(PDF) f(x) introduced in the next section for the continuous random
variables. For those familiar with the concept of “-function”, this relation
can be presented as
f ( x )   p( x ) ( x  x )
i
i
All others can simply ignore this formula.
(1.14)
Uniform and non-uniform distributions.
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If all the outcomes of an experiment are equally probable, the
corresponding probability function is called uniform. If the
contrary is true, the probability function is non-uniform.
Example On the face of a die with 6 dots, 5 dots are filled, so that only the
central one is left. What is the probability function for this case? The value
X=1 will occur in average 2 times more often than in the balanced die. As a
result, p(1)=1/3, p(2)=p(3)=p(4)=p(5) = 1/6 .
Working together
TryIt Suppose we pick a letter at random from the word TENNESSEE. What is
the sample space and what is the probability function for the outcomes?
Challenge: For two dice experiment, find the probability function for
X=“ Sum of two throws”.
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Continuous distribution (preliminary remarks)
 f ( xi )  1
i
P (E ) 
(1.5.1)
 f ( xi )
(1.5.2)
i E
Suppose that the circle has a unit circumference (we simply use
units in which 2 Pi R=1).
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Continuous distribution. Probability density function (PDF).
Suppose that every point on the circle is labeled by its distance x
from some reference point x=0. The experiment consists of
spinning the pointer and recording the label of the point at the tip
of the pointer. Let X is the corresponding random variable. The
sample space is the interval [0,1). Suppose that all values of X are
equally possible. We wish to describe it in terms of probability. If
it was a discrete variable (such as a dice), we would simply
assign to every outcome a fixed value of probability to all
outcomes.. p(xi)=const.
However, for a continuous variable we must assign to each
outcome a probability p(x )=0. Otherwise, we would not be able
to fulfill the requirement 1.12.
Something is obviously wrong!
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Continuous distribution and the probability density function
Dealing with the infinitesimal numbers is a tricky business indeed. Those who
studied calculus aware of this.
A random variable X is said to have a continuous distribution
with density function f(x) if for all a b we have
b
P (a  X  b )   f ( x )dx
(1.15)
a
The analogs of Eqs. 1.12 and 1.13 for the continuous distributions would be
(1.16)
 f (x)  1

P (E )   f ( x )dx
E
(1.17)
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P(E) is a probability that X belongs to E.
f(x)
P(a<X<b)
a
b
Geometrically, P(a<X<b) is the area under the curve f(x) between a and b.
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Examples:
1. The uniform distribution on (a,b):
We are picking a value at random from (a,b).
1
,
 ba
f (x)  
0

axb
(1.18)
otherwise
By direct integration you can verify that (1.18) satisfies the condition (1.16).
We can now find PDF which describes the experiment with the spinner in
which case b-a=2
1
,
 2
f ( )  
0

0  x  2
otherwise
The probability that the arrow will stop in the rage
between  and  +  equals /2.
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2. The exponential distribution
  e  x ,
f ( x )  
0

x0
otherwise
(1.19)
Those who know how to integrate can verify that (1.19) satisfies (1.16)
(the total area under the curve f(x) equals 1.
Note: In Matematica, the integral of a function f[x] (notice that […]
rather than (…) is used) can be found as:
Integrate[f[x],{x,x1,x2}] , Shift+Enter.
Here x1 and x2 are the limits of integration.
Please, practice with this at home.
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3. The standard normal distribution
f ( x)  (2 ) 1 / 2 Exp( x 2 / 2)
(1.20)
Using Mathematica check that this (corrected) formula
satisfies the second Axiom of probability
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Self-Test
1.
For what number of students n the probability of at least two having the same
birthday reaches 0.7?
2.
A card is drawn at random from an ordinary deck of 52 playing cards. Describe
the sample space if consideration of suits (hearts, spades, etc) (a) is not, (b) is,
taken in the account.
3.
Referring to the previous problem, let A="a king is drawn" and B="a club is
drawn". Describe the events (a)AUB, (b) AB, (c) ABc (d) Ac  Bc (d) (AB)
(ABc)
4.
Describe in words the events specified by the following subsets of
 = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}
(a) E = {HHH,HHT,HTH,HTT}; (b) E = {HHH,TTT}; (c) E= {HHT,HTH,THH};
(d) E = {HHT,HTH,HTT,THH,THT,TTH,TTT}
5. A die is loaded in such a way that the probability of each face turning up is
proportional to the number of dots on that face (for instance, six is three times as
probable as two). What is the probability of getting an even number in one throw?
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6. Let A and B be events such that
P( A B) 1/ 4, P( A) 1/ 3, and P(B) 1/ 2.
What is the P(AUB)?
7. A student must choose one of the subjects: art, geology, or psychology as an
elective. She is equally likely to choose art or psychology, and twice as likely
to choose geology.
What are the respective probabilities for each of three subjects?