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16

Chapter 2: Probability

The aim of this chapter is to revise the basic rules of probability. By the end of this chapter, you should be comfortable with: conditional probability, and what you can and can"t do with conditional expressions; the Partition Theorem and Bayes" Theorem; First-Step Analysis for finding the probability that a process reaches some state, by conditioning on the outcome of the first step; calculating probabilities for continuous and discrete random variables.

2.1Sample spaces and events

Definition:Asample space, Ω, is aset of possible outcomes of a random experiment. Definition:Anevent,, is asubset of the sample space. This means that eventis simplya collection of outcomes.

Example:

Random experiment:Pick a person in this class at random.

Sample space:Ω =

all people in class

Event:=

all males in class. Definition:Eventoccursifthe outcome of the random experiment is a member of the set In the example above, eventoccurs ifthe person we pick is male. 17 2.2

Probability Reference List

The following properties hold for all events,.

P() = 0.

0P()1.

Complement:

P() = 1?P().

Probability of a union:

P() =P() +P()?P().

For three events,,:

P() =P()+P()+P()?P()?P()?P()+P()

Ifandaremutually exclusive

, thenP() =P() +P().

Conditional probability:

P() =P()P()

Multiplication rule:

P() =P()P() =P()P()

The Partition Theorem:

if12mform a partitionof Ω, then

P() =m

i=1P(i) =m i=1P(i)P(i) for any event.

As a special case,and

partition Ω, so:

P() =P() +P(

=P()P() +P( )P() for any,.

Bayes" Theorem:

P() =P()P()P().

More generally, if12mform a partition

of Ω, then

P(j) =P(j)P(j)

mi=1P(i)P(i)for any.

Chains of events:

for any events12n,

P(12n) =P(1)P(21)P(321)P(nn?11)

2.3Conditional Probability

Suppose we are working with sample space

Ω =people in class. I want to find the

proportion of people in the class who ski. What do I do? Count up the number of people in the class who ski, and divide by the total number of people in the class. P(person skis) =number of skiers in classtotal number of people in class Now suppose I want to find the proportion offemalesin the class who ski.

What do I do?

Count up the number of females in the class who ski, and divideby the total number of females in the class.

P(female skis) =number of female skiers in class

total number of females in class By changing from asking about everyone to asking about females only, we have: restricted attention to the set of females only, or:reduced the sample spacefrom the set of everyone to the set of females, or:conditionedon the eventfemales.

We could write the above as:

P(skisfemale) =number of female skiers in classtotal number of females in class Conditioning is like changing the sample space: we are now working in a new sample space of females in class. 19 In the above example, we could replace 'skiing" withanyattribute. We have:

P(skis) =# skiers in class

# class;P(skisfemale) =# female skiers in class# females in class; so: P() = #"s in class total # people in class and:

P(female) =

# female"s in class total # females in class # in class who areand female # in class who are female Likewise, we could replace 'female" with any attribute: P() = number in class who areand number in class who are This is how we get the definition of conditional probability: P() =

P(and)

P()=P()P()

By conditioning on event, we havechanged the sample space to the set of"s only. Definition:Letandbe events on the same sample space: soΩand Theconditional probability of event, given event, is

P() =P()P()

20

Multiplication Rule:

(Immediate from above). For any eventsand,

P() =P()P() =P()P() =P()

Conditioning as 'changing the sample space"

The idea that"conditioning" = "changing the sample space"can be very helpful in understanding how to manipulate conditional probabilities. Any 'unconditional" probability can be written as a conditional probability:

P() =P(Ω)

WritingP() =P(Ω) just means that we are looking for the probability of event, out of all possible outcomes in the set Ω. In fact, the symbolPbelongsto the set Ω: it hasno meaning withoutΩ.

To remind ourselves of this, we can write

P=PΩ

ThenP() =P(Ω) =PΩ()

Similarly,P() means that we are looking for the probability of event, out of all possible outcomes in the set A. Sois just another sample space. Thuswe can manipulate conditional probabilitiesP()just like any other probabilities, as long as we always stay inside the same sample space. The trick:Because we can think ofas just another sample space, let"s write

P() =PA()

Note: NOTstandard notation!

Thenwe can usePAjust likeP, as long as we remember to keep the subscript on EVERYPthat we write. 21
This helps us to make quite complex manipulations of conditional probabilities without thinking too hard or making mistakes. There is only one rule you need to learn to use this tool effectively:

PA() =P()for any,,.

(Proof: Exercise).

The rules:

P() =PA()

P

A() =P() for any,,.

Examples:

1. Probability of a union. In general,

P() =

P() +P()?P()

So,PA() =PA() +PA()?PA()

Thus,P() =P() +P()?P()

2. Which of the following is equal toP()?

(a)P()(c)P()P() (b) P()

P()(d)P()P()

Solution:

P() =PA()

=PA()PA() =P()P()

Thus the correct answer is (c).

22

3. Which of the following is true?

(a)P( ) = 1?P()(b)P() =P()?P()

Solution:

P() =PA() = 1?PA() = 1?P()

Thus the correct answer is (a).

4. Which of the following is true?

(a)P( ) =P()?P()(b)P() =P()?P()

Solution:

P() =P()P() =PA()P()

1?PA()

P() =P()?P()P() =P()?P()

Thus the correct answer is (a).

5. True or false:P() = 1?P()?

Answer:

False.P() =PA()Once we havePA, we are stuck with

it! There is no easy way of converting fromPAtoP

A: or anything

else. Probabilities in one sample space (PA) cannot tell us anything about probabilities in a different sample space (P A). Exercise:if we wish to expressP() in terms of onlyand, show that

P() =P()?P(

)P()

1?P(). Note that this does not simplify nicely!

23
2.4

The Partition Theorem (Law of Total Probability)

Definition:Eventsandaremutually exclusive, ordisjoint, if= This means events A and B cannot happen together. If A happens, it excludes

B from happening, and vice-versa.

Ifandare mutually exclusive,P() =P() +P()

For all otherand,P() =P() +P()?P()

Definition:Any number of events12karemutually exclusiveif every pair of the events is mutually exclusive: ie.ij= for allwith=. Definition:Apartitionof Ω is acollection of mutually exclusive events whose union isΩ.

That is, sets12kform a partition of Ω if

ij=for allwith= and k i=1 i=12k= Ω

1kform a partition of Ω if theyhave no overlap

and collectively cover all possible outcomes. 24

Examples:

Partitioning an eventA

Any set A can be partitioned: it doesn"t have to be Ω. In particular, if1kform a partition of Ω, then (1)(k) form a partition of. Theorem 2.4:The Partition Theorem (Law of Total Probability) Let1mform a partition ofΩ. Then for any event A,

P() =m

i=1P(i) =m i=1P(i)P(i) Both formulations of the Partition Theorem are very widely used, but especially the conditional formulationmi=1P(i)P(i) 25

Intuition behind the Partition Theorem:

The Partition Theorem is easy to understand because it simply states that "the whole is the sum of its parts." 12 34

P() =P(1) +P(2) +P(3) +P(4)

2.5Bayes" Theorem: inverting conditional probabilities

Bayes" Theorem allows us to "invert" a conditional statement, ie.to express

P()in terms ofP().

Theorem 2.5:Bayes" Theorem

For any events A and B:P() =P()P()P()

Proof:

P() =P()

P()P() =P()P() (multiplication rule)

P() =P()P()

P() 26

Extension of Bayes" Theorem

Suppose that12mform a partition of Ω. By the Partition Theorem,

P() =m

i=1P(i)P(i) Thus, forany single partition memberj, put=jin Bayes" Theorem to obtain:

P(j) =P(j)P(j)P()=P(j)P(j)mi=1P(i)P(i)

12 34

Special case:m= 2

Given any event, the eventsandform a partition of Ω. Thus:

P() =P()P()P()P() +P()P()

Example:In screening for a certain disease, the probability that a healthy person wrongly gets a positive result is 0.05. The probability thata diseased person wrongly gets a negative result is 0.002. The overall rate of the disease in the population being screened is 1%. If my test gives a positive result, what is the probability I actually have the disease? 27

1. Define events:

=have disease=do not have the disease =positive test= =negative test

2. Information given:

False positive rate is 0.05P() = 005

False negative rate is 0.002P() = 0002

Disease rate is 1%P() = 001.

3. Looking forP():

We haveP() =P()P()P()

NowP() = 1?P(

= 1?P() = 1?0002 = 0998

AlsoP() =P()P() +P(

)P() = 0998001 + 005(1?001) = 005948 Thus

P() =0998001

005948= 0168

Given a positive test, my chance of having the disease is only 16.8%. 28
2.6 First-Step Analysis for calculating probabilities in a process In a stochastic process, what happens at the next step depends upon the cur- rent state of the process. We often wish to know the probability ofeventually reaching some particular state, given our current position. Throughout this course, we will tackle this sort of problem using a technique called

First-Step Analysis.

The idea is to consider all possible first steps away from the current state. We derive a system of equations that specify the probabilityofthe eventual outcome given each of the possible first steps. We then try to solve these equations for the probability of interest. First-Step Analysis depends uponconditional probabilityand thePartition Theorem.Let1kbe thepossible first steps we can take away from our current state. We wish to find the probability that eventhappens eventually.

First-Step Analysis calculatesP() as follows:

P() =P(1)P(1) ++P(k)P(k)

Here,P(1)P(k) give the probabilities of taking the different first steps 12.

Example:Tennis game at Deuce.

Venus and Serena are playing tennis, and have reached the score Deuce (40-40). (Deucecomes from the French wordDeuxfor 'two", meaning that each player needs to win two consecutive points to win the game.)

For each point, let:

=P(Venus wins point) = 1?=P(Serena wins point)

Assume that all points are independent.

Letbe the probability that Venus wins the game eventually, starting from

Deuce. Find.

29

VENUSWINS (W)VENUSAHEAD (A)

VENUS

BEHIND (B)

pq p p q qVENUSLOSES (L)DEUCE (D) Use First-step analysis. The possible steps starting from Deuce are:

1. Venus wins the next point (probability): move to state A;

2. Venus loses the next point (probability): move to state B.

Letbe the event that Venus wins EVENTUALLY starting from Deuce, so=P(). Starting from Deuce (D), the possible steps are to states A and B. So: =P(Venus wins) =P() =PD() =PD()PD() +PD()PD() =P()+P()() Now we need to findP(), andP(), again using First-step analysis:

P() =P()+P()

= 1+

Similarly,

P() =P()+P()

= 0+ 30

Substituting (a) and (b) into(),

=2+ 2 (1?2) =2 =2 1?2

Note:Because+= 1, we have:

1 = (+)2=2+2+ 2

So the final probability that Venus wins the game is: =21?2=22+2 Note how this result makes intuitive sense. For the game to finish from Deuce, either Venus has to win two points in a row (probability2), or Serena does (probability2). The ratio2(2+2) describes Venus"s 'share" of the winning probability.

First-step analysis as the Partition Theorem:

Our approach to finding=P(Venus wins) can be summarized as:

P(Venus wins) ==

first stepsP(first step)P(first step)

First-step analysis is just thePartition Theorem:

The sample space isΩ =all possible routes from Deuce to the end

An example of a sample point is:

Another example is:

Thepartitionof the sample space that we use in first-step analysis is:

1=all possible routes from Deuce to the end that start with

2=all possible routes from Deuce to the end that start with

31

Then first-step analysis simply states:

P() =P(1)P(1) +P(2)P(2)

=PD()PD() +PD()PD() Notation for quick solutions of first-step analysis problems Defining a helpful notation is central to modelling with stochastic processes. Setting up well-defined notation helps you to solve problemsquickly and easily. Defining your notation is one of the most important steps in modelling, because it provides the conversion from words (which is how your problem starts) to mathematics (which is how your problem is solved). Several marks are allotted on first-step analysis questionsfor setting up a well-defined and helpful notation.

VENUSWINS (W)VENUSAHEAD (A)

VENUS

BEHIND (B)

pq p p q qVENUSLOSES (L)DEUCE (D) Here is the correct way to formulate and solve this first-stepanalysis problem. Need the probability that Venus wins eventually, starting from Deuce.

1. Define notation: let

D=P(Venus wins eventuallystart at state D)

A=P(Venus wins eventuallystart at state A)

B=P(Venus wins eventuallystart at state B)

2. First-step analysis:

D=A+B()

A=1 +D()

B=D+0 ()

32

3. Substitute (b) and (c) in (a):

D=(+D) +(D)

D(1??) =2

D=2 1?2 as before.

2.7Special Process: the Gambler"s Ruin

This is a famous problem in probability. A gambler starts with $. She tosses a fair coin repeatedly. If she gets a Head, she wins $1. If she gets a Tail, she loses $1. The coin tossing is repeated until the gambler has either $0 or $, when she stops. What is the probability of the Gambler"s Ruin, i.e. that the gambler ends up with $0? 1/2

1/21/2

1/21/2

1/20 1 2 3

1/2 1/2

1/2 1/21/2N

1/2 x

Wish to find

P(ends with$0starts with$)

Define event

=eventual Ruin=ends with$0

We wish to findP(starts with$x)

Define notation:

x=P(Rcurrently has$x)for= 01 33

Information given:0=P(currently has$0) = 1

N=P(currently has$) = 0

First-step analysis:

x=P(has$) 1 2P has$(+ 1) 12P has$(?1) 1

2x+1+12x?1()

True for= 12?1, with boundary conditions0= 1,N= 0.

Solution of difference equation():

x=12x+1+12x?1for= 12?1;

0= 1()

N= 0 We usually solve equations like this using the theory of 2nd-order difference equations. For this special case we will also verify the answer by two other methods.

1. Theory of linear 2nd order difference equations

Theory tells us that the general solution of()isx=+for some constants ,and for= 01. Our job is to findandusing the boundary conditions: x=+for constantsandand for= 01 So

0=+0 = 1= 1;

N=+= 1 += 0=?1

34

So our solution is:

x=+ = 1?for= 01. For Stats 325, you will be told the general solution of the 2nd-order difference equation and expected to solve it using the boundary conditions. For Stats 721, we will study the theory of 2nd-order difference equations. You will be able to derive the general solution for yourself before solving it.

Question:

What is the probability that the gambler wins (ends with $N), starting with $x? P ends with$N = 1?P ends with$0 = 1?x=for= 01

2. Solution by inspection

The problem shown in this section is thesymmetricGambler"s Ruin, where the probability is 1

2of moving up or down on any step. For this special case,

we can solve the difference equation by inspection.

We have:x=12x+1+12x?1

1

2x+12x=12x+1+12x?1

Rearranging:x?1?x=x?x+1Boundaries:0= 1N= 0

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