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Discrete MathCS 2800

Prof. Bart Selmanselman@cs.cornell.edu

Module Probability --- Part b)

Bayes’ RuleRandom Variables

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Bayes’ Theorem

How to assess the probability that a particular event will occur on the basis of partial evidence?

Examples:

What is the likelihood that people who test positive to a particular disease

(e.g., HIV), actually have the disease?

What is the probability that an e-mail message is spam?

Key idea: one should factor in additional information regarding occurrence of events.

Assume that with respect to events F and E (“E” for “Evidence”):

We know P(F) – probability that event F occurs

(e.g. probability that email message is spam;

this is given by what fraction of email is spam)

We also know event E has occurred.

(e.g., email message contains words “sale” and “bargain”)

Therefore the probability conditional probability that F occurs given

that E occurs, P(F|E), is a more realistic estimate that F occurs than P(F).

How do we compute P(F|E)?

E.g., based on P(F), P(E|F), and P(E| ¬F)

Note: ¬F is also referred to as complement of F (FC or F).

BayesianBayesianInferenceInference

EvidenceEvidence

Original BeliefOriginal Belief(Prior Probability)(Prior Probability)

ModifiedModifiedBeliefBelief

HypothesisHypothesis TheoryTheory

P(F)

E

P(F|E)

Experiment: Pick one box at random (p = 0.5) and than a ball at random from that box.

Assume you picked a red ball.

What’s the probability that it came form the left box? > 0.5 ?

Define:E – you choose a red ball. (therefore ¬ E – you choose the green ball)F – you choose the left box. (therefore ¬ F– you choose the right box)

We want to know P(F|E)

Box A Box B

Why?

What we know:

P(E|F) = and P(E|¬F) =

Given that the boxes are selected at random: P(F) = P(¬F)=1/2

P(F|E) = P(E∩F)/P(E) so we need to compute P(E∩F) and P(E).

P(F|E)?

We know P(E|F) = P(E∩F)/P(F). So, P(E∩F) = P(E|F) P(F) = 7/9 * 1/2 = 7/18.

What about P(E)? Note that P(E) = P(EF) +P (E∩ ¬F). Why?Note also that P (E ∩ ¬F) = P(¬F) P(E|¬F) = 1/2 * 3/7 = 3/14So, P(E) = P(EF) +P (E∩ ¬F) = 7/18 + 3/14 = 38/63

And therefore P(F|E) = P(E∩F)/P(E) = (7/18) / (38/63) = 49/76 0.645

7/9 3/7

E – red colorF – left box

Indeed > 0.5 phew!

BayesianBayesianInferenceInference

Concrete (new) EvidenceConcrete (new) EvidenceRed ball picked (E)Red ball picked (E)

Original BeliefOriginal Belief there is a 0.5 that you will pick left box there is a 0.5 that you will pick left box

(P(F)).(P(F)).

Modified BeliefModified BeliefIncreased to 0.65 Probability Increased to 0.65 Probability

(P(F|E))(P(F|E))

( | ) ( )( | )

( | ) ( ) ( | ) ( )C CP E F P F

P F EP E F P F P E F P F

Theorem: Bayes’ Theorem

Suppose that E and F are events from a sample space S such that P(E) ≠0 and P(F)≠0. Then

Proof:

, ( ) ( | ) ( )

exp ( )

( ) ( ) ( )

( ) ( ) ( ) (

( | ) ( )( | )

(

( | ) ( ) / ( )

( | ) (

| ) ( ) ( | ) ( )

( |

)

) / (

)

)

( |

C C

C C C

So P E F P E F P F therefore

We only need an ression for P E

E E S E F F E F E F

So

P E P E F

P F E P

P

P E F P FP

E F

E F P E

P E F P

F EP

P E F P F P E F P F

So

P EP F E

F P

E

E F

( | ) ( )

( | ) ( ) ( | )

) ( )

( )( ) C C

P E F P F

P E F

F

P F

P

P E F P

F

P E F

( | ) ( )( | )

( )

P E F P FP F E

P E

Example --- A Classic!

Suppose that 1 person in 100,000 has a particular rare disease. There is an

good test for the disease that is correct in 99% of the time when given

to someone with the disease; it is correct in 99.5% of the time when given

to someone without the disease.

Find:

a) Probability that someone who tests positive has the disease.

b) Probability that someone who tests negative does not have the disease.

Solution:

a)

Always start by defining the events!

F – the person has the diseaseE – the person tests positive to the diseaseP(F|E) – probability of having the disease given positive test

P(F)=1/100,000 = 0.00001; P(FC) = 0.99999P(E|F) = 0.99; P(EC|F) = 0.01P(E|FC) = 0.005

002.0)99999.0)(005.0()00001.0)(99.0(

)00001.0)(99.0(

)()|()()|(

)()|()|(

CC FPFEPFPFEP

FPFEPEFP

Only 0.2% of people who test positive actually have the disease!!!

Why these probabilities?Most easily measured!

1 person in 100,000 has rare disease. Test correct in 99% of the time when given to someone with the disease; correct in 99.5% of the time when given to someone without the disease.

Counter –intuitive. That’s why it’s a classic! (note: test is good the disease is rare; but context matters.)

b) F – the person has the disease

E – the person tests positive to the disease

P(FC|EC) – probability of not having the disease given negative test

P(F) = 1/100,000 = 0.00001; P(FC) = 0.99999

P(E|F) = 0.99; P(EC|F) = 0.01

P(E|FC) = 0.005

9999999.0)00001.0)(01.0()99999.0)(995.0(

)999991.0)(995.0(

)()|()()|(

)()|()|(

FPFEPFPFEP

FPFEPEFP

CCCC

CCCCC

That’s… pretty good!

Generalized Bayes’ Theorem

Suppose that E is an event from a sample space S and F1, F2 ,…, Fn are

mutually exclusive events such that

Asume that P(E) ≠ 0 and P(Fi) ≠ 0 for i=1, 2,…, n. Then

SFini 1

)

1

( | ( )( | )

( | ) ( )

j jj n

i ii

P E F P FP F E

P E F P F

P(E)

( | ) ( )( | )

( | ) ( ) ( | ) ( )C C

P E F P FP F E

P E F P F P E F P F

Compare:

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Bayesian Spam Filters

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Applying Bayes’ Theorem:SPAM or HAM?

Let our sample space or universe be the set of emails. (So, we’re sampling fromLet our sample space or universe be the set of emails. (So, we’re sampling from

the space of possible emails.)the space of possible emails.)

Let S be the event a message is spam; hence is the event a message is not spamLet S be the event a message is spam; hence is the event a message is not spam

Let E be the event a message contains a word Let E be the event a message contains a word ww. .

Since we have no idea of likelihood of SPAM, we assume P(S)=P(SC)=1/2.

Can we do better?

How do we get and ? ( | )p E S ( | )p E S

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Estimations

Note these are estimates based on frequencies in samples.

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Estimation Continued

Note P(S) = P(SC) = ½ divides out.

So,

becomes

So, a quite straightforward formula for our first Bayesian spam filter!

So, what do we want forp(w) and q(w) ??What’s the max prob andwhat’s the min? Whendo we get 0.5?

( | ) ( )p E S p w

( | ) ( )p E S q w

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Spam based on single words?

Probabilities based on single words: Bad IdeaProbabilities based on single words: Bad Idea

– False positives AND false negatives a plentyFalse positives AND false negatives a plenty

Calculate based on Calculate based on n n words, assuming each event Ewords, assuming each event Eii|S (E|S (Eii|S|SCC) is ) is

independent (not true but reasonable approximation)independent (not true but reasonable approximation)

P(S) = P(SP(S) = P(SCC).).

Derivation see Sect. 6.3.

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Final Approximation

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Compare to single word:

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How do we use this?

User must train the filter based on messages in his/her inbox to estimate User must train the filter based on messages in his/her inbox to estimate probabilities.probabilities.

The program or user must define a threshold probability The program or user must define a threshold probability rr: :

If , the message is considered spam.If , the message is considered spam.

Gmail: Train on all users! (note: report spam button)Gmail: Train on all users! (note: report spam button)

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Example

Suppose the filter has the following dataSuppose the filter has the following dataThreshold Probability: .9Threshold Probability: .9““Nigeria” occurs in 250 of 2000 spam messages Nigeria” occurs in 250 of 2000 spam messages ““Nigeria” occurs in only 5 of 1000 non-spam messagesNigeria” occurs in only 5 of 1000 non-spam messagesLet’s try to estimate the probability, using the process we just definedLet’s try to estimate the probability, using the process we just defined

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Example Cont.

Step 1: Find the probability that the message has the Step 1: Find the probability that the message has the word “Nigeria” in it and is spam. word “Nigeria” in it and is spam. – p(p(NigeriaNigeria) = 250 / 2000 = 0.125) = 250 / 2000 = 0.125

Step 2: Find the probability that the message has the Step 2: Find the probability that the message has the word “Nigeria” in it and is not spam.word “Nigeria” in it and is not spam.– q(q(NigeriaNigeria) = 5 / 1000 = 0.005) = 5 / 1000 = 0.005

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Since we are assuming that it is equally likely that an Since we are assuming that it is equally likely that an incoming message is or is not spam, we can incoming message is or is not spam, we can estimate the probability with this equation:estimate the probability with this equation:

r(Nigeria) = r(Nigeria) = p(Nigeria) p(Nigeria)

p(Nigeria) + q(Nigeria)p(Nigeria) + q(Nigeria)

Example Cont.

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= = 0.1250.125

0.1300.130

= 0.962= 0.962

Since Since r(Nigeria)r(Nigeria) is greater than the threshold of 0.9, we can reject is greater than the threshold of 0.9, we can reject

this message as spam.this message as spam.

Example Cont.

0.125____0.125____0.125 + 0.0050.125 + 0.005

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Multiple Words

2000 Spam messages; 1000 real messages2000 Spam messages; 1000 real messages

““Nigeria” appears in 400 spam messagesNigeria” appears in 400 spam messages

““Nigeria” appears in 60 real messagesNigeria” appears in 60 real messages

““bank” appears in 200 spam and 25 real messagesbank” appears in 200 spam and 25 real messages

Threshold Probability: .9Threshold Probability: .9

Let’s calculate the probability that message with “Nigeria” and “bank” is Let’s calculate the probability that message with “Nigeria” and “bank” is spam.spam.

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Example Cont.

Step 1: Find the probability that the message has the word “Nigeria” in it Step 1: Find the probability that the message has the word “Nigeria” in it and is spam.and is spam.– p(Nigeria) = 400 / 2000 = 0.2p(Nigeria) = 400 / 2000 = 0.2

Step 2: Find the probability that the message has the word “Nigeria” and is Step 2: Find the probability that the message has the word “Nigeria” and is not spam.not spam.– q(Nigeria) = 60 / 1000 = 0.06q(Nigeria) = 60 / 1000 = 0.06

Step 3: Find the probability that the message contains the word “bank” and Step 3: Find the probability that the message contains the word “bank” and is spam.is spam.– p(bank) = 200 / 2000 = 0.1p(bank) = 200 / 2000 = 0.1

Step 4: Find the probability that the message contains the word “bank” and Step 4: Find the probability that the message contains the word “bank” and is not spam.is not spam.– q(bank) = 25 / 1000 = 0.025q(bank) = 25 / 1000 = 0.025

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Example Cont

Using our approximation, we have:Using our approximation, we have:

r(Nigeria,bank) = r(Nigeria,bank) = p(Nigeria) * p(bank) p(Nigeria) * p(bank) p(Nigeria) * p(bank) + q(Nigeria) * q(bank)p(Nigeria) * p(bank) + q(Nigeria) * q(bank)

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Example Cont.

Using our approximation, we have:Using our approximation, we have:

r(Nigeria,bank) = r(Nigeria,bank) = p(Nigeria) * p(bank) p(Nigeria) * p(bank) p(Nigeria) * p(bank) + q(Nigeria) * q(bank)p(Nigeria) * p(bank) + q(Nigeria) * q(bank)

r(Nigeria,bank)r(Nigeria,bank) = = (0.2)(0.1) (0.2)(0.1)

(0.2)(0.1) + (0.6)(0.025)(0.2)(0.1) + (0.6)(0.025)

= = 0.9300.930

This message will be rejected however since we set the threshold probability at 0.9.This message will be rejected however since we set the threshold probability at 0.9.

Concludes Bayes Reasoning

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Probability Paradox I

Magic Dice: Or How to Win Every Time!

a) You select any one of the four dice (A, B, C, or D). b) I’ll select another.Both dice are thrown, highest number wins throw. Do series of 10 throws. The person with the most highest throws wins the series. (I.e. die “more likely to get a higher number” wins.)

Claim: In a game of 'The Best of Ten Throws’, I will almost certainly win --- no matter which die you pick!!

Why is this strange? Say, you pick die A. Let’s assume, die B is better. So, I pick B. But, then, next game & next person picks B. Let’s assume C is better. I’ll select C. Next person, will pick C. I’ll pick D. Next person, will pick D… Hmm… ?? I’ll pick A and will win!!A < B < C < D …. < A !! Failure of transitivity!

But, could such a set ofdice exist?

Surprisingly, yes!

AB

C

D

Magic Dice

http://www.sciencenews.org/20020420/mathtrek.asp

D

C

B

A

Prob(D wins over C) = 2/32/6 + (4/6)* (1/2) = 4/6

Prob(C wins over B) = 2/3since3/6 + (3/6)* (2/6) = 4/6Prob(B wins over A) = 4/6 = 2/3

(i.e. Prob(A wins over B) = 1/3)

Prob(A wins over D) = 4/6 = 2/3

A < B < C < D …. < A !!

How about expected value of dice throw?Transitive or not?? E.g. E[B] = 16/6.

YES!!E[B] < E[A] = E[C] < E[D] 16/6 < 18/6 = 18/6 < 20/6

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Random Variables and Distributions

Random Variables

For a given sample space S, a random variable (r.v.) is any real valued

function on S, i.e., a random variable is a function that assigns a real

number to each possible outcome

Suppose our experiment is a roll of 2 dice. S is set of pairs.

Example random variables:

S0 2-2

X = sum of two dice. X((2,3)) = 5

Y = difference between two dice. Y((2,3)) = 1

Z = max of two dice. Z((2,3)) = 3

Sample spaceNumbers

Random variableSuppose a coin is flipped three times. Let X(t) be the random variable

that equals the number of heads that appear when t is the outcome.

X(HHH) = 3

X(HHT) = X(HTH) = X(THH) = 2

X(TTH) = X(THT) = X(HTT) = 1

X(TTT) = 0

Note: we generally drop the argument! We’ll just say the “random variable X” (even though it’s technically a function).

And write e.g. P(X = 2) for “the probability that the random variable X(t) takes on the value 2”.

Or P(X=x) for “the probability that the random variable X(t) takes on the value x.”

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Distribution of Random Variable

Definition:

The distribution of a random variable X on a sample space S is the set of

pairs (r, p(X=r)) for all r X(S), where p(X=r) is the probability that X

takes the value r.

A distribution is usually described specifying p(X=r) for each r X(S).

A probability distribution on a r.v. X is just an allocation of

the total probability mass, 1, over the possible values of X.

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The Birthday Paradox

Birthdays

How many people have to be in a room to assure that the probability that at least two of them have the same birthday is greater than 1/2?

a) 23b) 183c) 365d) 730

Let pn be the probability that no people share a birthday among n people in a room.

We want the smallest n so that 1 - pn > 1/2.

Then 1 - pn is the probability that 2 or more share a birthday.

Hmm. Why does such an n exist? Upper-bound?

For L optionsanswer is inthe orderof sqrt(L) ?

Informally, why??

A: 23

Birthdays

Assumption:

Birthdays of the people are independent.

Each birthday is equally likely and that there are 366 days/year

Let pn be the probability that no-one shares a birthday among n people in a room.

Assume that people come in certain order; the probability that the second person has a birthday. Different than the first is 365/366; the probability that the third person has a different birthday. Form the two previous ones is 364/366.. For the jth person we have (366-(j-1))/366.

What is pn? (“brute force” is fine)

After several tries, when n=22 1= pn = 0.475.

n=23 1-pn = 0.506

So,366

367

366

363

366

364

366

365 npn

366

367

366

363

366

364

366

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npn

Relevant to “hashing”. Why?

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From Birthday Problem to Hashing FunctionsProbability of a Collision in Hashing Functions

A hashing function h(k) is a mapping of the keys (or records, e.g., SSN, around 300x 106 in the US) to a much smaller storage location. A good hashing fucntion yields few collisions. What is the probability that no two keys are mapped to the same location by a hashing function?

Assume m is the number available storage locations, so the probability of mapping a key to a location is 1/m.Assuming the keys are k1, k2, kn, the probability of mapping the jth record to a free location is after the first (j-1) records is (m-(j-1))/m.

m

nm

m

m

m

mp

m

nm

m

m

m

mp

n

n

12111

121

Given a certain m, find the smallest nSuch that the probability of a collision is greater than a particular threshold p.

It can be shown that for p>1/2,

n 1.177 m

m = 10,000, gives n = 117. Not that many!